MHB How can you prove that a matrix is equal to zero if its trace is always zero?

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Here is this week's problem!

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Let $\Bbb k$ be a field. Suppose $A \in M_n(\Bbb k)$ such that $\operatorname{trace}(AM) = 0$ for all $M \in M_n(\Bbb k)$. Prove $A = 0$. Furthermore, show that the linear transformation $L_B : M_n(\Bbb k) \to M_n(\Bbb k)$ given by $L_B(X) = BX$ is an isometry with respect to the inner product $\langle X,Y\rangle = \operatorname{trace}(XY^T)$ if and only if $B$ is orthogonal.

Note: Due to the fact there there are essentially two problems this week, I have simplified this problem so that you may, if you wish, rely on the theory of eigenvalues and eigenvectors of matrices over fields.
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No answered this week's problem. You can view my solution below.

Spoiler

First Part
Let $A = (a_{ij})$. For fixed indices $i$, $j$, let $E_{ji}$ be the matrix with entry $1$ in position $(j,i)$ and zeros everywhere else. Then

$$(AE_{ji})_{\mu\nu} = \sum_k a_{\mu k} (E_{ji})_{k\nu} = \begin{cases}a_{\mu j}& i = \nu\\0&i = \nu\end{cases}.$$

That is, $(AE_{ji})_{\mu \nu} = a_{\mu j} \delta_{i\nu}$. Hence

$$0 = \operatorname{trace}(AE_{ji}) = \sum_\mu a_{\mu j} \delta_{i\mu} = a_{ij}.$$

Consequently, $A = 0$.
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Second Part
If $B$ is orthogonal, then $$\langle L_B(X),L_B(Y)\rangle = \langle BX, BY\rangle = \operatorname{trace}[(BX)(BY)^T] = \operatorname{trace}(BXY^TB^T) = \operatorname{trace}(B^TBXY^T) = \operatorname{trace}(IXY^T) = \operatorname{trace}(XY^T) = \langle X,Y\rangle$$

for all $X,Y \in M_n(\Bbb R)$. So $L_B$ is an isometry.

Conversely, suppose $L_B$ is an isometry. Then $\langle L_B(X), L_B(I) \rangle = \langle X,I\rangle$ for all $X \in M_n(\Bbb R)$. Thus $\operatorname{trace}(BXB^T) = \operatorname{trace}(X)$ $\implies$ $\operatorname{trace}(B^TBX) = \operatorname{trace}(X)$ $\implies$ $\operatorname{trace}(B^TBX) = \operatorname{trace}(X)$ $\implies$ $\operatorname{trace}((B^TB - I)X) = 0$ for all $X \in M_n(\Bbb R)$. By the first part, $B^TB - I = 0$, i.e., $B^TB = I$. Hence, $B$ is orthogonal.