Decomposition of the same matrix with different unitary matrices

Question

Let matrix $M$ be Hermitian

$$M = U_1\Lambda U^*_1 \qquad \qquad (1)$$ $$M = U_2\Lambda U^*_2 \qquad \qquad (2)$$

$U_1$ and $U_2$ are unitary matrices.

How can we prove that we can have $U_1 \ne U_2$ , with exactly the same diagonal $\Lambda$?

Example

MATLAB code:

M = [4 3-i; 3+i 10];       %Hermitian
[U1, diag]= eig(M);        % i.e. M = U1*diag*U1'
U2 = U1; 
U2(:,1) = -U2(:,1);

U1*diag*U1';               % gives M
U2*diag*U2';               % also gives M 

% U1 is not equal to U2

% U1 =
%    0.8716 - 0.2905i   0.3746 - 0.1249i
%   -0.3948 + 0.0000i   0.9188 + 0.0000i

% U2 =                                      
%   -0.8716 + 0.2905i   0.3746 - 0.1249i
%    0.3948 + 0.0000i   0.9188 + 0.0000i

% U1*diag*U1' and U2*diag*U2'     % they both return M
% ans =
%    4.0000 - 0.0000i   3.0000 - 1.0000i
%    3.0000 + 1.0000i  10.0000 + 0.0000i

Answer 1

The eigenvectors of a matrix a determined up to a phase $e^{i\phi}$, indeed if ${\bf u}$ is an eigenvector of $A$ with eigenvalue $\lambda$ then

\begin{eqnarray} A {\bf u} &=& \lambda {\bf u} \\ \Rightarrow~~~A(e^{i\phi}{\bf u}) &=& \lambda (e^{i\phi}{\bf u}) \end{eqnarray}

that means that $e^{i\phi}{\bf u}$ is also an eigenvector of $A$ with eigenvalue $\lambda$. Also note both ${\bf u}$ and $e^{i\phi}{\bf u}$ have the same norm

$$ \require{cancel} |e^{i\phi}{\bf u}| = \cancelto{1}{|e^{i\phi} |}| {\bf u}| = | {\bf u}| $$

So if the eigenvectors are normalized (which is actually your case), then after multiplying them by an arbitrary phase they remain normalized!