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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.

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  • $\begingroup$ Your answer for part (i) seems like a complete description to me. You should probably mention that $x,y \in \mathbb R^n$. $\endgroup$ Mar 26, 2021 at 3:32
  • $\begingroup$ not exactly the same problem, but similar: math.stackexchange.com/questions/1567957/… $\endgroup$
    – 311411
    Mar 26, 2021 at 3:41
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    $\begingroup$ The matrix $$\pmatrix{1 & 1\cr 0 &0}$$ is a counter example to (ii). $\endgroup$
    – Ruy
    Mar 26, 2021 at 4:48
  • $\begingroup$ OTOH, $\mathrm{rank}(M) \ge \mathrm{rank}(A) + \mathrm{rank}(C)$. $\endgroup$ Mar 26, 2021 at 4:51
  • $\begingroup$ @AdamZalcman Yes it should be rank$(C)$, my apologies. $\endgroup$
    – CBBAM
    Mar 26, 2021 at 6:05

1 Answer 1

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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 $\lambda_1 = \dots = \lambda_r = 0$.

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