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While studying matrices I came across this question:

Let M be this matrix:$\begin{bmatrix}A&B\\C&I\\\end{bmatrix}$

Where all four blocks are $n·n$ matrices, and I is an identity matrix.

Question: Show that $M$ is nonsingular iff $A-BC$ = nonsingular.

I did some crosswise multiplication and came to $A·I - B·C$ ...

Can I set the value of $I$ equal to 1? And how do I prove $A-BC$ = nonsingular?

Can someone explain?

Math420
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1 Answers1

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Notice that

$$ \begin{bmatrix} A & B \\ C & I \end{bmatrix}\begin{bmatrix} I & 0 \\ -C & I \end{bmatrix} = \begin{bmatrix} A - BC & B \\ 0 & I \end{bmatrix}. $$

(This is nothing but LU factorization in block form.) Recall that a block triangular matrix is invertible iff its diagonal elements are invertible. (See, for example this.) Since $\begin{bmatrix} I & 0 \\ -C & I \end{bmatrix}$ has diagonal elements $I$ and $I$, it is clearly invertible with inverse

$$ \begin{bmatrix} I & 0 \\ C & I \end{bmatrix}. $$

Thus

$$ \begin{bmatrix} A & B \\ C & I \end{bmatrix} = \begin{bmatrix} A - BC & B \\ 0 & I \end{bmatrix}\begin{bmatrix} I & 0 \\ C & I \end{bmatrix}. $$

The product of two matrices is invertible iff both are. The right matrix in the product is invertible iff $A-BC$ is invertible, as desired.

eepperly16
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