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I got the following question to solve:

Given the lower triangular matrix

\begin{bmatrix} A_{11} & 0 \\ A_{21} & A_{22} \end{bmatrix}

of size $n \times n$ (n is a power of 2) where $A_{11}$, $A_{21}$ and $A_{22}$ are matrices of size $(n/2) \times (n/2)$, show that the inverse is,

\begin{bmatrix} A_{11}^{-1} & 0 \\ -A_{22}^{-1}A_{21}A_{11} & A_{22}^{-1} \end{bmatrix}

how do I go about to solve this problem?

Edit: the matrix is invertible.

Edit: the second matrix should be changed to: \begin{bmatrix} A_{11}^{-1} & 0 \\ -A_{22}^{-1}A_{21}A_{11}^{\color{red}{-1}} & A_{22}^{-1} \end{bmatrix}
The inverse was missing.

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  • $\begingroup$ @Insertnamehere it feels like it's a problem for induction. n=2 I verified. But I have no idea how to go about the induction step. $\endgroup$ – PPGoodMan Nov 20 '17 at 4:53
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    $\begingroup$ Perhaps start at Wikipedia's page on Block Matrix. $\endgroup$ – StephenG Nov 20 '17 at 5:13
  • $\begingroup$ You can derive the invese from the fact that $\begin{bmatrix}I \\ X & I\end{bmatrix}^{-1} = \begin{bmatrix}I \\ -X & I\end{bmatrix}$, as explained in my answer to this question: math.stackexchange.com/questions/1377634/… $\endgroup$ – Nick Alger Nov 20 '17 at 22:39
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Clearly, you are having some trouble evaluating, and I think this is because of a typo! Evaluating with normal matrix multiplication I got $$ \begin{bmatrix} A_{11} & 0 \\ A_{21} & A_{22} \end{bmatrix}\begin{bmatrix} A_{11}^{-1} & 0 \\ -A_{22}^{-1}A_{21}A_{11} & A_{22}^{-1} \end{bmatrix} = \begin{bmatrix} A_{11}A_{11}^{-1} & 0 \\ A_{21} A_{11}^{-1} -A_{22}A_{22}^{-1}A_{21}A_{11} & A_{22}A_{22}^{-1} \end{bmatrix} $$ Everything evaluates trivially except for this term $$A_{21} A_{11}^{-1} - A_{21}A_{11}$$ Which clearly does not equal $0$ all the time. Thus I believe there was a typo made here and that the $A_{11}$ at the end should be an $A_{11}^{-1}$, as then the above expression reduces to the identity matrix, $$ \begin{bmatrix} I & 0 \\ 0 & I \end{bmatrix} $$

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  • $\begingroup$ Thank you very much! That was my typo and was not in the original. The problem I was having was that I did not know how to evaluate block matrices. Now I've learned it. $\endgroup$ – PPGoodMan Nov 21 '17 at 13:03
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Guide:

Evaluate the following quantity$$ \begin{bmatrix} A_{11} & 0 \\ A_{21} & A_{22} \end{bmatrix}\begin{bmatrix} A_{11}^{-1} & 0 \\ -A_{22}^{-1}A_{21}A_{11}^{\color{red}{-1}} & A_{22}^{-1} \end{bmatrix} .$$

Remark:

As pointed out by Isaac Browne, there is indeed a typo in the question.

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  • $\begingroup$ I'm sorry it's not clear to me. Since the elements are not single elements but rather matrices themselves, how do I evaluate it? $\endgroup$ – PPGoodMan Nov 20 '17 at 5:01
  • $\begingroup$ references for you: mathworld.wolfram.com/BlockMatrix.html $\endgroup$ – Siong Thye Goh Nov 20 '17 at 5:03
  • $\begingroup$ If you evaluate it, this does not evaluate to the identity matrix. $\endgroup$ – Isaac Browne Nov 20 '17 at 21:37
  • $\begingroup$ yes, you are right, there is indeed a typo. $\endgroup$ – Siong Thye Goh Nov 20 '17 at 22:30
  • $\begingroup$ Thanks for fixing that! $\endgroup$ – Isaac Browne Nov 21 '17 at 2:55

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