# A linear space of degenerate matrices.

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.

• 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? – jmerry Mar 5 at 11:40
• @jmerry it means the determinant of it is $0$ – Zero Mar 5 at 11:42
• 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 $\det A=0$? – dan_fulea Mar 5 at 11:43
• @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? – Zero Mar 5 at 11:56

I claim that the maximum dimension of a subspace of singular matrices of $$M_n(\mathbb{R})$$ is $$n^2 - n$$.
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.
• 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...) – dan_fulea Mar 5 at 18:33
• @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 $2^2$ – Sorin Tirc Mar 5 at 19:06