Prove that a normal matrix can be completely determined by its eigen values and a unitary matrix.

Question

Prove that a normal matrix can be completely determined by its eigenvalues and a unitary matrix.
I tried using the Spectral decomposition theorem: Is states that a normal matrix N with spectrum set {$k_1, k_2,.., k_n$} can be represented as N = $k_1P_1 + k_2P_2 + ... + k_nP_n$ where for i,j belonging to [n] and $P_iP_j$ = P_i if i = j, else 0.
I tried making it all a matrix with multiplication of a diagonal matrix containing all the eigen values and another matrix containing all the $P_is$, but it didn't work.

Answer 1

You have the right idea. A consequence of the spectral decomposition theorem is the that any normal matrix is unitarily diagonalizable: $$A=U^{-1}DU$$ Where $U$ is unitary and $D$ is diagonal. This means that given the corresponding unitary diagonalizing matrix and the eigenvalues of $A$ (which give you its diagonal form), you can find $A$ using this exact formula.