Multilinear polynomials of small degree evaluated on matrices over a unital algebra

Linear Algebra and its Applications 496 (2016) 262–287

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Linear Algebra and its Applications www.elsevier.com/locate/laa

Multilinear polynomials of small degree evaluated on matrices over a unital algebra Katherine Cordwell a , George Wang b,∗ a b

University of Maryland, College Park, United States University of Southern California, United States

a r t i c l e

i n f o

Article history: Received 24 July 2015 Accepted 30 December 2015 Available online 10 February 2016 Submitted by P. Semrl MSC: 15A54 16S50

a b s t r a c t Let R be a unital associative algebra over a field K of characteristic zero, and let f be a multilinear polynomial of degree m over K. If m ≤ 3, we prove that all traceless matrices can be written as the sum of two values of f evaluated over Mn (R) with n ≥ 2. If m = 4, we prove the same result for n ≥ 3. © 2016 Elsevier Inc. All rights reserved.

Keywords: Multilinear polynomial Unital algebra Matrix Trace

1. Introduction Throughout this paper, we refer to a matrix whose elements along the main diagonal sum to zero as a traceless matrix. In 1936, Shoda [9] proved that over a field of characteristic zero, any traceless matrix can be written as a single commutator. This result was later extended in 1957 to arbitrary fields by Albert and Muckenhoupt [1]. In * Corresponding author. E-mail address: [email protected] (G. Wang). http://dx.doi.org/10.1016/j.laa.2015.12.033 0024-3795/© 2016 Elsevier Inc. All rights reserved.

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2000, Rosset and Rosset [8] demonstrated that, over commutative unital rings, not every traceless matrix can be written as a single commutator. However, Mesyan [6] proved in 2006 that any traceless matrix over a unital associative ring R can be written as the sum of at most two commutators. Let us consider the multilinear degree two polynomial f (x1 , x2 ) = [x1 , x2 ] = x1 x2 − x2 x1 evaluated over Mn (R), the ring of n × n matrices with entries in R for n ≥ 2. Mesyan’s results tell us that any traceless matrix can be written as the sum of two values of f , or equivalently that any traceless matrix is contained in im f + im f , where im f + im f = {U + V | U, V ∈ im f }. More recently, in 2012, Khurana and Lam [5, Theorem 5.2] proved that for n ≥ 2, any matrix in the ring of n × n matrices over an arbitrary unital ring can be written as the sum of two generalized commutators. In particular, this means that for the multilinear degree three polynomial f (x1 , x2 , x3 ) = x1 x2 x3 − x3 x2 x1 , any traceless matrix over an arbitrary unital ring is contained in im f + im f . Another way to see this is to set x3 = 1, and the result will follow from the result quoted above. Our goal is to prove the following results for all multilinear polynomials f of degree ≤ 4 over K and evaluated on Mn (R), where R is a unital associative algebra over a field K of characteristic zero. If f has degree ≤ 3, we prove that all traceless matrices are contained in im f + im f , where f is over K and evaluated on Mn (R) for n ≥ 2. If f has degree 4, we prove that all traceless matrices are contained in im f + im f , where f is over K and evaluated on Mn (R) for n ≥ 3. This is not true for n = 2; a well-known counterexample is the polynomial f = [x1 , x2 ][x3 , x4 ] + [x3 , x4 ][x1 , x2 ], which is central over M2 (K). This paper is motivated by an open question posed by Lvov and Kaplansky. As reproduced in [7]: Question 1 (Lvov–Kaplansky). Let f be a multilinear polynomial over a field K. Is the set of values of f on the matrix algebra Mn (K) a vector space? A natural variation of this question for multilinear polynomials over unital associative algebras is: Question 2. Let R be a unital associative algebra over a field K and f be a multilinear polynomial over K and evaluated on Mn (R). Given any matrix in the vector space generated by the set of values of f , what is the minimal number of terms required to represent this matrix as a sum of values of f ? Our results provide a partial answer to this question when deg f ≤ 4 and char K = 0. For commutative unital associative algebras in particular, evidence thus far suggests that the minimal number required is two. Our work also suggests the following conjecture for polynomials of higher degree, which is a modified version of [7, Conjecture 11]:

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Conjecture 1. Let R be a unital associative algebra over a field K of characteristic 0, n ≥ 2 and m ≥ 1 integers, and f (x1 , . . . , xm ) a nonzero multilinear polynomial in Kx1 , . . . , xm . If n ≥ m − 1, then all traceless n × n matrices with entries in R are contained in im f + im f . Note that this conjecture does not hold in general for nonunital algebras, as we will show in the last section. 2. Preliminaries Let R be a unital associative algebra over a field K of characteristic zero. A multilinear polynomial over a field K is defined to be a polynomial of the form f =  σ∈Sm aσ xσ(1) · · · xσ(m) , with aσ ∈ K. Denote the set of traceless n × n matrices over R by Mn0 . Denote the matrix diagonal starting at the (1, k + 1) term as the k-diagonal, and the matrix diagonal starting at the (k+1, 1)-term as the (−k)-diagonal, for 0 ≤ k ≤ n −1. We will make repeated use of some important facts and theorems, which we reproduce here. The first remark, which we restate in the context of a unital associative algebra, can be found in [3, Remark 3] and [7, Lemma 6 and Corollary 8]. Remark 1. Let R be a unital associative algebra over a field K, m a positive integer,  and f (x1 , . . . , xm ) = σ∈Sm aσ xσ(1) · · · xσ(m) ∈ Kx1 , · · · , xm  be a nonzero multilinear polynomial of degree m evaluated on Mn (R). Then the following hold.  (i) If σ∈Sm aσ = 0, then im f = Mn (R). (ii) If A ∈ Mn (R) is an invertible matrix, then Af (x1 , . . . , xm )A−1 = f (Ax1 A−1 , . . . , Axm A−1 ). The next result is due to Mesyan [6, Theorem 15]. Theorem 1. Let R be a unital ring and n a positive integer. Then there exist matrices X, Y ∈ Mn (R) such that for all A ∈ Mn (R) having trace 0, A ∈ [X, Mn (R)] +[Y, Mn (R)]. n−1 Specifically, writing Eij for the matrix units, one can take X = i=1 Ei+1,i and Y = Enn . We also reproduce [7, Lemma 16]. Lemma 1. Let R be a unital ring, let n ≥ 2 be an integer, and let A = (aij ) ∈ Mn (R) be n−j such that for each j ∈ {0, 1, . . . , n−1} we have i=1 ai,i+j = 0 (i.e., sums of the elements along the diagonals above and including the main diagonal are 0). Then A = DX − XD n−1 for some D ∈ Mn (R) and X = i=1 Ei+1,i , where Eij are the matrix units. Next, we introduce a modified version of [2, Lemma 1.2]. As the proof is very short, we provide it for the sake of completeness.

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Lemma 2. Let R be a unital associative algebra over a field K of characteristic 0. If A = diag{a1 , a2 , . . . , an } with a1 , a2 , . . . , an distinct and ai ∈ K for 1 ≤ i ≤ n, then any zero-diagonal matrix in Mn (R) can be written as [A, B] for some zero-diagonal B ∈ Mn (R). Proof. Take B to be the zero-diagonal matrix with entries (ai − aj )−1 bij , where bij are arbitrary elements of R. Then the commutator of A with B has zeroes on the main diagonal and entries (ai − aj )−1 (ai − aj )bij = bij where i = j. 2 Finally, we prove a technical lemma regarding the rank of various matrices that we will encounter. The reader may notice that the statement of Lemma 3, part (iii), may look unusual due to the use of . However, it will be applied in Lemma 7 exactly as presented. Lemma 3. Entries in the matrices below are from a field K of characteristic 0. Empty entries indicate zero. (i) Let n ≥ 4 and z = 0 or z = n. Then ⎛ −(z − 2) −1 ⎜ −1 2 ⎜ ⎜ −1 ⎜ A=⎜ ⎜ ⎜ ⎝

the (n − 1) × n matrix A has rank n − 1. (z − 1) ⎞ −1 2 ...

