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iliketopology

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the eigen values of a matrix (that is unknown) and I have the eigen vectors.

I made some small changes to the diagonal eigen value matrix and would like to obtain

a full rank matrix based on the small change to eigen value that I made. I am not sure

how to proceed although I do know the following works.

Given D = diagonal eigen value matrix and V = eigen vectors. If C is the matrix

that we need to know, then C.V= D.V . I need to know C and found this solution

in a book:

C = transpose(Sqrt(D)*V )* (Sqrt(D)*V)

Note that the D here is positive diagonal matrix and that sqrt(D)*V has been normalized

such that the length of the all the rows is a unit vector.

Can someone point to the proof of this equality. I know that for an orthogonal matrix such as V, the inverse is the transpose. But, I still don't get it as to why to introduce the transpose and not use the inverse. And if I do that I just get D as

D*V*inv(V)=D.

Can someone comment.

thank