IMC 2015 · Day 2 · P9

Answer: The maximum dimension of such a space is $\frac{n(n+1)}{2}$.

The number $\frac{n(n+1)}{2}$ can be achieved, for example the symmetric matrices are obviously t-normal and they form a linear space with dimension $\frac{n(n+1)}{2}$. We shall show that this is the maximal possible dimension.

Let $M_n$ denote the space of $n \times n$ complex matrices, let $S_n \subset M_n$ be the subspace of all symmetric matrices and let $A_n \subset M_n$ be the subspace of all anti-symmetric matrices, i.e. matrices $A$ for which $A^t = -A$.

Let $V \subset M_n$ be a linear subspace consisting of t-normal matrices. We have to show that $\dim(V) \le \dim(S_n)$. Let $\pi : V \to S_n$ denote the linear map $\pi(A) = A + A^t$. We have $$\dim(V) = \dim(\operatorname{Ker}(\pi)) + \dim(\operatorname{Im}(\pi))$$ so we have to prove that $\dim(\operatorname{Ker}(\pi)) + \dim(\operatorname{Im}(\pi)) \le \dim(S_n)$. Notice that $\operatorname{Ker}(\pi) \subseteq A_n$.

We claim that for every $A \in \operatorname{Ker}(\pi)$ and $B \in V$, $A \pi(B) = \pi(B) A$. In other words, $\operatorname{Ker}(\pi)$ and $\operatorname{Im}(\pi)$ commute. Indeed, if $A, B \in V$ and $A = -A^t$ then $$\begin{gather*} (A + B)(A + B)^t = (A + B)^t (A + B) \Leftrightarrow \\ \Leftrightarrow A A^t + A B^t + B A^t + B B^t = A^t A + A^t B + B^t A + B^t B \Leftrightarrow \\ \Leftrightarrow A B^t - B A = -A B + B^t A \Leftrightarrow A (B + B^t) = (B + B^t) A \Leftrightarrow \\ \Leftrightarrow A \pi(B) = \pi(B) A. \end{gather*}$$ Our bound on the dimension on $V$ follows from the following lemma:

Lemma. Let $X \subseteq S_n$ and $Y \subseteq A_n$ be linear subspaces such that every element of $X$ commutes with every element of $Y$. Then $$\dim(X) + \dim(Y) \le \dim(S_n)$$

Proof. Without loss of generality we may assume $X = Z_{S_n}(Y) := \{ x \in S_n : xy = yx \ \forall y \in Y \}$. Define the bilinear map $B : S_n \times A_n \to \mathbb{C}$ by $B(x, y) = \operatorname{tr}(d [x, y])$ where $[x, y] = xy - yx$ and $d = \operatorname{diag}(1, \dots, n)$ is the matrix with diagonal elements $1, \dots, n$ and zeros off the diagonal. Clearly $B(X, Y) = \{0\}$. Furthermore, if $y \in Y$ satisfies that $B(x, y) = 0$ for all $x \in S_n$ then $\operatorname{tr}(d [x, y]) = -\operatorname{tr}([d, x], y]) = 0$

for every $x \in S_n$.

We claim that $\{ [d, x] : x \in S_n \} = A_n$. Let $E_i^j$ denote the matrix with 1 in the entry $(i, j)$ and 0 in all other entries. Then a direct computation shows that $[d, E_i^j] = (j - i) E_i^j$ and therefore $[d, E_i^j + E_j^i] = (j - i)(E_i^j - E_j^i)$ and the collection $\{ (j - i)(E_i^j - E_j^i) \}_{1 \le i < j \le n}$ span $A_n$ for $i \ne j$.

It follows that if $B(x, y) = 0$ for all $x \in S_n$ then $\operatorname{tr}(y z) = 0$ for every $z \in A_n$. But then, taking $z = \bar{y}$, where $\bar{y}$ is the entry-wise complex conjugate of $y$, we get $0 = \operatorname{tr}(y \bar{y}) = -\operatorname{tr}(y \bar{y}^t)$ which is the sum of squares of all the entries of $y$. This means that $y = 0$.

It follows that if $y_1, \dots, y_k \in Y$ are linearly independent then the equations $$B(x, y_1) = 0, \ \dots, \ B(x, y_k) = 0$$ are linearly independent as linear equations in $x$, otherwise there are $a_1, \dots, a_k$ such that $B(x, a_1 y_1 + \dots + a_k y_k) = 0$ for every $x \in S_n$, a contradiction to the observation above. Since the solution of $k$ linearly independent linear equations is of codimension $k$, $$\begin{align*} \dim(\{ x \in S_n : [x, y_i] = 0, \text{ for } i = 1, .., k \}) &\le \\ \le \dim( x \in S_n : B(x, y_i) = 0 \text{ for } i = 1, \dots, k ) &= \dim(S_n) - k. \end{align*}$$ The lemma follows by taking $y_1, \dots, y_k$ to be a basis of $Y$.

Since $\operatorname{Ker}(\pi)$ and $\operatorname{Im}(\pi)$ commute, by the lemma we deduce that $$\dim(V) = \dim(\operatorname{Ker}(\pi)) + \dim(\operatorname{Im}(\pi)) \le \dim(S_n) = \frac{n(n+1)}{2}.$$