6
$\begingroup$

$1.$ Are trace $0$ matrices always of the form $AB-BA$?

$2.$ Is a trace $0$ matrix over the complex field always similar to a matrix with $0$ as a diagonal element?

$3.$ Is a trace $0$ matrices over any field always similar to a matrix with $0$ as a diagonal element?

$4.$ Is a trace $0$ matrix not invertible if it is upper triangular.?

I solved one problem in hoffman kunze saying : $W$ be the span of $n\times n$ matrices over the field $F$ and $W_0$ be the subspace spanned by the matrices $C$ where $C=AB-BA$. Then we proved there that $W_0$ is the exactly subspace of matrices which have trace $0$, so from this result can we say $1$ is true?

$\endgroup$
  • $\begingroup$ 4. No: Think of a $2\times 2$ with $+1$ and $-1$ on the diagonal. $\endgroup$ – Hagen von Eitzen Jun 5 '13 at 18:33
  • $\begingroup$ 3. fails as well; take the same matrix as Hagen's but over $\mathbb{F}_2$. $\endgroup$ – fuglede Jun 5 '13 at 18:37
  • $\begingroup$ Your argument doesn't prove (1) because you've only proved that $C$ is a linear sum of matrices of the form $AB-BA$. $\endgroup$ – Thomas Andrews Jun 5 '13 at 18:50
  • $\begingroup$ @SamiBenRomdhane: $\begin{pmatrix}-1 & 1 \\ 1 & 1\end{pmatrix}^{-1} \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}-1 & 1 \\ 1 & 1\end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0\end{pmatrix}$ $\endgroup$ – fuglede Jun 5 '13 at 18:57
4
$\begingroup$

For (1), see the citation in my answer to a previous question. In particular, yes, the set of all traceless matrices are precisely the set of all commutators, regardless of the underlying field.

The exercise in Hoffman and Kunze asks whether the subspace of all traceless matrices is equal to the subspace spanned by all commutators. This is different from asking whether the subspace of all traceless matrices is equal to the set of all commutators. Put it another way, the exercise in Hoffman and Kunze evades the question of whether all commutators form a matrix subspace.

For (2), see my aforementioned answer again.

For (3) and (4), consider $I_2$ over $\mathbb{F}_2$.

$\endgroup$
1
$\begingroup$

Actually, written like this, the answer to (1) could be negative.

Consider the ring of polynomials in the indeterminates $x, y, z$ over a field $F$, $R= F[x,y,z]$, and $M=\begin{pmatrix}x&y\\z&-x\end{pmatrix} \in M_2(R)$.

There is no $A, B\in M_2(R)$ such that $M=AB-BA$.

$\endgroup$
  • 1
    $\begingroup$ Might I ask why the claim is true? I find it not so obvious. Thanks. $\endgroup$ – awllower Jul 14 '13 at 6:06
-1
$\begingroup$
  1. If it is the matrix algebra $\mathfrak{sl}_n$ over a field then yes, because $\mathfrak{sl}_n$ is a simple Lie algebra and $[\mathfrak{sl}_n,\mathfrak{sl}_n]$ is a nontrivial ideal of $\mathfrak{sl}_n$ thus equal to it.

  2. No, look at the matrix with diagonal $(1, -1)$.

  3. See 2.

  4. See 2.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.