# In the Cholesky Factorization, does the spectrum of the individual factors L and L^t tell us anything about the spectrum of the original matrix A?

I computed a (the?) Cholesky factorization of a 3x3 real, symmetric matrix. No problem, just a pretty standard computation.

However part (b) asks how could I determine that the original matrix $A$ is positive-definite, by inspecting individual entries of $L$, where $A= LL^t$.

Well, both factors, $L$ and $L^t$, that I computed have positive diagonal entries; of course, taking the transpose keeps the diagonal fixed. So, I know that both factors have the same, positive eigenvalues, counting multiplicity, etc.

Is this enough to conclude that $A$ is positive-definite? I.e., is the product of two positive-definite matrices again positive definite? Basically I can't say 100%, because, although I know, by the multiplicativity of the determinant, $A$ will have positive determinant, but I feel like it's possible for $A$ to have pairs of negative eigenvalues, too.

Unless...the spectrum of $L$ (and of $L^t$) somehow also determines the spectrum of $A$? Then we could say for sure, that by inspecting the factors $L$ and $L^t$, we could conclude that $A$ is positive-definite.

What do you think?

Thanks,

• It is not clear what you are asking. If $A=L L^T$ then $x^T Ax = (L^T x)^2$ so $A$ is positive semidefinite and $A$ is positive definite iff $L$ (or $A$) is invertible. – copper.hat Dec 28 '15 at 22:35
• Yes, I arrived at the same calculation, using just the definitions, too. But now I see it, by the way you put it. $L^t$ can't kill any non-vectors, so it must have trivial null-space, i.e., full-rank. So $L^t$ must be invertible. But does this fact also make $L$ invertible? (And consequently A invertible ...) @copper.hat, thanks, – user301446 Dec 28 '15 at 22:42
• O yes, their spectrum is the same, since they are similar to one another, I think...@copper.hat, thanks so much, – user301446 Dec 28 '15 at 22:43
• One quick follow-up question, before moving on, if you don't mind @copper.hat: can I give a uniqueness argument for the factor $L$, in this case where $A$ is positive-definite? (So then $L$, lower triangular, has positive diagonal entries.) If it's a bit more work, then I will think about it more, but perhaps there's a quick way to show the uniqueness. What do you think? Thanks, – user301446 Dec 28 '15 at 22:49