# Need Help with positive definite matrices

1. Aug 8, 2013

### Basil4000

1. The problem statement, all variables and given/known data
If A is positive definite, show that $A = C C^T$ where $C$ has orthogonal columns.

3. The attempt at a solution

So, I've got the first part figured out. Because $A$ is symmetric, an orthogonal matrix $P$ exists such that $P^TA P = D = diag(\lambda_1,...,\lambda_n)$ where $\lambda_i > 0$ because A is positive definite. Next I've defined $B = diag(\sqrt{\lambda_1},...,\sqrt{\lambda_n})$ Then I wrote $C = P^TB P$ then $C C^T = (P^T B P) (P^T B P)^T = (P^T B P) P^T B^T P = P^T B B P = P^T D P = A$

So I'm at the last step and I'm stuck on how to show that C has orthogonal columns. Any hints?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 8, 2013

### Office_Shredder

Staff Emeritus
C has orthogonal columns if and only if CtC is a diagonal matrix (you should be able to check that this claim is true easily- the off diagonal entries of Ct C are the inner products of the columns of C).

Also PDPt = A, not PtDP.

3. Aug 9, 2013

### Basil4000

Right, that's just sloppyness on my part typing that out. I haven't learned anything about inner products yet. Is there a way to look at this differently?

4. Aug 9, 2013

### Office_Shredder

Staff Emeritus
Inner product is the same as dot product. The typical definition of orthogonal vectors is that their dot product is zero, if you're working with a different definition you'll have to say what it is

5. Aug 9, 2013

### Basil4000

I just read your first reply again. I get it now. I was pretty tired last night. Thanks a lot!