# Proving that any unitary matrix can be diagonalised by a similar matrix

I'm having struggles with understanding important facts about spectral theorem in finite dimensional spaces. For hermitian matrices, I saw in classes that the similarity matrix that diagonalises any nxn hermitian matrix is unitary. In fact, it's constructed explicitly using the eigenvalues of the hermitian matrix (A of nxn).

If S is the matrix that has eigenvectors as columns, by multiplying S* S (S* is the bedaggered matrix -conjugated transposed matrix-), following the kronecker delta, we will get S* S = I Identity matrix.

Until there, everything is OK. Then, the eigenvectors of A have an associated eigenvalue. So, if we multiply A and S, we should get the S matrix with every eigenvector multiplied by the associated eigenvalue, but why does this happen? I'm guessing everything has to do with the eigenvalues problem itself, but don't we need S as a vector, and not a matrix?

Then, finally, in the lecture the proof follows multiplying S*(by the left) and AS, obtaining a diagonal matrix of the n eigenvalues. So, in conclusion, this hermitian matrix can be diagonalised by a unitary matrix.

S* A S = D, then A = S D S*.

What happens when the eigenvalues are repeated, or the eigenvectors aren't linearly independent? My professor said "then apply GrammSchmidt..." but here I am.

What I need to proof, actually, is that a nxn unitary matrix can be diagonalised using the same relation that we can derive from a hermitian matrix, over a finite dimension space. I know that a unitary matrix is normal too, and I guess I can apply the same process, but wouldn't that be the same proof ? what changes in normal (or unitary) matrices?

Let us consider $$T$$ normal, since it's basically the same proof. What you want to show is two things:
• that eigenspaces are not only invariant for $$T$$ but actually reducing, in that they are also invariant for $$T^*$$;
For the first part one uses a slightly stronger result: if $$T$$ is normal, then $$Tw=\mu w$$ if and only if $$T^*w=\bar\mu w$$. To see this, suppose $$\|w\|=1$$. Then \begin{align} \|T^*w-\bar\mu w\|^2&=\langle T^*w,T^*w\rangle +|\mu|^2-2\operatorname{Re}\mu\langle T^*w,w\rangle\\ &=\langle TT^*w,w\rangle +|\mu|^2-2\operatorname{Re}\mu\langle w,Tw\rangle\\ &=\langle T^*Tw,w\rangle +|\mu|^2-2\operatorname{Re}\bar\mu\langle Tw,w\rangle\\ &=\langle Tw,Tw\rangle +|\mu|^2-2\operatorname{Re}\langle Tw,\mu w\rangle\\ &=\|Tw-\mu w\|^2. \end{align}
Now suppose that $$Tv=\lambda v$$, $$Tw=\mu w$$, with $$\|v\|=\|w\|=1$$. Then $$\lambda\langle v,w\rangle=\langle Tv,w\rangle=\langle v,T^*w\rangle=\langle v,\bar\mu w\rangle=\mu\langle v,w\rangle.$$ So, if $$\lambda\ne\mu$$, then $$\langle v,w\rangle=0$$.
With that, you can redo the same argument you did for selfadjoints, to find an orthonormal basis of eigenvectors for $$T$$.