Proof for the dimension of a subspace of M(n,R)

Question

First I will state the exercise and then show what I could get.

Let $A_{1},...,A_{k} \in M(n,R)$ such that for all $ 1 \le i \neq j \le k $: $ A_{i}^{2}=I$ and $A_{i}A_{j}+A_{j}A_{i}=0$ show that $k \le \frac{n(n+1)}{2}$

Ok, first of all we can see that the condition $ A_{i}^{2}=I$ means that all the matrices in this form are diagonalizable with eigenvalues $\in \big\{1,-1\big\} $, so we can state that $A_{i}$ are all invertible matrix and $A_{i}^{-1}=A_{i}$.

Now if I multiply the relation between the matrices for $A_{i}$ i get that $A_{i}A_{j}A_{i}=-A_{j}$ wich means that $A_{j}$ and $-A_{j}$ are similar and because they need to share determinant it has to be 0 if the dimension of the space is odd. I also know that they commute. I feel like I should be able to find some other relation (maybe about the Ker of the linear application) from $A_{i}A_{j}+A_{j}A_{i}=0$ but I don't know how.

Thanks in advance

Answer 1

I shall work with the assumption that the base field is any field $K$. Let $V:=\text{Mat}_{n\times n}(K)$. Suppose that $A_1,A_2,\ldots,A_k\in V$ are such that $A_i^2=1_n$ for every $i\in[k]$, and $A_iA_j+A_jA_i=0_n$ for all $i,j\in[k]$ with $i\neq j$. Here, $1_r$ and $0_r$ denote, respectively, the $r$-by-$r$ identity matrix and the $r$-by-$r$ zero matrix, and $[m]:=\{1,2,\ldots,m\}$ for all $m\in\mathbb{Z}_{>0}$.

First, we handle the case where $K$ is of characteristic not equal to $2$. It has been shown that, if $k>1$, then $A_i$ and $-A_i$ are similar matrices for all $i\in[k]$. That is, if $k>1$, then $$\det(A_i)=\det(-A_i)=(-1)^n\,\det(A_i)\,,$$ making $(-1)^n=1$, whence $n$ is even. Thus, if $n$ is odd, then $k\leq1$. From now on, we assume that $k>1$ and $n$ is even.

Consider the linear operator $T_X:V\to V$ defined for each $X\in V$ by $$T_X(Y):=XY+YX\text{ for any }Y\in V\,.$$ If $X$ is diagonalizable over $K$, then let $x_1,x_2,\ldots,x_n\in K^n$ be eigenvectors of $X$, with respect to eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_n\in K$, respectively, and let $\chi_1,\chi_2,\ldots,\chi_n\in K^n$ be eigenvectors of $X^\top$, with respect to eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_n\in K$, respectively. Then, the $n^2$ linearly independent matrices $x_\mu\,\chi_\nu^\top$ are eigenvectors of $T_X$, with eigenvalues $\lambda_\mu+\lambda_\nu$, for $\mu,\nu\in[n]$.

Without loss of generality, we suppose that $$A_1=\begin{bmatrix}+1_{\frac{n}{2}}&0_{\frac{n}{2}} \\0_{\frac{n}{2}}&-1_{\frac{n}{2}} \end{bmatrix}\,.$$ Then, $A_1$ is diagonalizable with eigenvectors $e_1,e_2,\ldots,e_{\frac{n}{2}}$ corresponding to the eigenvalue $+1$, and with eigenvectors $e_{\frac{n}{2}+1},e_{\frac{n}{2}+2},\ldots,e_n$ corresponding to the eigenvalue $-1$. Here, $e_1,e_2,\ldots,e_n\in K^n$ are the standard basis vectors of $K^n$. This shows that the nullspace of $T_{A_1}$ is of dimension $2\left(\dfrac{n}{2}\right)^2=\dfrac{n^2}{2}$, since it is spanned by $e_i\,e_j^\top$ and $e_j\,e_i^\top$ with $i\in\left\{1,2,\ldots,\dfrac{n}{2}\right\}$ and $j\in\left\{\dfrac{n}{2}+1,\dfrac{n}{2}+2,\ldots,n\right\}$. Thus, from David C. Ullrich's answer, $A_2,A_3,\ldots,A_k$ are linearly independent element of $\ker\big(T_{A_1}\big)$, so we get $$k-1\leq \frac{n^2}{2}\text{ or }k\leq \frac{n^2}{2}+1\leq \frac{n(n+1)}{2}\,.$$ I however believe that the maximum possible value of $k$ is at most $\dfrac{n^2}{4}+1$.

If $K$ is of characteristic $2$, then clearly the matrices $A_1,A_2,\ldots,A_k$ need to only commute. There is no bound on $k$ (for example, $A_1=A_2=\ldots=A_k=1_n$ works). However, if we demand that $A_1,A_2,\ldots,A_k$ be linearly independent, then a bound for $k$ can be determined. First, we may assume that $A_1,A_2,\ldots,A_k$ are upper triangular matrices each of whose diagonal entries is $1$. Then, $A_2-A_1,A_3-A_1,\ldots,A_k-A_1$ are strictly upper triangular matrices. The subspace of $V$ composed by strictly upper triangular matrices is of dimension $\dfrac{n(n-1)}{2}$. That is, $$k-1\leq \frac{n(n-1)}{2}\text{ or }k\leq \frac{n(n-1)}{2}+1\leq \frac{n(n+1)}{2}\,.$$ I conjecture that, with the extra assumption that the matrices $A_1,A_2,\ldots,A_k$ be linearly independent, $k\leq\left\lfloor\dfrac{n^2}{4}\right\rfloor+1$.

Answer 2

You say in a comment that $k$ is "supposed to be" the dimension of the span of the $A_j$, but you give no indication why this is actually so.

In fact it's easy to see that the $A_i$ are independent. Suppose $$\sum_ic_iA_i=0.$$Then for every $j$ we have $$\sum_ic_iA_iA_j=0$$and also $$\sum_i c_iA_jA_i=0.$$Now add those two equations; since $A_iA_j+A_jA_i=0$ for $i\ne j$ and $A_j^2=I$ you get $$2c_jI=0,$$hence $c_j=0$.

Answer 3

I see the bound is already proved. So i will give a different perspective. If you add the condition that $A_i^tA_j+A_j A_i^t = 0$,$\forall i,j$ (when $i=j$, this condition applies only when $A_i \neq A_i^t$), then the following is true. Now consider $A_1,...,A_k$ and now replace $A_i$ by $\frac{1}{\sqrt{2}} (A_i+A^t_i)$,$ \frac{1}{\sqrt{2}} (A_i-A^t_i)$ if $A_i \neq A^t_i$ and remove $A_i^t$ from the set if present or else if $A_i = A^t_i$ keep $A_i$ as it is. Let the replaced matrices be $B_1,...B_k$.

Now $B_1,...,B_{k}$ also satisfies the equations $B^2_i = I$ and $B_i B_j + B_j B_i = 0$.