# Rank and null space of a particular block matrix.

Let $D_1, D_2 \in \mathbb{R}^{N \times N}$ be diagonal matrices with diagonals that are linearly independent vectors. Let $A, B \in \mathbb{R}^{N \times N}$ be rank-deficient matries. Define $S = \mathrm{null}(A) \cap \mathrm{null}(B)$ and suppose $\mathrm{dim}(S) = p$.

I'm interested in the rank of the block matrix $\Phi = \begin{pmatrix} D_1 A & D_2 A \\ D_1 B & D_2 B \end{pmatrix} \in \mathrm{R}^{2N \times 2N}$.

It's easy to show $\mathrm{rank}(\Phi) \leq 2N - 2P$, as any block vector with with blocks taken from $S$ is in $\mathrm{null}(\Phi)$.

Question: does this hold with equality? That is, is $\mathrm{null}(\Phi) = \left\{ [u, v]^T | \ u, v \in S \right\}$?

I have a rough argument that is based on partitioning the matrix $\begin{pmatrix} A \\ B \end{pmatrix}$ as $\begin{pmatrix} A^\prime \\ B^\prime \\ \tilde{A} \\ \tilde{B} \\ C \end{pmatrix}$

where the rows $\begin{pmatrix} A^\prime \\ B^\prime \end{pmatrix}$ are linearly independent, $\tilde{A}$ (resp. $\tilde{B})$ depends only on rows of $A^\prime$ (resp. $B^\prime$), and $C$ are rows that are a linear combination of rows of $A^\prime$ and $B^\prime$. I then use the diagonal weighting as in my problem statement, and show that $C = 0$.

This approach seems crude. Is there a better way?

## 1 Answer

The answer to your question an obvious "no". It should be easy to generate a random counterexample. E.g. suppose $$A=B=\pmatrix{1&-1\\ 1&-1\\ &&1},\ D_1=\pmatrix{1\\ &1\\ &&0},\ D_2=\pmatrix{0\\ &1\\ &&1}.$$ Then $S=\operatorname{null}(A)$ consists of scalar multiples of $(1,1,0)^T$ and $p=\dim S=1$. However, $$\Phi=\pmatrix{1&-1&0&0&0&0\\ 1&-1&0&1&-1&0\\ 0&0&0&0&0&1\\ 1&-1&0&0&0&0\\ 1&-1&0&1&-1&0\\ 0&0&0&0&0&1}$$ has rank $3<2N-2p=4$ and its null space consists of vectors of the form $(x,x,z,y,y,0)^T$, rather than only vectors of the form $(x,x,0,y,y,0)^T$.

Edit. The answer is still no if $D_1$ and $D_2$ are invertible. Consider the same $A$ and $B$ in the above, but replace $D_1$ by $I_3$ and $D_2$ by $\operatorname{diag}(1,1,2)$. Again, $S=\operatorname{null}(A)$ consists of scalar multiples of $(1,1,0)^T$ and $p=\dim S=1$. However, $$\Phi=\pmatrix{1&-1&0&1&-1&0\\ 1&-1&0&1&-1&0\\ 0&0&1&0&0&2\\ 1&-1&0&1&-1&0\\ 1&-1&0&1&-1&0\\ 0&0&1&0&0&2}$$ has only rank $2<2N-2p=4$ and its null space consists of vectors of the form $(x,y,-2w,z,x-y+z,w)^T$, rather than only vectors of the form $(x,x,0,y,y,0)^T$.

• Great, thanks! I should have have added an extra condition : that the diagonal matrices are full rank; so no zeros on the diagonal. – lp333 Jun 27 '16 at 23:07
• @LukeP The answer is still no. See my new edit. – user1551 Jun 28 '16 at 2:01