Suppose A, B are invertible matrices of the same size. Show that $$M = \begin{bmatrix} 0& A\\ B& 0\end{bmatrix}$$ is invertible.

I don't understand how I could show this. I have learned about linear combinations and spanning in my college class, but I don't know how that would help in this case.


You can check directly that $$ M^{-1}=\begin{bmatrix} 0&B^{-1}\\ A^{-1}&0\end{bmatrix}. $$


Note that $M \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} w \\ z \end{bmatrix}$ iff $A y = w$ and $Bx = z$.

From this you can compute an explicit inverse.


Since $A$ and $B$ are invertible, we have

$M \begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix} \iff Ay=0$ and $Bx=0 \iff x=y=0.$


Since you mentioned linear combinations, you can also follow that path and prove that all the matrix rows are linearly independent, which is equivalent to invertibility.

For that, let $a_i$ and $b_i$ be the $i$-th rows of $A$ and $B$, respectively. Now assume that we have a linear combination of the rows of $M$. These rows have the form $(0|a_i)$ (some zeros followed by a row of $A$) or $(b_i|0)$ (a row of $B$ followed by some zeros). A linear combination of the rows of $M$ is therefore written as

$$ \sum_i \alpha_i(0|a_i) + \sum_i \beta_i(b_i|0) $$

where $\alpha_i$ and $\beta_i$ are some scalar coefficients. Assume the sum above is the null vector, and let's prove that the coefficients must be zero.

Now since the number of added zeros in the vectors $(0|a_i)$ and $(b_i|0)$ is exactly the size of $A$ and $B$, the above sum can be written as

$$ (\sum_i \alpha_i a_i | \sum_i \beta_i b_i) $$

This is the null vector only when we have both

$$ \sum_i \alpha_i a_i = 0 \qquad \sum_i \beta_i b_i = 0 $$

Since $A$ and $B$ are invertible, their rows are linearly independent, so in the sums above we must have $\alpha_i=0$ and $\beta_i=0$, for all $i$. QED.


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