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I came across a question on SE Math and was reading a proof on Sylvester's Determinant Theorem posted by anon. But I have a few doubts as I'm reading it. I will re-produce from the pdf file on the part that I don't understand so that it makes it easier for reference.

In Theorem 9, it says if matrix $A$ and $D$ are invertible, then in the block matrix below, $$ \begin{vmatrix} A_{m \times m} & B_{m \times n}\\ C_{n \times m} & D_{n \times n} \end{vmatrix} = \left | A \right |\left | D-CA^{-1}B \right | = \left | D \right |\left | A-BD^{-1}C \right | $$

The proof given to the claim was: $$\begin{align*} \begin{bmatrix} A_{m \times m} & B_{m \times n}\\ C_{n \times m} & D_{n \times n} \end{bmatrix} &= \begin{bmatrix} A_{m \times m} & 0_{m \times n}\\ C_{n \times m} & I_{n \times n} \end{bmatrix} \begin{bmatrix} I_{m \times m} & A^{-1}B_{m \times n}\\ C_{n \times m} & D-CA^{-1}B_{n \times n} \end{bmatrix}\\ &= \begin{bmatrix} I_{m \times m} & B_{m \times n}\\ 0_{n \times m} & D_{n \times n} \end{bmatrix} \begin{bmatrix} A-BD^{-1}C_{m \times m} & 0_{m \times n}\\ D^{-1}C_{n \times m} & I_{n \times n} \end{bmatrix} \end{align*}$$

First, I don't understand how $\begin{bmatrix} A_{m \times m} & 0_{m \times n}\\ C_{n \times m} & I_{n \times n} \end{bmatrix} \begin{bmatrix} I_{m \times m} & A^{-1}B_{m \times n}\\ C_{n \times m} & D-CA^{-1}B_{n \times n} \end{bmatrix}$ was derived from $\begin{bmatrix} A_{m \times m} & B_{m \times n}\\ C_{n \times m} & D_{n \times n} \end{bmatrix}$.

Second, when I did a multiplication between the matrices, $$ \begin{align*} &\begin{bmatrix} A_{m \times m} & 0_{m \times n}\\ C_{n \times m} & I_{n \times n} \end{bmatrix} \begin{bmatrix} I_{m \times m} & A^{-1}B_{m \times n}\\ C_{n \times m} & D-CA^{-1}B_{n \times n} \end{bmatrix} =\\ &\begin{bmatrix} A_{m \times m}I_{m \times m}+0_{m \times n}C_{n \times m} & A_{m \times m}A^{-1}B_{m \times n}+0_{m \times n}(D-CA^{-1}B_{m \times n})\\ C_{n \times m}I_{m \times m}+I_{m \times m}C_{n \times m} & C_{n \times m}A^{-1}B_{m \times n}+I_{n \times n}(D-CA^{-1}B_{m \times n}) \end{bmatrix}=\\ &\begin{bmatrix} A_{m \times m} & B_{m \times n}\\\ 2C_{n \times m} & D_{n \times n} \end{bmatrix}\neq \begin{bmatrix} A_{m \times m} & B_{m \times n}\\ C_{n \times m} & D_{n \times n} \end{bmatrix} \end{align*}$$ It's weird that in my multiplication doesn't get back the original matrix.

How was the first equation being derived by sort of "splitting" the matrix into 2 matrices? Also, why doesn't my multiplication get back the original matrix?

Thanks for any help.

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Does the decomosition come from Schur complement? You may refer wikipedia about it –  Yao Jin Sep 15 '11 at 8:28
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3 Answers

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You need three items to get the result and they are quite straightforward to see:

  1. $\det A \det B= \det AB$
  2. The determinant of a lower or upper triangular matrix is the product of its diagonal entries.

Hence if there is a matrix in the form of $$\begin{bmatrix}1&\sin(2.74)&8&-14\\&2&-3&\sqrt{2}\\&&3&\tan(\frac{\pi}{10})\\&&&4 \end{bmatrix} $$ and zero in the lower part. The determinant is simply $1\times 2\times 3\times 4$. You can see it by yourself if you run down along the first column to compute the determinant.

  1. Gauss Elimination method also works in block matrix form provided that fractions are replaced with the nonsingular matrices properly.

What this means is that multiplying a row by $\frac{b}{a}$ is replaced $BA^{-1}$ or $A^{-1}B$ depending on the particular case.

Then, you simply eliminate row and column entries just as you would do with two by two case. Let me give you the scalar case and you can fill up the rest for the block matrices: I have a matrix

$$ \begin{bmatrix} a&b\\c&d \end{bmatrix} $$

First row is OK but I want to get rid of $c$. So I take the first row $[1 \quad 0]$ and I take $-\frac{c}{a}$ multiple of the first row which gives me $[-c\quad -\frac{cb}{a}]$ and then add it to the second row.

$$ \begin{bmatrix} 1&0\\-\frac{c}{a}&1 \end{bmatrix} \begin{bmatrix} a&b\\c&d \end{bmatrix} = \begin{bmatrix} a&b\\0&d-\frac{cb}{a} \end{bmatrix} $$ Then perform a similar column operation, to get rid of $b$ $$ \begin{bmatrix} a&b\\0&d-\frac{cb}{a} \end{bmatrix} \begin{bmatrix} 1&\frac{-b}{a}\\0&1 \end{bmatrix} = \begin{bmatrix} a&0\\0&d-\frac{cb}{a} \end{bmatrix} $$ Regrouped together, $$ \begin{bmatrix} 1&0\\-\frac{c}{a}&1 \end{bmatrix} \begin{bmatrix} a&b\\c&d \end{bmatrix} \begin{bmatrix} 1&\frac{-b}{a}\\0&1 \end{bmatrix}= \begin{bmatrix} a&0\\0&d-\frac{cb}{a} \end{bmatrix} $$ Replace them with block matrices, respect the order and take the determinant of the last expression using the previous items and you are done.

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Just to clarify that this answer isn't in contradiction to the ones by Brian and me; by "the result", percusse is referring to the correct result without the typo. –  joriki Sep 15 '11 at 9:12
    
@joriki : Ah, yes thanks a bunch for the warning. Sorry for the sloppiness. –  user13838 Sep 15 '11 at 9:18
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Your multiplication is fine; the problem is a typo in the notes. They should read $$\left[\begin{matrix} A_{m\times m}&B_{m\times n}\\ C_{n\times n}&D_{n\times n} \end{matrix}\right]= \left[\begin{matrix} A_{m\times m}&0_{m\times n}\\ C_{n\times m}&I_{n\times n} \end{matrix}\right] \left[\begin{matrix} I_{m\times m}&A^{-1}B_{m\times n}\\ 0_{n\times m}&D-CA^{-1}B_{n\times n} \end{matrix}\right], $$ with a $0_{n\times m}$ in the lower left-hand corner of the second factor, just as the second factor in the next line of the notes has $0_{m\times n}$ in the upper right-hand corner.

Perhaps someone else has a feel for where the decomposition comes from; I’d be hard pressed to come up with it and can merely observe after the fact that it works.

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This is a typo in the proof. The second factor should be

$$\begin{bmatrix} I_{m \times m} & A^{-1}B_{m \times n}\\ 0_{n \times m} & D-CA^{-1}B_{n \times n} \end{bmatrix}$$

with a zero instead of a $C$, in analogy with the other product. Note that the claimed determinant wouldn't come out right either with the $C$ included.

The decomposition is related to the Schur complement; see also A geometric way to reason about Schur complements?

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Very helpful, that second link. –  Brian M. Scott Sep 15 '11 at 9:04
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