# Matrix inversion

1. Nov 3, 2008

### Frillth

1. The problem statement, all variables and given/known data

Suppose A is an invertible mxm matrix, B is an invertible nxn matrix, and C is an arbitrary mxn matrix. Is the matrix M =

A|C
----
O|B

invertible? Solve with proof.

Hint: Use block multiplication.

Note: I'm not quite sure how to draw this matrix on the forums. It should me an (m+n)x(m+n) matrix with the entries of A in the top left, the entries of C in the top right, zeros in the bottom left, and the entries of B in the bottom right. I hope that's clear.

Also, O is a matrix with all entries equal to zero.

2. Relevant equations

For some mxm matrix M:

M*M^-1 = M^-1*M = I_m

3. The attempt at a solution

I showed in a different problem that a matrix composed like in this problem, but with C being all zeros is invertible, so my approach for this problem is to assume that the matrix has an inverse, and then show that all of C's entries must be 0.

Let the matrix's inverse M^-1 be:
A|C
------
O|B
With A, C, O, and B being some matrices with the same dimensions as their non primed counterparts. Now multiply M^-1*M =
AA + CO|AC + CB
----------------------
OA + BO|OC + CB
Then multiply M*M^-1:
AA + CO|AC + CB
----------------------
OA + BO|OC + BB
This must be equal to the (m+n)x(m+n) identity matrix, which can be written as:
I_m| O
--------
O |I_n

With this knowledge and the fact that any matrix times O = O, we can write the following equations:

nxm matrices:
OA + BO = OA + BO = O
mxn matrices:
AC + CB = AC + CB = O
mxm matrices:
CO = O
nxn matrices:
OC = O

I also know:
A*A = A*A = I_m
B*B = B*B = I_n

I've tried rearranging these a little bit, but nothing has come of it so far. Am I on the right track? If so, could somebody nudge me in the right direction? If not, how should I attack this problem?

Thanks!

Edit: I just realized that I can do the following
OA = O
Multiply both sides by A = A^-1
OAA = OA
OI_m = O
O` = O
This doesn't really help me prove that C is all zeros though.

Last edited: Nov 4, 2008
2. Nov 4, 2008

### Dick

If you just multiply by (A^(-1)|B^(-1)), or however you write that, don't you get an upper triangular matrix with one's down the diagonal? Isn't the determinant of that nonzero? In fact, isn't it one?

3. Nov 4, 2008

### Frillth

Wait, how do determinants come into play here?

4. Nov 4, 2008

### Dick

A matrix is invertible if it's determinant is nonzero. Isn't it?

5. Nov 4, 2008

### Frillth

I actually didn't know that. Thank you.

However, I'm not sure that I'm supposed to use that method. I don't recall proving that in class, and we haven't learned how to find determinants of matrices larger than 3x3.

Is there another way that I could solve this?

6. Nov 4, 2008

### Dick

If you have an upper triangular matrix with ones down the diagonal, it's also pretty easy to see what row operations would reduce it to the identity. Is that proof enough?

7. Nov 4, 2008

### morphism

You can also pretty much write down the inverse of that matrix. It's easy to see that it has to be of the form

$$\begin{bmatrix}A^{-1} & X \\ 0 & B^{-1}\end{bmatrix}.$$

You only have to solve for X.