Cfrgtkky

So, long ago in a linear systems class I was taught that you can take a matrix A and decompose it into a matrix $T^{-1}Lambda T$, where $Lambda$ is a diagonal matrix of eigenvalues, and $T$ is a matrix composed of the corresponding eigenvectors.

A problem just presented itself to me where I want to take a known set of eigenvalues and generate a random-looking matrix $A$ from that.

I'd like to generate a $T$ matrix from a random-number generator, but I'm faced with two issues: first, I want $T$ to not only be non-singular, but I'd like it to be at least reasonably well-conditioned. Second, if I figure out how to generate such a $T$, do I know that it's the one and only $T$ matrix that will generate my $A$ matrix from a given $Lambda$?

It seems like there's too many free parameters in T for that to be the case, but perhaps the requirement that it be non-singular puts enough restrictions on it that it's unique.

Someone enlighten me, please.

It is never unique. You can always multiply $T$ by a nonzero scalar and get another $T$.
More generally, if $S$ is an invertible diagonal matrix, then
$T^{-1} Lambda T = (ST)^{-1} Lambda (ST)$.

It is not true, in general, that every $ntimes n$ matrix (over, say, the complex numbers) is diagonalizable. For instance, the matrix $left(begin{smallmatrix}0&1\0&0end{smallmatrix}right)$ is not diagonalizable.

On the other hand, if you obtain a $ntimes n$ matrix from a random number generator, you can be almost $100%$ sure that $T$ will be non-singular. So, generate your matrix, test to see whether it is singular and, in the unlikely case in which it is, change slightly one of its entries. You will quickly get a non-singular matrix.

But there will not be a unique such matrix. If you replace $T$ with $lambda T$, for some $lambdaneq0$, $(lambda T)^{-1}Lambda(lambda T)=T^{-1}Lambda T$.