# Help proving equality of matrix expressions

1. Jan 3, 2012

### Chuck37

I've been beating my head against this problem for hours. I see numerically that two expressions are equal, but I can't prove it:

(J$^{T}$R$^{-1}$J + P$^{-1}$)$^{-1}$J$^{T}$R$^{-1}$

=

PJ$^{T}$(JPJ$^{T}$ + R)$^{-1}$

J is arbitrary size, P and R are square though not necessarily equal. Can anyone help? I thought binomial inverse theorem would save me but I haven't been able to get rid of the extraneous terms.

2. Jan 3, 2012

### Chuck37

Well, right after I posted I tried a new tack of setting them equal and doing operations to both sides until it came out the same. Not quite as satisfying as transforming one into the other, but I guess it works. If someone can see how to make one into the other, I'd love to see it.

3. Jan 3, 2012

### AlephZero

This is equivalent to

J$^{T}$R$^{-1}$ JPJ$^{T}$ + J$^{T}$
=
J$^{T}$R$^{-1}$ JPJ$^{T}$ + J$^{T}$

Factorize it two different ways:

J$^{T}$R$^{-1}$ (JPJ$^{T}$ + R)
=
(J$^{T}$R$^{-1}$J + P$^{-1}$) PJ$^{T}$

And the final step to get to your equation should be obvious.

(Of course this was invented by working backwards.)

Last edited: Jan 3, 2012
4. Jan 4, 2012

### Fredrik

Staff Emeritus
As you just said, you don't actually have to "transform one into the other" to prove this result. However, it's not hard to just rewrite your left-hand side until you end up with the right-hand side. You just have to use the identity $(AB)^{-1}=B^{-1}A^{-1}$ repeatedly (and the fact that matrix multiplication is distributive over matrix addition). I recommend that you prove this identity first. It's not hard.

I can't tell you the complete solution because of the forum rules about textbook-style problems, so for now I will only suggest that you start by applying this identity to the factor $J^TR^{-1}$. To get more help, you will have to show your work up to the point where you get stuck.

By the way, you shouldn't put itex tags around each symbol. Instead, put an opening tag (itex or tex) before each formula and a closing tag after it. See the LaTeX guide for more information.

Edit: The method I suggested only works when J is invertible (and therefore square). When J isn't square, I think you have to do something like what AlephZero suggested.

Last edited: Jan 4, 2012