# Nullspace of a block matrix

Suppose I have $$M \in \mathbb{R}^{2n\times 2n}$$ such that $$M = \begin{bmatrix} A &B\\0 & C\end{bmatrix}$$ for some $$A,B,C \in \mathbb{R}^{n \times n}$$.

I wish to:

(i) Find the nullspace of $$M$$

(ii) Prove that rank$$(M) \geq \text{rank}(A) + \text{rank}(C)$$

For (i) it is clear that for some vector $$[x, y]^T$$ will be in the null space if and only if $$y \in \text{null}(C)$$ and that $$Ax + By = 0$$, but I am unsure how to proceed from here. For (ii) I can see why this is true, but am not sure how to approach proving it.

Thanks.

• Your answer for part (i) seems like a complete description to me. You should probably mention that $x,y \in \mathbb R^n$. Mar 26, 2021 at 3:32
• not exactly the same problem, but similar: math.stackexchange.com/questions/1567957/… Mar 26, 2021 at 3:41
• The matrix $$\pmatrix{1 & 1\cr 0 &0}$$ is a counter example to (ii).
– Ruy
Mar 26, 2021 at 4:48
• OTOH, $\mathrm{rank}(M) \ge \mathrm{rank}(A) + \mathrm{rank}(C)$. Mar 26, 2021 at 4:51
• @AdamZalcman Yes it should be rank$(C)$, my apologies. Mar 26, 2021 at 6:05

Let's set $$\operatorname{rank}(A) = r$$ and $$\operatorname{rank}(C) = s$$. Choose $$i_1, \dots, i_r$$ such that $$Ae_{i_1},\dots,Ae_{i_r}$$ (which are just the $$i_1,\dots,i_r$$-th columns of $$A$$) are linearly independent and similarly choose $$j_1, \dots, j_s$$ such that $$Ce_{j_1},\dots,Ce_{j_s}$$ are linearly independent. It is enough to show that the $$r + s$$ columns $$M \begin{bmatrix} e_{i_1} \\ 0 \end{bmatrix}, \dots, M \begin{bmatrix} e_{i_r} \\ 0 \end{bmatrix}, M \begin{bmatrix} 0 \\ e_{j_1} \end{bmatrix}, \dots, M \begin{bmatrix} 0 \\ e_{j_s} \end{bmatrix}$$ of $$M$$ are linearly independent. And indeed, assume that $$\lambda_1 M \begin{bmatrix} e_{i_1} \\ 0 \end{bmatrix} + \dots + \lambda_r M \begin{bmatrix} e_{i_r} \\ 0 \end{bmatrix} + \mu_1 M \begin{bmatrix} 0 \\ e_{j_1} \end{bmatrix} + \dots + \mu_s M \begin{bmatrix} 0 \\ e_{j_s} \end{bmatrix} = \\ \lambda_1 \begin{bmatrix} Ae_{i_1} \\ 0 \end{bmatrix} + \dots + \lambda_r \begin{bmatrix} Ae_{i_r} \\ 0 \end{bmatrix} + \mu_1 \begin{bmatrix} Be_{k_1} \\ Ce_{k_1} \end{bmatrix} + \dots + \mu_s \begin{bmatrix} Be_{k_s} \\ Ce_{k_s} \end{bmatrix} = \\ \begin{bmatrix} \lambda_1 Ae_{i_1} + \dots + \lambda_r Ae_{i_r} + \mu_1 Be_{j_1} + \dots + \mu_s Be_{j_s} \\ \mu_1 Ce_{j_1} + \dots \mu_s Ce_{j_s} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$
Since $$Ce_{j_1},\dots,Ce_{j_s}$$ are linearly independent this implies that $$\mu_1 = \dots = \mu_s = 0$$ but then we get the equation $$\lambda_1 Ae_{i_1} + \dots + \lambda_r Ae_{i_r} = 0$$ and since $$Ae_{i_1},\dots,Ae_{i_r}$$ are also linearly independent we get that $$\lambda_1 = \dots = \lambda_r = 0$$.