In the space of all real matrices with dimension $n$, find the maximal possible dimension of a subspace $V$ such that $\forall X,Y\in V,\, \operatorname{tr}(XY)=0$.


If $V$ is a subspace of real matrices of size $n$ with the property that $\operatorname{tr}(XY)=0$ for all $X,Y\in V$, then $\operatorname{tr}(X^2)=0$ for all $X\in V$.

Let $W$ be a subspace of real matrices of size $n$ with the property that $\operatorname{tr}(X^2)=0$ for all $X\in W$. Then $\operatorname{tr}((X+Y)^2)=0$ for all $X,Y\in W$. But $\operatorname{tr}((X+Y)^2)=\operatorname{tr}(X^2)+\operatorname{tr}(Y^2)+2\operatorname{tr}(XY)$. It follows that $\operatorname{tr}(XY)=0$ for all $X,Y\in W$.

This shows that the problem reduces to the following: Investigations about the trace form which is already solved.

  • $\begingroup$ its all very nice! $\endgroup$ – dineshdileep Jan 8 '13 at 8:10

(This is not a difficult problem, but it is also not an easy homework problem. I wonder if the original problem is much simpler and the OP has copied the problem statement wrongly.)

Note that $\operatorname{tr}(XY)=\operatorname{vec}(Y^T)^T\operatorname{vec}(X)= \operatorname{vec}(Y)^TP\operatorname{vec}(X)$, where $P$ is the symmetric permutation matrix defined by $P_{(j-1)n+i,\ (i-1)n+j}=P_{(i-1)n+j,\ (j-1)n+i}=1$ for $i,j\in\{1,\ldots,n\}$ and other entries zero. Since the matrix $P$ is orthogonally similar to $I_p\oplus (-I_q)$, where $q=n(n-1)/2$ and $p=n^2-q$, a maximal matrix subspace $V$ such that $\operatorname{tr}(XY)=0$ whenever $X,Y\in V$ can be identified with a maximal subspace $W\subset\mathbb{R}^{n^2}$ such that $u^T(I_p\oplus(-I_q))v=0$ whenever $u,v\in W$.

We now claim that $\dim W\le q$. Suppose the contrary. Then there exists a set of $q+1$ linearly independent vectors $u_1,\ldots,u_{q+1}\in\mathbb{R}^{n^2}$ such that $u_i^T(I_p\oplus(-I_q))u_j=0$ for all $i,j=1,\ldots,q+1$. Hence there exists a nonzero linear combination $x=\sum_i c_iu_i\not=0$ such that the last $q$ entries of $x$ are zero. Write $x^T=(\tilde{x}^T,0_{1\times q})$. Then $x^T(I_p\oplus(-I_q))x=\|\tilde{x}\|^2\not=0$ because $\tilde{x}\not=0$, but we also have $x^T(I_p\oplus(-I_q))x=0$ because $x$ is a linear combination of $u_1,\ldots,u_{q+1}$. So we arrive at a contradiction. Therefore $\dim W\le q$.

Finally, the dimension $q$ is attainable. For example, consider the set of all strictly upper triangular matrices. Hence the required maximal dimension is $q=n(n-1)/2$.

  • $\begingroup$ @YACP The OP has asked several problems in last few hours, and all but this one are typical homework problems. $\endgroup$ – user1551 Jan 7 '13 at 16:19
  • $\begingroup$ seems very elegant $\endgroup$ – dineshdileep Jan 8 '13 at 8:52

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.