−1 ... −1

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

... −1 2

2 −1

(ii) The n × n matrix B has rank n − 1, where n ≥ 6. ⎛ −1 1 ⎜ 3 −1 ⎜ ⎜ −3 3 −1 ⎜ ⎜ 3 −1 ⎜ 1 −3 B=⎜ ⎜ 1 −3 3 −1 ⎜ ⎜ ... ... ... ... ⎜ ⎜ ⎝ 1 −3 3 1 −3

−3 1

−1 3

−1 ⎞ 3 −3 ⎟ ⎟ 1 ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ −1

(iii) Let n ≥ 7. The (n − ) × (n − ) matrix C has rank n −  for 3 ≤  ≤ n − 4. ⎛ 2 ⎞ −1 ⎜ −1 ⎜ ⎜ ⎜ C=⎜ ⎜ ⎜ ⎝

2 ...

−1 ... −1

... 2 −1

−1 2 −1

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ (n − 2) ⎠ −(n − 3)

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(iv) The n × (n − 1) matrix D has rank n − 1, where n ≥ 6. ⎛

1 −1

⎜ ⎜ ⎜ ⎜ D=⎜ ⎜ ⎜ ⎜ ⎝ −(n − 2) (n − 2)

⎞

−1 2 ...

−1 ...

...

... ...

−1 −(n − 2) ...

2 −(n − 1) (n − 2)

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ −1 ⎟ −(2n − 5) ⎠ (2n − 4)

Proof. In each part below, we consider linear combinations of rows to determine the rank of each matrix. Within these linear combinations, ak denotes an element from the field K which is the coefficient of the kth row. (i) It is clear that rows 2, . . . , n − 1 of A are independent, since the kth row has a nonzero entry in a column for which all previous rows from 2, . . . , k − 1 have a zero entry. Then it suffices to show that the first row cannot be written as a linear combination of the others. In order to seek a contradiction, we will assume it can be. First let us consider the case where z = 0. Let a2 be fixed. By observation, the kth row has coefficient ak = 2ak−1 − ak−2 . This forces an−1 = (n − 2)a2 . Then since z = 0, we must have a2 = −2, and therefore an−1 = −2(n − 2). From this, 2(n − 2) = (z − 1) = −1. However, since we assumed n ≥ 4, the left-hand side of the equation is always positive and the right-hand side is always negative. This produces the desired contradiction. Next, let us consider the case where z = n. Once again, fixing a2 forces an−1 = (n−2)a2 . Then since z = n, we must have a2 = (n−2), and therefore an−1 = (n−2)(n−2). This forces −(n − 2)(n − 2) = (z − 1) = (n − 1), or that n2 − 3n + 3 = 0. The solutions of this quadratic are neither integers, nor greater than 4, and so this produces the desired contradiction. (ii) For ease of calculation, we begin by taking the first three rows and moving them to the bottom with elementary row operations. Then our matrix B is row equivalent to ⎛

1

−3 1

⎜ ⎜ ⎜ ⎜ ⎜ ⎜  B =⎜ ⎜ ⎜ ⎜ ⎜ −1 ⎜ ⎝ 3 −1 −3 3

3 −3 ...

−1 3 ... 1

−1

⎞ −1 ... −3 1

... 3 −3 1

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ −1 ⎟ ⎟ 3 −1 ⎟ ⎟ −3 3 ⎟ ⎟ 1 −3 ⎠ 1

The first n − 3 rows of B  are linearly independent, since the kth row has a nonzero entry in a column for which all previous rows from 1, . . . , k − 1 have a zero entry.

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As in (i), a recurrence relation shows that row n − 2 is linearly independent of the first n − 3 rows. In order to seek a contradiction, we will assume it can be written as a linear combination of the first n − 3 rows. Let a1 = −1. Then by inspection, we find that the (n − k)th row is forced to have a coefficient of ak = 3ak−1 − 3ak−2 + ak−3 . We would like to prove by induction that ak = ak−1 − k. It is easy to see by the recurrence relation that a1 = −1, a2 = −3, and a3 = −6. Then assume ak = ak−1 − k and ak−1 = ak−2 − (k − 1). This means that ak−1 = ak + k and ak−2 = ak−1 + k − 1. Then ak+1 = 3ak − 3ak−1 + ak−2 = 3ak − 2(ak + k) − ak−1 + ak−2 = ak − 2k − ak−1 + ak−1 + k − 1 = ak − k − 1 = ak − (k + 1) Due to the entries of row n − 2, an−3 = −3 is forced. However, the recurrence relation n−3 forces an−3 = i=1 −i, which is never equal to −3 when n ≥ 6. This produces the desired contradiction, and thus the first n − 2 rows are linearly independent. Now we wish to show that row n − 1 is linearly independent from the first n − 2 rows; as above, we assume that it is not and seek a contradiction. Let an−1 = 1. First fix the coefficient of row n − 2 to be an−2 = m. Then a1 − an−2 = 3an−1 , so a1 = 3an−1 + an−2 = 3 + m. By observation, a2 = 8 + 3m and a3 = 15 + 6m. We would like to show that in general, the kth row has a coefficient of

ak =

k

(2i + 1) + m

i=1

k

i

i=1

Proceeding by induction, assume that this holds for ak−2 , ak−1 , and ak . Then ak+1 = 3ak − 3ak−1 + ak−2 k k k−1 k−1 k−2 k−2 = 3( (2i + 1) + m i) − 3( (2i + 1) + m i) + ( (2i + 1) + m i) i=1

i=1

= 3(2k + 1 + mk) +

k−2

i=1

(2i + 1) + m

i=1

i=1 k−2

i=1

i

i=1

= (2(k − 1) + 1 + 2k + 1 + 2(k + 1) + 1) + (m(k − 1) + mk + m(k + 1)) +

k−2

(2i + 1) + m

i=1

=

k+1 i=1

(2i + 1) + m

k−2 i=1

k+1 i=1

i

i

i=1

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In the (n − 2)th row, we find that an−2 = m =

n−2

n−2

i=1

i=1

(2i + 1) + m

i

Solving for m gives us that n−2 m=

(2i + 1) n−2 1 − i=1 i i=1

(1)

However, the terms in the rightmost column force 3an−2 − an−3 = −3, which gives 3m −

n−3

n−3

i=1

i=1

(2i + 1) − m

i = −3

Solving for m tells us that m=

−3 +

n−3

i=1 (2i + 1)  n−3 3 − i=1 i

(2)

Let the numerator of equation (1) equal x and the denominator equal y. Then comparing equations (1) and (2), we find that x + 2n x = y y−n This implies that −x = 2y. Substituting the numerator and denominator of (1) back in for x and y, we find that 2−

n−2 i=1

2i = −

n−2

(2i + 1)

i=1

Canceling terms leaves us with 2 = −(n − 2), which is the desired contradiction. Therefore, the matrix B  has rank at least (n − 1). However, the matrix B  cannot have full rank since the last row is the negative sum of all previous rows. Thus, B  has rank (n −1), and therefore B has rank (n − 1). (iii) Starting from the (n − )th row and moving upwards, the kth row of C, for 2 ≤ k ≤ n − , has a nonzero entry in a column for which the rows k + 1, . . . , n −  have zero entries. Therefore, the rows excluding the first are linearly independent. To write the first row in terms of the others, the second row must have a coefficient of −2, the third row must have a coefficient of −3, and in general row k must have a coefficient of −k. When we reach the last two rows, the terms in the rightmost column force (n −  − 1)(n − 2) − (n − )(n − 3) = 0 This produces a contradiction whenever  = 2. Thus the rank of C is n − .

