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Denote by $M_n(\mathbb{R})$ the set of all $n$ by $n$ real matrices. It's a linear space over $\mathbb{R}$.

  1. Can we find a nontrivial vector subspace in which all the matrices are degenerate?

  2. If possible, what is the maximum possible dimension of this subspace?

I thought about the set of upper triangular matrices, where all the the numbers at a certain position on the diagonal line of all matrices are always zero. This meets the requirements, but I don't how to deal with the 'maximum' problem.

Any hints? Thank you in advance.

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  • $\begingroup$ By "degenerate", do you mean nilpotent? That for each such $X$, there is some $k$ with $X^k=0$? Or is it singular - that the determinants are all zero? $\endgroup$ – jmerry Mar 5 at 11:40
  • $\begingroup$ @jmerry it means the determinant of it is $0$ $\endgroup$ – Zero Mar 5 at 11:42
  • $\begingroup$ A "subspace" is a vectorial subspace? And we do not need the group / ring structure of the matrices? And a matrix $A$ is degenerate iff $\det A=0$? $\endgroup$ – dan_fulea Mar 5 at 11:43
  • $\begingroup$ @dan_fulea Yes, it's a vectorial subspace. I think that a square matrix is degenerate if and only if its determinant is $0$ is a common conclusion, isn't it? $\endgroup$ – Zero Mar 5 at 11:56
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I claim that the maximum dimension of a subspace of singular matrices of $M_n(\mathbb{R})$ is $n^2 - n$.

This number is achieved, for example, for the subspace of matrices having the first row equal to the zero vector.

Now assume by contradiction that there were a subspace $S$ of singular matrices of dimension $n^2 - n + 1$. Firstly, there must be an element of $S$ having a nonzero entry in the first row and having all rows from $2$ to $n$ equal to $0$, by dimension considerations. To see why this is true, let A be the subspace of matrices of the form: \begin{bmatrix} * & * & * & \dots & * \\ 0 & 0 & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} Then $$\dim(A \cap S) = \dim(A) + \dim(S) - \dim(A + S) \geq n + n^2 - n + 1 - n^2 \geq 1$$ so there must be a nonzero matrix $X$ in $A \cap S$. Thus $X$ looks like: \begin{bmatrix} a_{11} & * & * & \dots & * \\ 0 & 0 & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} WLOG assume $a_{11} \neq 0$ (otherwise permute the elements of the first row and apply below verbatim). Consider next the subspace $M$ of $M_n(\mathbb{R})$ of matrices of the form: \begin{bmatrix} 0 & 0 & 0 & \dots & 0 \\ 0 & * & * & * & * \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & * & * & * & * \end{bmatrix} (it's like an embedding of $M_{n-1}(\mathbb{R})$ into $M_n(\mathbb{R})$). Now $$\dim(S \cap M) = \dim(S) + \dim(M) - \dim(S + M) \geq n^2-n+1 + (n-1)^2 - n^2 = n^2 - 3n + 2 > (n-1)^2 -(n-1) + 1$$ so by our induction hypothesis it follows that $S \cap M$ contains a matrix $Y$ which when viewed as an $(n-1)$ by $(n-1)$ matrix is invertible (and also Y has the first row equal to $0$ in $M_n(\mathbb{R})$. But then $X+Y$ will have a nonzero determinant, contradiction.

Also, after solving this problem I found a more general problem (that implies it) here.

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  • $\begingroup$ Let us consider the space $$\begin{bmatrix}2 & 1\\1&2\end{bmatrix}\begin{bmatrix} 0 & 0 \\*&*\end{bmatrix}$$ of $M_{2\times 2}(\Bbb R)$. It has the dimension $2^2-2$, well not $\ge 2^2-2+1$ since we cannot produce such a case... but it is not having a non-empty intersection with $$\begin{bmatrix} *&*\\0 & 0 \end{bmatrix}\ .$$ In view of this, why "there must be an element of $S$ having a nonzero entry in the first row and having all rows from $2$ to $n$ equal to $0$"? (Well, we may arrange...) $\endgroup$ – dan_fulea Mar 5 at 18:33
  • $\begingroup$ @dan_fulea : $2^2-2+1$ is essential here and you can produce such a space because the total dimension of your space of matrices is $2^2$ $\endgroup$ – Sorin Tirc Mar 5 at 19:06
  • $\begingroup$ Then please arrange for details. "One can produce" and "there must be" is not enough for beginning an argumentation. If there is a simple argument, just insert it. It is easier then for me to upvote the answer.... $\endgroup$ – dan_fulea Mar 5 at 23:14
  • $\begingroup$ @dan_fulea : I have edited the answer, I hope it's clear now. I was being a bit terse, you are right. $\endgroup$ – Sorin Tirc Mar 6 at 7:51

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