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Let $M_1$ and $M_2$ be two matrices of order $3\times 3$ with real entries and $M_1^2+M_2^2=O$, where $O$ is the null matrix.

Show that there exist infinitely many matrix $M$ of order $3\times 3$ such that $M(M_1+M_2)$ is non-invertible.

My Attempt

$M_1^2=-M_2^2\Rightarrow \operatorname{det}(M_1^2)=(-1)^3\operatorname{det}(M_2^2)\Rightarrow (\operatorname{det}(M_1))^2=-(\operatorname{det}(M_2))^2\Rightarrow (\operatorname{det}(M_1))^2+(\operatorname{det}(M_2))^2=0$

$\Rightarrow \operatorname{det}(M_1)=\operatorname{det}(M_2)=0$

So, $M_1$ and $M_2$ are not invertible.

But what can we say about invertibility of $M_1+M_2$

Sigur
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Maverick
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2 Answers2

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Any non-invertible $M$ will cause the product to be non-invertible, since $\det(MN) = \det (M) \det (N)$. Is the question missing anything?

Javier Lim
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  • Can we say anything about invertibility of $M_1+M_2$ – Maverick Sep 27 '24 at 01:50
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    @Maverick No. If we let $M_1 = \pmatrix{1 & 0 \ 0 & 1}$ and $M_2 = \pmatrix{0&-1\1&0}$, then $M_1^2 + M_2^2 = O$, and their sum is invertible. But, if we embed $M_1$ and $M_2$ into $3 \times 3$ matrices, adding $0$s in the third row/column, they won't be invertible. – Theo Bendit Sep 27 '24 at 05:56
  • @TheoBendit that isn't right. OP has stated that we are in an odd dimensional (real) vector space and that forces $\big(M_1 +M_2\big)$ to be singular. – user8675309 Oct 05 '24 at 19:30
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But what can we say about invertibility of $M_1+M_2$

OP has already shown $\text{rank}\big(M_1\big)\in \big\{0,1,2\big\}$. If $\text{rank}\big(M_1\big)\leq 1$ then
$\text{rank}\Big(\big(M_1+M_2\big)^2\Big)=\text{rank}\Big(M_1M_2+M_2M_1\big)\leq 1+1\lt 3$
$\implies \det\big(M_1+M_2\big)^2=0=\det\big(M_1+M_2\big)$. So it remains to consider the case where $\text{rank}\big(M_1\big)=2\implies \text{rank}\big(M_1^2\big)\in \big\{1,2\big\}$ (e.g. by Sylvester's Rank Inequality).

By rank-nullity, one of $\ker M_1^2 \big(=\ker M_2^2\big)$ or $\text{image } M_1^2 \big(= \text{image } M_2^2\big)$ is 1 dimensional and label that $W$. Then $M_1 W\subseteq W\text{ and }M_2 W\subseteq W\implies$ for unit length $\mathbf w\in W$ that $M_1 \mathbf w = \lambda_1 \mathbf w$ and $M_2 \mathbf w = \lambda_2 \mathbf w$
$\implies 0\leq \lambda_1^2 = \mathbf w^T (M_1^2 \mathbf w) = -\mathbf w^T (M_2^2 \mathbf w) = -\lambda_2^2 \leq 0$, i.e. the eigenvalues are zero.

Thus $\big(M_1+M_2\big)\mathbf w =M_1\mathbf w + M_2\mathbf w = \mathbf 0+ \mathbf 0$; conclude $\big(M_1+M_2\big)$ is not invertible.

remark:
The above argument generalizes to $M_1, M_2 \in \mathcal M_{2k+1}\big(\mathbb R\big)$. With the trivial Base Case at $k=0$, proceed by strong induction. By OP's argument $\det\big(M_1\big)=0=\det\big(M_2\big)$. The case of $\text{rank}\big(M_1\big) \leq k \implies \det\big(M_1 +M_2\big) =0$ by the first paragraph's argument (i.e. sub-multiplicativity and sub-additivity of rank). It remains to consider $\text{rank}\big(M_1\big) \geq k+1$; rank-nullity tells us one of $\ker M_1^2$ and $\text{image } M_1^2$ is even dimensional and the other odd dimensional. The even dimensional one cannot be zero dimensional (per Sylvester's Rank Inequality and the fact that $\dim \ker M_1 \gt 0$), so label the odd dimensional subspace $W$, noticing $\dim W\lt 2k+1$, and look at the restrictions $M^{(1)}_{\vert W}$ and $M^{(2)}_{\vert W}$ where $\det\Big(M^{(1)}_{\vert W}+M^{(2)}_{\vert W}\Big)=0$ per strong induction hypothesis, which gives the result.

user8675309
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  • The above argument also gives the result for https://math.stackexchange.com/questions/4833103/if-a2b2-0-n-and-ab-ba-is-invertible-then-n-is-even/ (over $\mathbb R$). I.e. Suppose $n=2k +1$; if $\text{rank}(A) \leq k$ then $\text{rank}(AB-BA)\lt 2k+1$. Otherwise $\text{rank}(A) \geq k+1$ and one of $\ker A^2$ or $\text{image } A^2$ has odd dimension $\lt 2k +1$, call that $W$ which is $C$-invariant, for the commutator $C=AB-BA$. Then consider the restriction $C_{\vert W}$ and by induction hypothesis $C_{\vert W}$ is singular, forcing the conclusion that $n$ is even. – user8675309 Oct 05 '24 at 20:49