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(iv) The first n − 2 rows of D are clearly linearly independent. To show that the rank of D is at least n − 1, we will ignore the (n − 1)th row and instead show that the last row is linearly independent from the first n − 2 rows. For ease of calculation, we can fix an = (n −2)−1 . Then we proceed from row n −2 upwards. We can see that −an−2 = 2an , so that an−2 = −2. Similarly, by observation, an−3 = −5 and an−4 = −9. In general, we have a recurrence relation of an−k = 2an−(k−1) − an−(k−2) − 1, with the −1 due to the entries in row n. We would like to prove by induction that an−k = an−(k−1) − k. Assume that this holds for an−k and an−(k−1) , then an−(k+1) = 2an−k − an−(k−1) − 1 = 2an−k − (an−k + k) − 1 = an−k − (k + 1) The leftmost column forces a1 − a2 = (n − 2)an . The recurrence relation forces a1 − a2 = an−(n−1) − an−(n−2) = (an−(n−2) − (n − 1)) − an−(n−2) = 1 − n, and from before, (n − 2)an = (n − 2)(n − 2)−1 = 1. But 1 − n = 1 since n ≥ 6, so we have the desired contradiction. Thus the rank of D is at least n − 1. To show that the rank is not n, observe that the first row is the negative sum of all other rows. 2 3. Polynomials of degree 1 to 3 We begin with an obvious case. Proposition 1. If f is a degree 1, nonzero, multilinear polynomial over K and evaluated on Mn (R) with n ≥ 2, then Mn0 ⊂ im f . Proof. As in [7, Lemma 7], we have f = ax1 for some a ∈ K. Take U ∈ Mn (R), then set x1 = a−1 U to show that U ∈ im f . Thus, Mn0 ⊂ im f . 2 Proposition 2. If f is a degree 2, nonzero, multilinear polynomial over K and evaluated on Mn (R) with n ≥ 2, then Mn0 ⊆ im f + im f . Proof. Again, as in [7, Lemma 7], if deg f = 2, then f = ax1 x2 +bx2 x1 for some a, b ∈ K. If a + b = 0, then by Remark 1, Mn0 ⊂ im f . If a + b = 0, then f = ax1 x2 − ax2 x1 = a(x1 x2 − x2 x1 ) = a[x1 , x2 ]. Let U ∈ Mn0 . By Theorem 1, there exist matrices X, Y ∈ Mn (R) such that U ∈ [X, Mn (R)] + [Y, Mn (R)] ∈ im f + im f . 2 Theorem 2. If f is a degree 3, nonzero, multilinear polynomial over K and evaluated on Mn (R) with n ≥ 2 and R is a unital associative algebra over a field K of characteristic 0, then Mn0 ⊆ im f + im f . Proof. If the sum of coefficients is nonzero, then by Remark 1, Mn0 ⊂ im f + im f . If the sum of coefficients is zero but f (x1 , x2 , I) = 0, then f (x1 , x2 , I) = ax1 x2 + bx2 x1 for

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some a, b ∈ K, then the result follows from Proposition 2. Similarly, if f (x1 , I, x3 ) = 0 or f (I, x2 , x3 ) = 0, we are done. The remaining possible multilinear polynomials are of the form f (x1 , x2 , x3 ) = [x1 , [x3 , x2 ]] + z[x3 , [x1 , x2 ]], for some z ∈ K. The proof is due to [7, Theorem 13], and we reproduce it here. Let us write the polynomial as f (x1 , x2 , x3 ) = ax1 x2 x3 + bx1 x3 x2 + cx2 x1 x3 + dx2 x3 x1 + ex3 x1 x2 + gx3 x2 x1 with a, b, c, d, e, g ∈ K. From above, we may assume a + b + c + d + e + g = 0, so f (x1 , x2 , x3 ) = a(x1 x2 x3 − x3 x2 x1 ) + b(x1 x3 x2 − x3 x2 x1 ) + c(x2 x1 x3 − x3 x2 x1 ) + d(x2 x3 x1 − x3 x2 x1 ) + e(x3 x1 x2 − x3 x2 x1 ) Also from above, we may assume that 0 = f (I, x2 , x3 ) = a(x2 x3 − x3 x2 ) + c(x2 x3 − x3 x2 ) + d(x2 x3 − x3 x2 ) = (a + c + d)(x2 x3 − x3 x2 ) which implies that a+c+d = 0. We may also assume f (x1 , I, x3 ) = 0 and f (x1 , x2 , I) = 0, so that a + b + c = 0 and a + b + e = 0. This in turn implies that b = d, c = e, and a = −b − c. Then f (x1 , x2 , x3 ) = (−b − c)(x1 x2 x3 − x3 x2 x1 ) + b(x1 x3 x2 − x3 x2 x1 + x2 x3 x1 − x3 x2 x1 ) + c(x2 x1 x3 − x3 x2 x1 + x3 x1 x2 − x3 x2 x1 ) = (x1 x3 x2 − x3 x2 x1 + x2 x3 x1 − x1 x2 x3 ) + c(x2 x1 x3 − x3 x2 x1 + x3 x1 x2 − x1 x2 x3 ) = b(x1 [x3 , x2 ] − [x3 , x2 ]x1 ) + c(x3 [x1 , x2 ] − [x1 , x2 ]x3 ) = b[x1 , [x3 , x2 ]] + c[x3 , [x1 , x2 ]] Observe that b−1 f (x1 , x2 , x3 ) = f (b−1 x1 , x2 , x3 ), so these polynomials have the same image. Then we can rewrite the polynomial as f (x1 , x2 , x3 ) = [x1 , [x3 , x2 ]] +z[x3 , [x1 , x2 ]] for some z ∈ K. Given a matrix U ∈ Mn0 , we can rewrite it as U = V + W where V is a traceless diagonal matrix with vii = uii and W is a zero-diagonal matrix with wij = uij for i = j. We wish to prove that V, W ∈ im f so that U ∈ im f + im f . We begin with V . n−1 n−1 Consider the three matrices A = En,1 + i=1 Ei,i+1 , B = E1,n + i=1 Ei+1,i , and n C = i=1 ci Ei,i , where Ei,j denotes a matrix unit. Then f (A, B, C) = [A, [C, B]] + z[C, [A, B]] = [A, [C, B]] since A = B −1 .

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271

If n = 2,

[A, [C, B]] =

2c2 − 2c1 0

0 2c1 − 2c2



Letting c1 = 0 and c2 = u11 /2 shows that V ∈ im f . If n ≥ 3, ⎛ ⎜ ⎜ C=⎜ ⎝

⎞

c1 ..

. cn

⎛

0 ⎜ c2 − c1 ⎜ ⎜ ⎜ [C, B] = ⎜ ⎜ ⎜ ⎝ ⎛ ⎜ ⎜ [A, [C, B]] = ⎜ ⎝

⎟ ⎟ ⎟ ⎠

c2

c1 − cn 0 c3 − c2

... ...

...

cn − cn−1

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

0

cn − 2c1 + c2

⎞ ⎟ ⎟ ⎟ ⎠

c1 − 2c2 + c3 ...

cn−1 − 2cn + c1 When n = 3, we have the system of equations M c = V : ⎛

−2 1 ⎝ 1 −2 1 1

⎞⎛ ⎞ ⎛ ⎞ 1 c1 v11 ⎠ 1 ⎠ ⎝ c2 ⎠ = ⎝ v22 −2 c3 −v11 − v22

where v33 = −v11 − v22 since U is traceless. A quick calculation shows that this system has a solution. When n ≥ 4, we have the system of equations M c = V : ⎛

−2 1 ⎜ 1 −2 ⎜ ⎜ 1 ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 1

1 −2 ...

1 ... 1

... −2 1

⎞

⎛ ⎞ ⎛ ⎞ v11 c1 ⎟ ⎟ ⎜ c2 ⎟ ⎜ ⎟ v22 ⎟⎜ ⎟ ⎜ ⎟ .. ⎟ ⎜ .. ⎟ ⎜ ⎟ ⎟⎜ . ⎟ ⎜ ⎟ . ⎟⎜ . ⎟ = ⎜ ⎟ .. ⎟⎜ . ⎟ ⎜ ⎟ . ⎟⎜ . ⎟ ⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ 1 ⎟ ⎝ cn−1 ⎠ ⎝ vn−1,n−1 ⎠ −2 1 ⎠ n−1 cn − i=1 vii 1 −2 1

n−1 Note that as in the n = 3 case, we have written vnn as − i=1 vii because U is traceless. Letting z = 0 in case (i) of Lemma 3, we find that the rank of M is n − 1, and so the

