This is a qualifying exam question which I have little experience approaching. This is certainly an exercise in linear algebra so I would like to obtain a method in solving similar questions.
Let $M_n(\mathbb{R})$ be the ring of $n\times n$ matrices over $\mathbb{R}$. If $A\in M_n(\mathbb{R})$, then there exists a $B\in M_n(\mathbb{R})$ s.t. $\operatorname{rank}(A)+\operatorname{rank}(B)=n$ and $AB=0$. Also does there exist a $B$ with the same properties such that $BA=0$ as well?
As far as I can tell, there is a similar question, but the question is over $\mathbb{C}$ coefficients so this is not a repeat question.
My approach is as follows. Since we work over $\mathbb{R}$, we should focus on the Rational Canonical Form. So write $A$ as a direct sum of companion matrices. The rank is preserved under similarity classes. The rank of $A$ is the sum of the ranks of the companion matrices. So the problem reduces to proving the result for a companion matrix.
Let $A=\begin{pmatrix}
0 & \dots & \dots & -a_1\\
1 & 0 & \dots &-a_2\\
\ddots &\ddots & \ddots & \ddots \\
0 & \dots & 1 &-a_{n-1}
\end{pmatrix}$ be a companion matrix. This has either rank $n-1$ or $n-2$. Then there should be choices of $B$ which work. Of course, this would be a tedious computation.
My question. Is there a better approach to this question? For example, some way to relate $M_n(\mathbb{R})$ to $M_n(\mathbb{C})$ so that I can use Jordan canonical forms instead?
Best Answer
The kernel, or nullspace, of $A$ is a subspace of $\mathbb{R}^n$. Let $\{b_1, \ldots, b_k\}$ be a basis for the kernel. Take $B$ to be the matrix representing the linear transformation mapping the standard basis of $\mathbb{R}^k$ to the basis of the kernel. Then $AB = 0$, and, due to the Rank-Nullity Theorem, you have the result.