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 $\operatorname{dim}(V) \leq$ $\operatorname{dim}\left(S_{n}\right)$. Let $\pi: V \rightarrow S_{n}$ denote the linear map $\pi(A)=A+A^{t}$. We have $$\operatorname{dim}(V)=\operatorname{dim}(\operatorname{Ker}(\pi))+\operatorname{dim}(\operatorname{Im}(\pi))$$ so we have to prove that $\operatorname{dim}(\operatorname{Ker}(\pi))+\operatorname{dim}(\operatorname{Im}(\pi)) \leq \operatorname{dim}\left(S_{n}\right)$. 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 $$(A+B)(A+B)^{t}=(A+B)^{t}(A+B) \Leftrightarrow$$ $$\Leftrightarrow A A^{t}+A B^{t}+B A^{t}+B B^{t}=A^{t} A+A^{t} B+B^{t} A+B^{t} B \Leftrightarrow$$ $$\Leftrightarrow A B^{t}-B A=-A B+B^{t} A \Leftrightarrow A\left(B+B^{t}\right)=\left(B+B^{t}\right) A \Leftrightarrow$$ $$\Leftrightarrow A \pi(B)=\pi(B) A$$ 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 $$\operatorname{dim}(X)+\operatorname{dim}(Y) \leq \operatorname{dim}\left(S_{n}\right)$$ Proof. Without loss of generality we may assume $X=Z_{S_{n}}(Y):=\left\{x \in S_{n}: x y=y x \quad \forall y \in Y\right\}$. Define the bilinear map $B: S_{n} \times A_{n} \rightarrow \mathbb{C}$ by $B(x, y)=\operatorname{tr}(\mathrm{d}[\mathrm{x}, \mathrm{y}])$ where $[x, y]=x y-y x$ and $d=\operatorname{diag}(1, \ldots, n)$ is the matrix with diagonal elements $1, \ldots, n$ 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 $\left.\operatorname{tr}(\mathrm{d}[\mathrm{x}, \mathrm{y}])=-\operatorname{tr}([\mathrm{d}, \mathrm{x}], \mathrm{y}]\right)=0$ for every $x \in S_{n}$. We claim that $\left\{[d, x]: x \in S_{n}\right\}=A_{n}$. 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 $\left[d, E_{i}^{j}\right]=(j-i) E_{i}^{j}$ and therefore $\left[d, E_{i}^{j}+E_{j}^{i}\right]=$ $(j-i)\left(E_{i}^{j}-E_{j}^{i}\right)$ and the collection $\left\{(j-i)\left(E_{i}^{j}-E_{j}^{i}\right)\right\}_{1 \leq i<j \leq n} \operatorname{span} A_{n}$ for $i \neq j$. It follows that if $B(x, y)=0$ for all $x \in S_{n}$ then $\operatorname{tr}(\mathrm{yz})=0$ 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 $0=\operatorname{tr}(\mathrm{y} \overline{\mathrm{y}})=-\operatorname{tr}\left(\mathrm{y} \overline{\mathrm{y}}^{\mathrm{t}}\right)$ which is the sum of squares of all the entries of $y$. This means that $y=0$. It follows that if $y_{1}, \ldots, y_{k} \in Y$ are linearly independent then the equations $$B\left(x, y_{1}\right)=0, \quad \ldots, \quad B\left(x, y_{k}\right)=0$$ are linearly independent as linear equations in $x$, otherwise there are $a_{1}, \ldots, a_{k}$ such that $B\left(x, a_{1} y_{1}+\ldots+\right.$ $\left.a_{k} y_{k}\right)=0$ for every $x \in S_{n}$, a contradiction to the observation above. Since the solution of $k$ linearly independent linear equations is of codimension $k$, $$\begin{gathered}\operatorname{dim}\left(\left\{x \in S_{n}:\left[x, y_{i}\right]=0, \text { for } i=1, \ldots, k\right\}\right) \leq \\ \leq \operatorname{dim}\left(x \in S_{n}: B\left(x, y_{i}\right)=0 \text { for } i=1, \ldots, k\right)=\operatorname{dim}\left(S_{n}\right)-k\end{gathered}$$ The lemma follows by taking $y_{1}, \ldots, y_{k}$ to be a basis of $Y$. Since $\operatorname{Ker}(\pi)$ and $\operatorname{Im}(\pi)$ commute, by the lemma we deduce that $$\operatorname{dim}(V)=\operatorname{dim}(\operatorname{Ker}(\pi))+\operatorname{dim}(\operatorname{Im}(\pi)) \leq \operatorname{dim}\left(S_{n}\right)=\frac{n(n+1)}{2}$$