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rank of the augmented matrix M |V is at least n − 1. Observe that the last row of the augmented matrix is equal to the negative sum of the first n − 1 rows, so we have rank(M ) = rank(M |V ) = n − 1. Since the last row is determined by the negative sum of the previous rows, we can focus on the first n − 1 rows and remove the last row from the linear system. Then we may proceed by Gaussian elimination to put the system into row-echelon form. Then because the coefficient matrix only contains elements from the field K, we can use back substitution of the row-echelon form to give us a solution for the system. Next, to show W ∈ im f , we show that any zero-diagonal matrix is in the image of f . Let S be a diagonal matrix with distinct elements. By Lemma 2, we can find a zero-diagonal matrix H with [S, H] = W . We can also find a zero-diagonal matrix G with [G, S] = H. Then f (S, S, G) = [S, [G, S]] + z[G, [S, S]] = [S, [G, S]] = [S, H] = W , and thus W ∈ im f . 2 4. Polynomials of degree 4 Now we will prove that for any multilinear polynomial of degree 4 and n ≥ 3, ⊆ im f + im f . If the sum of coefficients is nonzero, the result follows from Remark 1. If the sum of coefficients is zero but at least one of f (I, x3 , x3 , x4 ), f (x1 , I, x3 , x4 ), f (x1 , x2 , I, x4 ), or f (x1 , x2 , x3 , I) is nonzero, then the result follows from Theorem 2. Thus we only need to consider when Mn0

f (I, x3 , x3 , x4 ) = f (x1 , I, x3 , x4 ) = f (x1 , x2 , I, x4 ) = f (x1 , x2 , x3 , I) = 0 Applying a result by Falk [4] as in [3], we have that the form of these degree 4 polynomials is f (x1 , x2 , x3 , x4 ) = z1 [[[x2 , x1 ], x3 ], x4 ] + z2 [[[x3 , x1 ], x2 ], x4 ] + z3 [[[x4 , x1 ], x2 ], x3 ] + c1234 [x1 , x2 ][x3 , x4 ] + c1324 [x1 , x3 ][x2 , x4 ] + c1423 [x1 , x4 ][x2 , x3 ] + c2314 [x2 , x3 ][x1 , x4 ] + c2413 [x2 , x4 ][x1 , x3 ] + c3412 [x3 , x4 ][x1 , x2 ] (3) where z1 , z2 , z3 , c1234 , c1324 , c1423 , c2314 , c2413 , c3412 ∈ K. It is important to note that although the results of [3] require the field in question to be algebraically closed, the result by Falk [4] only requires that the field has characteristic 0, so we may apply it in the same way. As in [3], we separate this into several cases. 4.1. Case 1 Lemma 4. Let f be a degree 4 multilinear polynomial of the form (3) over K and evaluated on Mn (R), with n ≥ 3. If any of z1 , z2 , z3 in (3) are nonzero, then Mn0 ⊆ im f + im f .

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Proof. Consider U ∈ Mn0 . Rewrite U so that U = V + W where as before V is a traceless diagonal matrix and W is a zero-diagonal matrix. We can choose S = diag{s1 , s2 , . . . , sn } a diagonal matrix with distinct elements along the diagonal. At least one of z1 , z2 , or z3 is nonzero, so without loss of generality, we assume z1 = 0. We may then reduce our equation to f (S, x2 , S, S) = z1 [[[x2 , S], S], S]. By Lemma 2, there exist zero-diagonal matrices H, Q, and T such that [H, S] = z1−1 W , [Q, S] = H, and [T, S] = Q. Then we can write f (S, T, S, S) = z1 [[[T, S], S], S] = z1 [[Q, S], S] = z1 [H, S] = W . Thus W ∈ im f . (Note that if we instead assume z2 = 0, we would consider f (S, S, x3 , S), and if z3 = 0, we would consider f (S, S, S, x4 ) to obtain similar results.) To prove that V ∈ im f , we start with the special cases for 3 × 3, 4 × 4, and 5 × 5 matrices, which are easily checked by a computer. Then we move to the general case for 3 n × n matrices with n ≥ 6. For 3 × 3, consider A = i=1 ai Eii and B = E12 + E23 + E31 , where Eij denotes a matrix unit. Without loss of generality, assume z1 = 0. Then we have ⎛ f (B, A, B, B) = z1 [[[A, B], B], B] = z1 ⎝

⎞

−3a2 + 3a3

⎠

3a1 − 3a3 −3a1 + 3a2

and it is easy to see that all 3 × 3 traceless diagonal matrices are contained in im f . For 3 3 the 4 × 4 case, consider A = a4 E14 + i=1 ai Ei+1,i and B = E14 + i=1 Ei+1,i . Then we have f (B, A, B, B) = z1 [[[A, B], B], B] ⎛ −a + 3a − 3a + a 1

2

3

⎞

4

a1 − a2 + 3a3 − 3a4

= z1 ⎝

−3a1 + a2 − a3 + 3a4

⎠ 3a1 − 3a2 + a3 − a4

We must solve a system of equations M a = V to show that any 4 ×4 traceless diagonal matrix can be written in the above form: ⎛

−1 ⎜ 1 ⎜ ⎝ −3 3

3 −3 −1 3 1 −1 −3 1

⎞⎛ ⎞ ⎛ a1 1 ⎜ a2 ⎟ ⎜ −3 ⎟ ⎟⎜ ⎟ = ⎜ ⎜ 3 ⎠ ⎝ a3 ⎠ ⎝ −1

a4

−

u11 u22 u33 3 i=1

⎞ ⎟ ⎟ ⎟ ⎠ uii

The rank of M and M |V are both 3, and thus we can get any 4 × 4 traceless diagonal 5 matrix in im f . For 5 × 5 matrices, we choose A = a1 E41 + a2 E52 + i=3 ai Ei−2,i 4 and B = E5,1 + i=1 Ei,i+1 . Taking f (B, A, B, B) again, we obtain another system of equations for the diagonal elements, M a = V :

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⎛

−1 0 ⎜ 3 −1 ⎜ ⎜ −3 3 ⎜ ⎝ 1 −3 0 1

1 −3 0 1 −1 0 3 −1 −3 3

⎞ ⎞⎛ ⎞ ⎛ u11 3 a1 ⎟ u22 ⎜ ⎟ ⎜ −3 ⎟ ⎟ ⎟ ⎜ a2 ⎟ ⎜ ⎜ ⎟ ⎜ a3 ⎟ = ⎜ u 1 ⎟ 33 ⎟ ⎟⎜ ⎟ ⎜ ⎟ u44 ⎠ 0 ⎠ ⎝ a4 ⎠ ⎝ 4 −1 a5 − i=1 uii

As above, this system of equations has a solution, and all 5 × 5 traceless diagonal matrices are contained in im f . Now we move to the general case of an n × n matrix, n−3 n n−1 n ≥ 6. Let A = i=1 ai Ei+3,i + i=n−2 ai Ei−(n−3),i and B = En,1 + i=1 Ei,i+1 . Then ⎛

0 0

⎜ ⎜ ⎜ ⎜a − a ⎜ n 1 [A, B] = ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛

an−2 − an−1 0

0 0

...

0 a1 − a2

...

...

...

...

0 0

an−3 − an−2

⎞ 0 an−1 − an ⎟ ⎟ ⎟ ⎟ 0 ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 0 an−2 − 2an−1 + an

0

⎜ ⎜ an−1 − 2an + a1 ⎜ [[A, B], B] = ⎜ ⎜ ⎜ ⎝

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

... an − 2a1 + a2

... ...

⎞

... an−3 − 2an−2 + an−1

0

And finally we have that f (B, A, B, B) = z1 [[[A, B], B], B] = ⎛ ⎜ z1 ⎜ ⎝

−a1 + an−2 − 3an−1 + 3an

⎞

3a1 − a2 + an−1 − 3an

⎟ ⎟ ⎠

...

an−3 − 3an−2 + 3an−1 − an

Once more we have a system of equations using the terms along the diagonal, M a = V : ⎛

−1 ⎜ 3 −1 ⎜ ⎜ −3 3 ⎜ ⎜ ⎜ 1 −3 ⎜ ⎜ 1 ⎜ ⎜ ⎜ ⎜ ⎝

1 −1 3 −3 ...

−1 3 ... 1

−1 ...

...

