Maximal dimension of subspace of matrices whose products have zero trace

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

-

2 Answers

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.

-

its all very nice! – dineshdileep Jan 8 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 $q=n(n-1)/2$.

-

@YACP The OP has asked several problems in last few hours, and all but this one are typical homework problems. – user1551 Jan 7 at 16:19
seems very elegant – dineshdileep Jan 8 at 8:52
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 49, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9235427975654602, "perplexity_flag": "head"}