Consider the following theorem:

Let $\mathrm{\mathbf{A}}\in\mathbb{C}^{n\times n}$ be an upper triangular block matrix with $2\times2$ blocks ($p=2$), i. e.,

$\mathrm{\mathbf{A}}=\left(\begin{array}{cc} \mathrm{\mathbf{A}_{11}} & \mathrm{\mathbf{A}_{12}}\\ 0 & \mathrm{\mathbf{A}_{22}} \end{array}\right)$.

Then $\mathrm{\mathbf{A}}$ is invertible if and only if $\mathbf{A}_{11}$ and $\mathbf{A}_{22}$ are invertible matrices.

(a) Prove the theorem.

(b) Generalize the theorem to $\mathrm{\mathbf{A}}\in\mathbb{C}^{n\times n}$ upper triangular block matrix with $p\times p$ blocks, i. e.,

$\mathrm{\mathbf{A}}=\left(\begin{array}{cccc} \mathrm{\mathbf{A}_{11}} & \mathrm{\mathbf{A}_{12}} & \cdots & \mathrm{\mathbf{A}}_{1n}\\ & \mathrm{\mathbf{A}_{22}} & \cdots & \mathrm{\mathbf{A}}_{2n}\\ & & \ddots & \vdots\\ & & & \mathrm{\mathbf{A}}_{nn} \end{array}\right)$,

and prove. (Note that it's not necessary to calculate the inverse matrix)


(a) In this case $\mathrm{\mathbf{A}^{-1}}\in\mathbb{C}^{n\times n}$ is

$\mathrm{\mathbf{A}}^{-1}=\left(\begin{array}{cc} \mathrm{\mathbf{A}_{11}^{-1}} & \mathrm{-\mathrm{\mathbf{A}_{11}^{-1}}\mathbf{A}_{12}}\mathrm{\mathbf{A}_{22}^{-1}}\\ 0 & \mathrm{\mathbf{A}_{22}^{-1}} \end{array}\right)$. Therefore, $\mathrm{\mathbf{A}}$ is invertible if and only if $\mathbf{A}_{11}$ is invertible and $\mathbf{A}_{22}$ is invertible. $\square$

(b) I think I have to use induction on the number of blocks ($p$), but I don't know how to do

Can you help me? Thanks.


1 Answer 1


a/ -- You have only shown "if $\mathbf{A}_{11}$ and $\mathbf{A}_{22}$ are invertible then $\mathbf{A}$ is invertible". It still remains to show the other implication.

b/ A matrix $\mathbf{A}$ is invertible if and only if the equation $$ \mathbf{A}x = 0$$ has a unique (zero) solution.

$\mathbf{A}_{kk}$ is invertible for all $k = 1 ... n$

We are solving the equation $$\mathbf{A}x = 0.$$

As $\mathbf{A}_{nn}$ is invertible then $x[(n-1)p + 1:np] = 0$ is the unique solution of the equation $$\mathbf{A}_{nn} x[(n-1)p +1: np] = 0$$ i.e. the last $p$ elements of $x$ are zero. Thus, the rightmost column $[\mathbf{A}_{1n};...;\mathbf{A}_{nn}]$ has no effect (it is multiplied by 0). It means that $x[1:(n-1)p]$ is determined by the equation $$ \left(\begin{array}{ccc} \mathbf{A}_{11} & \ldots & \mathbf{A}_{1,n-1} \\ & \ddots & \vdots \\ \mathbf{0} & & \mathbf{A}_{n-1,n-1} \end{array}\right) x[1:(n-1)p] = \mathbf{0}. $$

Now repeat the same argument inductively for matrices $\mathbf{A}_{n-1,n-1}, ..., \mathbf{A}_{11}$.

$\mathbf{A}$ is invertible

Suppose that $\mathbf{A}_{11}$ is not invertible and there exists $z$ such that $\mathbf{A}_{11} z = 0$. Then a vector $y = [z;0;0;...;0]$ is a solution to $\mathbf{A} y = 0$ while $y \neq 0$. This is a contradiction to assumption that $\mathbf{A}$ is invertible and thus $\mathbf{A}_{11}$ is invertible too. That means you can express $$x[1:p] = - \mathbf{A}_{11}^{-1}[\mathbf{A}_{12},...,\mathbf{A}_{1n}]x[p+1:np],$$ and the rest of $x$ is solution to $$\left(\begin{array}{ccc} \mathbf{A}_{22} & \ldots & \mathbf{A}_{2,n} \\ & \ddots & \vdots \\ \mathbf{0} & & \mathbf{A}_{n,n} \end{array}\right) x[p+1:np] = \mathbf{0}.$$ You can now proceed recursively.


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