−3 1

3 −3

⎞ ⎞⎛ a ⎞ ⎛ u11 1 −3 3 ⎜ a2 ⎟ ⎜ ⎟ u22 1 −3 ⎟ ⎟ ⎜ ⎟ ⎟⎜ ⎜ ⎟ ⎜ ⎟ u33 a ⎜ ⎟ ⎜ ⎟ 3 1 ⎟ ⎟⎜ ⎟ ⎜ ⎟ .. ⎟ ⎜ .. ⎟ ⎜ ⎟ . ⎟⎜ . ⎟ ⎜ ⎟ ⎟⎜ . ⎟ = ⎜ . ⎟ .. ⎟⎜ . ⎟ ⎜ ⎟ ⎟⎜ . ⎟ ⎜ ⎟ .. ⎟⎜ . ⎟ ⎜ ⎟ ⎟ ⎜ .. ⎟ ⎜ . ⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎠⎝a −1 u ⎠ ⎝ ⎠ n−1,n−1 n−1 n−1 3 −1 an − uii i=1

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By part (ii) of Lemma 3, M has rank n − 1. The last row of M is the negative sum of all previous rows, so the augmented matrix M |V also has rank n − 1. Therefore, this system has a solution and we can choose elements such that an arbitrary traceless diagonal matrix is in the image of f . Thus, U = V + W ∈ im f + im f . 2 4.2. Case 2 Lemma 5. If f (x1 , x2 , x3 , x4 ) = [x1 , x2 ][x1 , x3 ] + λ[x1 , x3 ][x1 , x2 ] with λ ∈ K, λ = −1 and f evaluated on Mn (R) with n ≥ 3, then Mn0 ⊆ im f + im f . n−1 Proof. Let X = i=1 Ei+1,i . By Lemma 1, we can choose C such that [X, C] = −(n − n 1)E11 + i=2 Eii , where Eij denotes a matrix unit, and we can choose B such that we have a lower triangular matrix ⎛

0

⎜ b2,1 ⎜ b3,1 ⎜ [X, B] = ⎜ ⎜ ... ⎜ ⎝ bn−1,1 bn,1

0 (λ + 1)−1 a2 b3,2 .. . bn−1,2

0 0 (λ + 1)−1 a3 .. . bn−1,3

... ... ... ...

0 0 0 .. . (λ + 1)−1 an−1

bn,2

bn,3

...

bn,n−1

...

⎞

0 0 0 .. . 0 −(λ + 1)−1

n−1 i=2

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ai

where each bi,j denotes an arbitrary element of R that can be chosen independently of any other element. Then let M1 = [X, B][X, C] + λ[X, C][X, B] so that ⎛

0 ⎜ (λ − (n − 1))b2,1 ⎜ (λ − (n − 1))b 3,1 ⎜ M1 = ⎜ .. ⎜ . ⎜ ⎝ (λ − (n − 1))bn−1,1 (λ − (n − 1))bn,1

0 a2

0 0

... ...

0 0

0 0

(λ + 1)b3,2 .. . (λ + 1)bn−1,2

a3 .. . (λ + 1)bn−1,3

...

0 .. .

...

an−1

0 .. . 0

(λ + 1)bn,2

(λ + 1)bn,3

...

(λ + 1)bn,n−1

...

−

n−1 i=2

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ai

Since we assumed λ = −1, if λ = n − 1, then the bi,j elements each remain unchanged except for a scalar multiplication, so they remain arbitrary. Assuming λ = n − 1, we can n−1 obtain any matrix with diagonal elements 0, a2 , a3 , . . . , − i=2 ai and arbitrary elements below the diagonal. By Remark 1 (ii), matrices similar to M1 are also in im f . Consider the similarity transformation M2 = P M1 P −1 where ⎛ P = P −1 = ⎝

.. 1

.

1

⎞ ⎠

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This gives the matrix ⎛ − n−1 a

(λ + 1)bn,n−1

(λ + 1)bn,n−2

...

(λ + 1)bn,2

an−1

(λ + 1)bn−1,n−2

...

(λ + 1)bn−1,2

0

0

an−2

...

1)bn−2,2

. .. 0 0

. .. 0 0

. .. 0 0

...

i=2

⎜ ⎜ ⎜ M2 = ⎜ ⎜ ⎜ ⎝

i

0

(λ +

... ...

. .. a2 0

(λ − (n − 1))bn,1

⎞

(λ − (n − 1))bn−1,1 ⎟

⎟ ⎟ ⎟ ⎟ ⎠

(λ − (n − 1))bn−2,1 ⎟ . .. (λ − (n − 1))b2,1 0

M2 has arbitrary elements of the ring above the main diagonal as well as a traceless main diagonal. Adding together M1 + M2 clearly gives us arbitrary elements of the ring in entries above and below the main diagonal. Since M1 and M2 are each traceless, their sum is also traceless. Let all the main diagonal entries of M2 to be set equal to zero except for the (1, 1) and (2, 2) entries. Since M2 is traceless, the (2, 2) entry will be the negative of the (1, 1) entry. By observation, the main diagonal elements of M1 and M2 can be chosen such that M1 + M2 can have any arbitrary traceless main diagonal. Thus, any traceless matrix in Mn (R) is contained in im f + im f . n−1 Now we examine the case where λ = n − 1. Again let X = i=1 Ei+1,i . Applying n−1 Lemma 1, we can choose C such that [X, C] = −(n − 1)Enn + i=1 Eii , and we can choose B such that  −(λ + 1)−1 n−1 i=2 ai ⎜ b 2,1 ⎜ ⎜ b3,1 ⎜ [X, B] = ⎜ .. ⎜ ⎜ . ⎜ ⎝ bn−1,1 bn,1 ⎛

0 (λ + 1)−1 a2 b3,2 .. . bn−1,2 bn,2

0 0 (λ + 1)−1 a3 .. . bn−1,3 bn,3

... ... ... ...

... ...

0 0 0 .. . (λ + 1)−1 an−1 bn,n−1

⎞ 0 0⎟ ⎟ 0⎟ ⎟ .. ⎟ ⎟ .⎟ ⎟ 0⎠ 0

where each bi,j denotes an arbitrary element of R that can be chosen independently of any other element. Then M1 = [X, B][X, C] + λ[X, C][X, B] so that ⎛ ⎜ ⎜ ⎜ M1 = ⎜ ⎜ ⎜ ⎝

−

n−1 i=2

ai

0

0

...

0

(λ + 1)b2,1

a2

0

...

0

(λ + 1)b3,1 . . . (λ + 1)bn−1,1

(λ + 1)b3,2 . . . (λ + 1)bn−1,2

a3 . . . (λ + 1)bn−1,3

...

... ...

0 . . . an−1

(1 − λ(n − 1))bn,1

(1 − λ(n − 1))bn,2

(1 − λ(n − 1))bn,3

...

(1 − λ(n − 1))bn,n−1

0

⎞

0⎟

⎟

0⎟ ⎟ .⎟ .⎟ . ⎠ 0 0

Since λ = −1 and λ(n − 1) = 1 for λ = (n − 1)−1 (recall that we assumed λ = n − 1), all of the bi,j terms are multiplied by a nonzero constant, so the terms below the diagonal remain arbitrary. Applying Remark 1 (ii) as before, we use a similarity transformation

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M2 = P M1 P −1 and add M1 + M2 . By the same reasoning as above, we find that any traceless matrix is contained in im f + im f . 2 4.3. Case 3 Lemma 6. If f (x1 , x2 , x3 , x4 ) = [x1 , x2 ][x1 , x3 ] − [x1 , x3 ][x1 , x2 ] = [[x1 , x2 ], [x1 , x3 ]] and is evaluated on Mn (R) with n ≥ 3, then Mn0 ⊆ im f + im f . n−1 n−2 Proof. Let A = E1,n + i=1 Ei+1,i and B = −(n − 1)En−1,1 + i=1 −iEi,i+2 so that n−1 [A, B] = −(n − 1)En,1 + i=1 Ei,i+1 , where Eij denotes a matrix unit. We take C with entries cij for 1 ≤ i, j ≤ n, so that D = [A, C] has entries di,j = ci−1,j − ci,j+1 . Here an index of 0 means an index of n, and an index of n + 1 means an index of 1, so that for example c0,n+1 is actually cn,1 . Then V = [[A, B], [A, C]] has the form ⎛ ⎜ ⎜ ⎜ ⎜ V =⎜ ⎜ ⎜ ⎝

d21

d22

...

d2n

d31 .. .

d32 .. .

...

d3n .. .

...

dn1 dn2 −(n − 1)d11 −(n − 1)d12 ⎛ −(n − 1)d1n d11 ⎜ −(n − 1)d d21 2n ⎜ ⎜ ⎜ .. .. −⎜ . . ⎜ ⎜ ⎝ −(n − 1)dn−1,n dn−1,1 −(n − 1)dnn

dn1

... ...

dnn −(n − 1)d1n ⎞ ... d1,n−1 ... d2,n−1 ⎟ ⎟ ⎟ ⎟ . ... .. ⎟ ⎟ ⎟ . . . dn−1,n−1 ⎠ ...

