# For any $A\in M_n(\mathbb{R})$, there is a $B$ s.t. $\operatorname{rank}(A)+\operatorname{rank}(B)=n$ and $AB=0$.

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?

• take a matrix $B$ whose columns generate the kernel of $A$. In general, you can solve it by considering the real Jordan form Apr 3, 2022 at 20:25
• @Exodd So the rank of $B$ is the nullity of $A$. Hence the ranks add up to $n$ as they should. On the other hand, the column space generating the kernel of $A$ implies $AB=0$ since each column $v_i$ of $B$ sits inside the kernel and so $Av_i=0$. Is that the correct line? Apr 3, 2022 at 20:31
• @Exodd Okay I get the argument in my previous comment. Can you provide details about the real Jordan form argument? Apr 3, 2022 at 20:42
• Apr 4, 2022 at 5:54
• There are proofs that work over every field, so this is a duplicate. Apr 4, 2022 at 5:55

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.

If $$\operatorname{rank}(A)=0$$ or $$\operatorname{rank}(A)=n$$, the result is obvious.

If $$\operatorname{rank}(A)=r$$ and $$0, then, we can find $$P$$ and $$Q$$ in $$GL_n(\mathbb R)$$ such that $$A = P \begin{pmatrix} I_r & 0 \\ 0 & 0\end{pmatrix}Q$$.

And we can choose $$\ \ \ B = Q^{-1} \begin{pmatrix} 0 & 0 \\ 0 & I_{n-r}\end{pmatrix}P^{-1}$$.

• This is a nice proof -- do you have a reference for the existence of the $P$ and $Q$ in $GL_n(\mathbb{R})$ satisfying said property? It looks to be some sort of SVD decomposition or spectral theorem result which I can't recall. Apr 3, 2022 at 23:28
• @AHappyMathematician In general, if you have a linear transformation $T : V \to W$ of rank $r$, there are bases $B$ and $B'$ of $V$ and $W$ respectively, such that $[T]_B^{B'}$ is of the given form. (Proof outline: take a basis $T x_1, \ldots, T x_r$ of the image of $T$, extend it to $B'$, and extend $x_1, \ldots, x_r$ with a basis of the kernel of $T$ to form $B$.) And then, for the given statement, $P$ and $Q$ are transition matrices between $B, B'$ and the standard bases of $\mathbb{R}^n$ in appropriate directions for the linear transformation of multiplying by $A$. Apr 8, 2022 at 23:41

Here is a constructive approach. Fix such $$n\times n$$ real matrix $$A$$. Let $$W$$ be its row space (span of the rows of $$A$$), namely $$W = im(A^T)$$. Now $$W$$ is a subspace in $$\mathbb R^n$$, so you can compute its orthogonal complement $$W^\perp$$. Let $$b_1,\ldots,b_k$$ be a basis of $$W^\perp$$, and consider a matrix $$B$$ whose columns are $$b_1,\ldots,b_k$$, and fill the rest with zeros.

Now imagine how matrix product is computed, $$AB$$ is the zero matrix, as each row of $$A$$ is orthogonal to each column of $$B$$. Further, rank of $$A$$ is the dimension of $$W$$ as $$rk(A)=rk(A^T)$$. Lastly, we have $$rk(B) = dim(W^\perp) = n - dim(W) = n- rk(A)$$.

(Note, $$W^\perp = im(A^T)^\perp = ker(A)$$. So you just need to take a basis for $$ker(A)$$ for $$b_1,\ldots,b_k$$.)

Note, this isn't a "canonical form" approach, but I'm illustrating this can be done in a more elementary brute way.

In order to address also the last part of the question (getting $$BA=0$$ as well), let the image of $$A$$ be generated by the $$k$$ column vectors $$a_1,...,a_k$$ and complement this (in an arbitrary way) with vectors $$c_1,...,c_{n-k}$$ so that the matrix $$P=[a_1,...,a_k,c_1,...,c_{n-k}]$$ has rank $$n$$, whence is invertible. Let the kernel of $$A$$ be generated by $$n-k$$ vectors $$z_1,...,z_{n-k}$$.

We want $$B$$ to map $${\rm Im}\,A$$ to zero vectors and the $$c_j$$'s to $$\ker A$$. This is fulfilled for example when $$B$$ satisfies $$BP= B ( a_1,...,a_k,c_1 ..., c_{n-k}) = {(0,...,0, z_1,...,z_{n-k})} =: Q$$ And the solution is simply to take $$B=QP^{-1}$$ which verifies $${\rm rank} A+{\rm rank} B=n$$ and both of $$AB=0$$, $$BA=0$$ at the same time.