Hello i have been trying to get the Spectral Decomposition of the following matrix:
$$ A =\begin{bmatrix} 2 & i & 1 \\- i & 2 & i \\ 1 & -i & 2 \end{bmatrix}$$
my efforts have been in noting that this matrix is Hermitian so i have found the eigenvalues $\lambda_1=0$ and $\lambda_2=3$ and $\lambda_3=3$ ; from them i have chosen the following eigenvectors :
$$\vec{v_{\lambda_1}} = \begin{bmatrix} 1 \\ i \\ -1 \end{bmatrix},\,\,\,\,\vec{v_{\lambda_2}} = \begin{bmatrix} 0 \\ 1 \\ -i \end{bmatrix} ,\,\,\,\,\vec{v_{\lambda_3}} = \begin{bmatrix} 1 \\ -i \\ 0 \end{bmatrix} $$
And from these eigenvectors i found their respective projectors by the use of the external product:
$$ P_k = \frac{\vec{v_{\lambda_k}} \otimes \,\vec{v_{\lambda_k}}}{\vec{v_{\lambda_k}}\cdot\vec{v_{\lambda_k}}} = \frac{\vec{v_{\lambda_k}}\cdot\left(\vec{v_{\lambda_k}}^{\dagger}\right)}{\left|\left| \vec{v_{\lambda_k}} \right|\right|^2}$$
for $k =1, 2, 3 $ where $P_1$ is the projector of $\, \vec{v_{\lambda_1}}$ and so on, where i got:
$$ P_1 = \begin{bmatrix} 1/3 & -i/3 & -1/3 \\ i/3 & 1/3 & -i/3 \\ -1/3 & i/3 & 1/3 \end{bmatrix}$$
$$ P_2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1/2 & i/2 \\ 0 & -i/2 & 1/2 \end{bmatrix} $$
$$ P_3 = \begin{bmatrix} 1/2 & i/2 & 0 \\ -i/2 & 1/2 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$
But when i try to write $$A = \lambda_1P_1 +\lambda_2P_2 + \lambda_3P_3$$
it doesn't work and i dont have the slightest idea why it doesn't, can someone help me please ?
Your eigenvector $v_{\lambda_3}$ is not orthogonal to $v_{\lambda_2}$.
Don't forget that $P_1+P_2+P_3$ must equal $I$.