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

dn,n−1

The entries on and above the main diagonal may be explicitly described as v11 = −(n − 2)c11 + (n − 1)cnn − c22 vnn = (n − 1)c11 − (n − 2)cnn − cn−1,n−1 vij = di+1,j − di,j−1 = 2ci,j − ci+1,j+1 − ci−1,j−1 for 1 ≤ i < n, 1 < j ≤ n, and j ≥ i. We wish to show that for some choice of C, we can construct a traceless matrix that has an arbitrary main diagonal with terms that sum to zero, as well as arbitrary elements in R above the main diagonal. Note that we are not worried about what the terms below the main diagonal are, as these terms can be corrected later using the second matrix we will construct. For the main diagonal, we must show that the following system of equations M c = V0 has a solution:

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278

⎛

−(n − 2) −1 ⎜ −1 2 ⎜ ⎜ −1 ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

n−1

−1 2 ...

−1 ... −1

... 2 −1

(n − 1)

−1 2 −1

⎞⎛

c11 c22 c33 c44 .. .

⎞

⎛

v11 v22 v33 v44 .. .

⎞

⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎟⎜ ⎟=⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎟⎝ −1 ⎠ cn−1,n−1 ⎠ ⎝ vn−1,n−1 ⎠ n−1 cnn −(n − 2) − i=1 vii

Letting z = n in part (i) of Lemma 3, M at least has rank n − 1. Since the last row of M is the negative sum of the previous rows, it is evident that M |V0 has rank n − 1, which implies that the system of equations does have a solution. For the 1-diagonal, we must show that the following system of equations M c = V1 has a solution: ⎛ 2 −1 ⎜ −1 2 ⎜ ⎜ −1 ⎜ ⎜ ⎜ ⎜ ⎝

−1 2 ...

−1 ... −1

... 2 −1

⎛ ⎞ ⎛ ⎞ v12 −1 ⎞ c12 ⎜ ⎟ ⎟⎜ c23 ⎟ ⎟ ⎜ v23 ⎟ ⎟⎜ ⎜ ⎜ ⎟ ⎟ ⎜ c34 ⎟ ⎜ v34 ⎟ ⎟ ⎟⎜ .. ⎟ =⎜ ⎟ ⎜ .. ⎟ ⎜ ⎟ ⎟⎜ . ⎟ ⎜ . ⎟ ⎟⎜ . ⎟ ⎜ . ⎟ ⎠ ⎝ .. ⎠ ⎝ .. ⎟ ⎠ −1 2 −1 cn1 vn−1,n

Letting z = 0, by part (i) of Lemma 3, this matrix has rank n −1. Thus the 1-diagonal can be chosen arbitrarily. Finally, we wish to consider the k-diagonals for 1 < k < n. The entries of each k-diagonal can be placed into an (n − k) × (n − k + 2) coefficient matrix, which gives the following system of equations M c = Vk : ⎛ −1 ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

2 −1

−1 2 −1

−1 2 ...

−1 ... −1

... 2 −1

⎛

⎞ v1,1+k v2,2+k ⎟ ⎟ ⎜ c1,1+k ⎟ ⎜ ⎟ ⎟ ⎜ ⎟⎜ ⎟ ⎜ c2,2+k ⎟ ⎜ v3,3+k ⎟ ⎜ ⎟ ⎟ ⎜ ⎟⎜ ⎟ = ⎜ .. ⎟ ⎟⎜ .. ⎟ ⎜ . ⎟ ⎟⎜ . ⎟ ⎜ . ⎟ ⎟⎜ ⎠ ⎝ . ⎟ ⎠⎝ c −1 . ⎠ n−k,n 2 −1 cn−k+1,1 vn−k,n ⎞⎛

cn,k

⎞

The rows of M are clearly all linearly independent by observation, so the k-diagonals can all be chosen arbitrarily. Thus, we have that for some choice of C, V can have any main diagonal with terms that sum to zero, as well as arbitrarily chosen elements in R above the main diagonal. As mentioned above, we are not concerned with the form of the elements below the main diagonal. The second matrix we construct is an arbitrary strictly lower triangular matrix. Consider a commutator of the form [[X, B], [X, C]], where B can be chosen by Lemma 1 such that [X, B] = S = diag{s1 , s2 , . . . , sn } is a traceless diagonal matrix with distinct

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n−1 elements si = i for 1 ≤ i ≤ n − 1 and sn = − i=1 i, and C can be chosen such that [X, C] = W is a lower-triangular matrix with arbitrary elements wij for i > j and wij = 0 for j ≥ i. The (i, j)th entry of the commutator [S, W ] is (si − sj )wij for i > j and 0 otherwise. Since si − sj is never 0, we can choose the elements of W such that any arbitrary strictly lower-triangular matrix is in the image of f . Fixing the main diagonal and upper triangular elements using V and then correcting the lower triangular elements using W , we find that we can obtain any traceless matrix in im f + im f . 2 4.4. Case 4 Lemma 7. If f (x1 , x2 , x3 , x4 ) = [x1 , [[x1 , x2 ], [x1 , x3 ]]] and is evaluated on Mn (R) with n ≥ 3, then Mn0 ⊆ im f + im f . Proof. We begin with some special cases for small values of n. If n = 3, let X = E21 +E32 , where Eij denotes a matrix unit. Then by Lemma 1, for a, b, c, d, e ∈ R, we can find matrices B and C such that ⎛ ⎞ ⎛ ⎞ a 0 0 1 1 0 [X, B] = ⎝ b c 0 ⎠ and [X, C] = ⎝ 1 1 −1 ⎠ d

e

−a − c

1 1

−2

Also, we may consider, for q, r, s, t ∈ R, ⎛

1 A = ⎝0 0

0 2 0

⎞ ⎛ 0 0 q 0⎠, D = ⎝s 0 3 0 0

⎞ ⎛ r 0 t ⎠ , E = ⎝ −1 − 12 0

−1 0 −1

⎞ − 12 −1 ⎠ 0

A computer calculation shows that ⎛

c−a M = [X, [[X, B], [X, C]]] + [A, [[A, D], [A, E]]] = ⎝ t − e − 2b 3c − 3e + 2s

−2r 2a + c b + 2e − q

⎞ 2q − 2t 2r − s ⎠ −a − 2c

We can then consider the system of 8 equations corresponding to all of the entries in M except for the lower rightmost entry. Another computer calculation shows that these equations are linearly independent, and thus any traceless 3 × 3 matrix is contained in im f + im f . The 4 × 4 and 5 × 5 cases are similar to the general solution, however we give them separately because these small cases involve slightly different coefficients. If n = 4, let 3 X = i=1 Ei+1,i , then apply Lemma 1 to find B and C such that ⎛

0 ⎜0 [X, B] = ⎜ ⎝0 0

1 0 0 0

⎛ ⎞ a1 0 0 ⎜ c1 1 0 ⎟ ⎟ , [X, C] = ⎜ ⎝ d1 0 −2 ⎠ 0 0 g1

0 a2 c2 d2

b1 0 a3 c3

⎞ 0 ⎟ −b1 ⎟ ⎠ 0 −a1 − a2 − a3

280

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Then we define M1 so that M1 = [X, [[X, B], [X, C]]] ⎛ a1 − a2 0 ⎜ 2c1 − c2 −a1 + 2a2 − a3 =⎜ ⎝ 2d1 + 2d2 −c1 + 2c2 + 2c3 −d1 − d2 −g1

−b1 0 −2a1 − 3a2 − 3a3 −c2 − 4c3

⎞ 0 ⎟ b1 ⎟ ⎠ 0 2a1 + 2a2 + 4a3

We then take a similarity transformation using the permutation matrix ⎛ P = P −1 = ⎝

..

1

.

⎞ ⎠

1 such that M2 = P M1 P −1 ⎛  2a1 + 2a2 + 4a3 ⎜ ⎜ 0 =⎜ ⎜ b1 ⎝ 0

−c2 − 4c3

−d1 − d2

−g1

−2a1 − 3a2 − 3a3

−c1 + 2c2 + 2c3

0

−a1 + 2a2 − a3

−b1

0

⎞

⎟ 2d1 + 2d2 ⎟ ⎟ 2c1 − c2 ⎟ ⎠   a1 − a2

M2 is in im f by Remark 1. Adding M1 + M2 and checking the systems of equations quickly gives us that any traceless 4 × 4 matrix is in im f . Note that here, considering M1 on its own, the (−2)-diagonal terms cannot be chosen arbitrarily, since one is a multiple of the others. However, upon adding the terms of the (−2)-diagonal of M2 to the terms of the (−2)-diagonal of M1 , we find that now the (−2)-diagonal of M1 + M2 can be chosen arbitrarily. Similarly, although the 2-diagonal of M2 cannot be chosen arbitrarily, the 2-diagonal of M1 + M2 can be chosen arbitrarily. 4 If n = 5, we take X = i=1 Ei+1,i , then apply Lemma 1 to choose B and C such that ⎛

0 ⎜0 ⎜ [X, B] = ⎜ ⎜0 ⎝0 0

1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

⎛ ⎞ a1 0 ⎜ c1 0 ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎟ , [X, C] = ⎜ d1 ⎝h ⎠ −3 1 g1 0

0 a2 c2 d2 h2

b1 0 a3 c3 d3

0 b2 0 a4 c4

⎞ 0 ⎟ 0 ⎟ ⎟ −b1 − b2 ⎟ ⎠ 0 −a1 − a2 − a3 − a4

so that M1 = [X, [[X, B], [X, C]]] has the form ⎛

a1 − a 2 ⎜ 2c1 − c2 ⎜ 2d1 − d2 ⎝ 2h1 + 3h2 −2g1

0 −a1 + 2a2 − a3 −c1 + 2c2 − c3 −d1 + 2d2 + 3d3 −h1 − 2h2

b1 − b2 0 2a3 − a2 − a4 −c2 + 2c3 + 3c4 −d2 − 2d3

0 −b2 0 −3a1 − 3a2 − 4a3 − 5a4 −c3 − 6c4

⎞

0 0 ⎟ ⎟ 2b2 − b1 ⎠ 0 3a1 + 3a2 + 3a3 + 6a4

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Then as above, we take the similarity transformation M2 = P M1 P −1 to get our second matrix. Once again, the (−2)-diagonal of M1 is not able to be chosen arbitrarily, but the extra terms provided after adding M2 solves this problem, and similarly, adding M1 provides extra terms for M2 in the 2-diagonal. Thus, adding M1 + M2 and checking the systems of equations, we find that any traceless 5 × 5 matrix is contained in im f + im f . The general solution for n ≥ 6 works almost identically. Applying Lemma 1, we n−1 let X = i=1 Ei+1,i and can pick B and C such that [X, B] = −(n − 2)En−1,n + n−2 i=1 Ei,i+1 , and ⎛ ⎞ a1 0 b1 0 ... 0 ⎜ .. ⎟ ... ... ... ... ⎜ c11 . ⎟ ⎟ ⎜ ⎜ ⎟ ... ... ... ... ⎜ c 0 ⎟ ⎟ ⎜ 21 [X, C] = ⎜ ⎟ .. ... ... ... ... ⎟ ⎜ ⎜ bn−2 ⎟ . ⎟ ⎜ ⎟ ⎜ .. ... ... ... ⎝ 0 ⎠ . an c(n−1)1 . . . . . . c2n−2 c1n−1 n n−2 where i=1 ai = 0, i=1 bi = 0, and cki indicates the ith term of the (−k)-diagonal and denotes an arbitrary element of R that can be chosen independently of any other element. Taking the commutator [X, [[X, B], [X, C]]], we have diagonals as follows: n−3 Along the 2-diagonal, we have the terms (where bn−2 = i=1 bi ) b1 − b2 −b1 + 2b2 − b3 .. . −bn−5 + 2bn−4 − bn−3 −(n − 3)bn−3 − bn−4 − bn−2 = −(n − 4)bn−3 + bn−5 + bn−6 + . . . + b1 bn−2 + (n − 2)bn−3 = (n − 3)bn−3 − bn−4 − bn−5 − . . . − b1 The main diagonal has terms (where an =

n−1 i=1

ai )

a1 − a2 −a1 + 2a2 − a3 .. . −an−3 + 2an−2 − an−1 (n − 2)an − an−2 − (n − 3)an−1 = −(2n − 5)an−1 − (n − 1)an−2 − (n − 2)(an−3 + . . . + a1 ) −(n − 2)an + (n − 2)an−1 = (2n − 4)an−1 + (n − 2)(an−2 + . . . + a1 )

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The (−1)-diagonal has terms 2c11 − c12 −c11 + 2c12 − c13 .. . −c1n−4 + 2c1n−3 − c1n−2 −c1n−3 + 2c1n−2 + (n − 2)c1n−1 −c1n−2 − 2(n − 2)c1n−1 The (−)-diagonal 2 ≤  ≤ (n − 4) has terms 2c1 − c2 −c1 + 2c2 − c3 .. . −cn−−3 + 2cn−−2 − cn−−1 −cn−−2 + 2cn−−1 + (n − 2)cn− −cn−−1 − (n − 3)cn− The −(n − 3)-diagonal has terms 2c(n−3)1 − c(n−3)2 −c(n−3)1 + 2c(n−3)2 + (n − 2)c(n−3)3 −c(n−3)2 − (n − 3)c(n−3)3 The −(n − 2)-diagonal has terms 2c(n−2)1 + (n − 2)c(n−2)2 −c(n−2)1 − (n − 3)c(n−2)2 The (n, 1) term will be −(n − 3)c(n−1)1 , which is arbitrary, and straightforward calculations show that the −(n − 2)-diagonal terms can be chosen independently of each other when n ≥ 5 and the −(n − 3)-diagonal terms can be chosen independently when n ≥ 6.

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The (−)-diagonals, for 2 ≤  ≤ n − 4, form systems of equations M k = V of the form ⎛ 2 ⎜ −1 ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

−1 2 ...

⎞⎛ −1 ... −1

... 2 −1

−1 2 −1

(n − 2) −(n − 3)

⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

⎞

⎛ v ⎞ 1 ⎟ ⎜ v ⎟ ⎟ ⎜ 2 ⎟ ⎟ ⎜ . ⎟ ⎟ = ⎜ .. ⎟ ⎟ ⎜ ⎟ ⎝ .. ⎟ ⎠ . ⎠ vn− cn− c1 c2 .. . .. .

When n = 6, the only possible value is  = 2. The case where  = 2 is addressed below. When n ≥ 7, by Lemma 3 (iii), M has rank n −  for 3 ≤  ≤ n − 4, and so M |V has rank n − . Then these systems of equations have solutions, and the (−)-diagonal can be chosen arbitrarily. The terms of the (−2)-diagonal are not able to be chosen independently, so we use the same similarity transformation as above to get M2 = P M1 P −1 . As in the 4 × 4 and 5 × 5 cases, the (−2)-diagonal of M2 is determined by the 2-diagonal of M1 . After adding M2 to M1 , the first, second, and third entries of the (−2)-diagonal of M1 have (n − 3)bn−3 , −(n − 4)bn−3 , and −bn−3 added, respectively. These are also the only rows that have bn−3 terms. A calculation shows that these new terms now make it possible to choose the (−2)-diagonal arbitrarily. M2 suffers the same issue except in the 2-diagonal, but as before, this is remedied by the entries of the 2-diagonal of M1 . The (−1)-diagonal forms a system of equations M k = V1 of the form ⎛ 2 ⎜ −1 ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

−1 2 ...

⎞⎛ −1 ... −1

... 2 −1

−1 2 −1

(n − 2) −2(n − 2)

⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

⎞

⎛ v ⎞ 1 ⎟ ⎜ v ⎟ ⎟ ⎜ 2 ⎟ ⎟ ⎜ . ⎟ ⎟ = ⎜ .. ⎟ ⎟ ⎜ ⎟ ⎝ .. ⎟ ⎠ . ⎠ vn−1 c1n−1 c11 c12 .. . .. .

We apply the same reasoning as in the proof of Lemma 3 (iii), except now the terms in the rightmost column force −(n − 2)(n − 2) + 2(n − 2)(n − 1) = n2 − 2n = 0 for which the only integer solutions are n = 0, 2. Since neither of these are possible by assumption, M must have rank n − 1, and so M |V1 must also have rank n − 1. Thus a solution exists for arbitrary V1 .

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Finally, we come to the terms of the main diagonal. Here we have another system of equations M a = V0 : ⎛

1 −1

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ −(n − 2) (n − 2)

⎞

−1 2 ...

−1 ...

...

... ...

−1 −(n − 2) ...

2 −(n − 1) (n − 2)

⎛ a ⎞ ⎛ ⎞ v1 1 ⎟ ⎟⎜ a ⎟ ⎜ ⎟ v2 ⎟⎜ 2 ⎟ ⎜ ⎟ ⎟⎜ . ⎟ ⎜ . ⎟ .. ⎟ ⎜ .. ⎟ = ⎜ ⎟ ⎟⎜ ⎜ ⎟ ⎟ ⎟ ⎝ .. ⎠ ⎝ v −1 ⎠ ⎟ n−1 . n−1 −(2n − 5) ⎠ an−1 − i=1 vi (2n − 4)

From Lemma 3 (iv), M has rank n − 1, and since the last row of M is the negative sum of all the previous rows, it is clear that the augmented matrix M |V also has rank n − 1 and this system has a solution. Thus, it follows that the terms of any diagonal can be chosen arbitrarily after adding M1 + M2 , except for the main diagonal, for which we can choose any diagonal with terms summing to zero. Therefore, any traceless matrix in Mn (R) is contained in im f + im f . 2 4.5. Main result We are now ready to prove the main result. Theorem 3. If f is a nonzero multilinear polynomial of degree 4 over K and evaluated on Mn (R), where R is a unital associative algebra over a field K of characteristic 0 and n ≥ 3, then Mn0 ⊆ im f + im f . Proof. As shown before, we can write the nontrivial cases of the polynomial in the form of Equation (3). If we assume that at least one of z1 , z2 , z3 is nonzero then the result follows from Lemma 4. Assume z1 = z2 = z3 = 0. If c1234 = c2314 = c3412 = c1423 = −c1324 = −c2413 holds, then as in [3] the polynomial takes the form f (x1 , x2 , x3 , x4 ) = c1234 ([x1 , x2 ][x3 , x4 ] + [x3 , x4 ][x1 , x2 ] + [x2 , x3 ][x1 , x4 ] + [x1 , x4 ][x2 , x3 ] − [x1 , x3 ][x2 , x4 ] − [x2 , x4 ][x1 , x3 ]) Let A, B, and C be elements of Mn (R). We can then take f (A, A2 , B, C) = c1234 ([A, A2 ][B, C] + [B, C][A, A2 ] + [A2 , B][A, C] + [A, C][A2 , B] − [A, B][A2 , C] − [A2 , C][A, B]) = c1234 ([A2 , B][A, C] + [A, C][A2 , B] − [A, B][A2 , C] − [A2 , C][A, B]) = c1234 ([A, [[A, B], [A, C]]])

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Then the result follows from Lemma 7. If c1234 = c2314 = c3412 = c1423 = −c1324 = −c2413 does not hold then, as in [3], at least one of the following expressions is not zero, where again A, B, and C are from Mn (R): f (A, A, B, C) = (c1324 + c2314 )[A, B][A, C] + (c1423 + c2413 )[A, C][A, B] f (A, B, A, C) = (c1234 − c2314 )[A, B][A, C] + (c3412 − c1423 )[A, C][A, B] f (A, B, C, A) = (−c1234 − c2413 )[A, B][A, C] + (−c1324 − c3412 )[A, C][A, B] f (B, A, A, C) = (−c1234 − c1324 )[A, B][A, C] + (−c2413 − c3412 )[A, C][A, B] f (B, A, C, A) = (−c1423 + c1234 )[A, B][A, C] + (c3412 − c2314 )[A, C][A, B] f (B, C, A, A) = (c1324 + c1423 )[A, B][A, C] + (c2314 + c2413 )[A, C][A, B] Thus the polynomial is reduced to the form f = [x1 , x2 ][x1 , x3 ] + λ[x1 , x3 ][x1 , x2 ]. If λ = −1, then the result follows from Lemma 5. If λ = −1, then the result follows from Lemma 6. All cases have been addressed, and the proof is complete. 2 5. Examples A natural question is whether Conjecture 1 will hold when there is no multiplicative identity. Our first example shows that this is not the case, even for polynomials of degree 2. Example 1. Let R be the commutative polynomial ring over a field K that contains only polynomials with constant term equal to zero in the variables xi , yi for 1 ≤ i ≤ 6. Consider the matrices   



0 x1 0 x3 0 x5 , A2 = , A3 = −y2 0 −y4 0 −y6 0   



0 x2 0 x4 0 x6 , B2 = , B3 = B1 = y1 0 y3 0 y5 0

A1 =

Then the upper leftmost entry of [A1 , B1 ] + [A2 , B2 ] + [A3 , B3 ] is x1 y1 + x2 y2 + x3 y3 + x4 y4 + x5 y5 + x6 y6 . Also consider the matrices

A4 =

∗ −β2

α1 ∗



, A5 =

∗ −β4

α3 ∗



, B4 =

∗ β1

α2 ∗



, B5 =

∗ β3

α4 ∗



where each ∗ denotes an arbitrary element of R that can be chosen independently of any other element and αi , βi ∈ R for 1 ≤ i ≤ 4. The upper leftmost entry in the matrix [A4 , B4 ] + [A5 , B5 ] is α1 β1 + α2 β2 + α3 β3 + α4 β4 .

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We wish to show that there is no way to choose αi , βi ∈ R, for 1 ≤ i ≤ 4, such that x1 y1 + x2 y2 + x3 y3 + x4 y4 + x5 y5 + x6 y6 = α1 β1 + α2 β2 + α3 β3 + α4 β4

(4)

First note that it is impossible to choose α and β to make x1 y1 + x2 y2 = αβ. In order to seek a contradiction, assume that this is possible. Then one of the terms α or β must contain a constant multiple of x1 . Without loss of generality, let α = k1 x1 + 1 , where k1 ∈ K and 1 ∈ R. We would like to replace x1 with −k1−1 1 . Formally, we consider the factor ring R → R/x1 + k1−1 1 . Under this mapping, we obtain −k1−1 1 y1 + x2 y2 = 0. Now map R/x1 + k1−1 1  → R/x1 + k1−1 1 , y1 . Under this mapping, our equation becomes x2 y2 = 0. In both of these homomorphic images, x2 and y2 are mapped to x2 and y2 respectively, thus x2 y2 = 0 and we have the desired contradiction. Now, in order to seek a contradiction, assume that we can pick αi βi such that Equation (4) holds. As above, the way to proceed is fairly straightforward. We take repeated homomorphic images to annihilate the αβ pairs on the right-hand side of the equation and ultimately come to a contradiction when the right-hand side is zero and the left-hand side is nonzero. Thus in general, the sum of three commutators cannot be written as the sum of two commutators. However, our second example gives a nonunital ring where this is possible. This suggests that Conjecture 1 may be true for certain types of nonunital rings. Example 2. Let R = kZ, for an integer k > 1. For any matrices Ai , Bi ∈ Mn (kZ), 1 ≤ i ≤ 3, let Ai = k1 Ai and Bi = k1 Bi for 1 ≤ i ≤ 3. Note that Ai , Bi ∈ Mn (Z). By Theorem 1, there exist A4 , A5 , B4 , B5 ∈ Mn (Z) such that [A1 , B1 ] + [A2 , B2 ] + [A3 , B3 ] = [A4 , B4 ] + [A5 , B5 ] Then, letting A4 = kA4 , A5 = kA5 , B4 = kB4 , and B5 = kB5 , we find that [A1 , B1 ] + [A2 , B2 ] + [A3 , B3 ] = [A4 , B4 ] + [A5 , B5 ] where A4 , A5 , B4 , B5 ∈ Mn (kZ). Acknowledgements We would like to thank our mentor, Dr. Mikhail Chebotar, for his guidance and support. We would also like to thank the Department of Mathematical Sciences at Kent State University for its hospitality. We are grateful to the referee for his/her useful comments and suggestions. The authors are supported in part by the NSF, grant DMS 1156798.

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