###### tags: `linear algebra`
# relationship between hermitian matrix and orthogonal matrix
#### hw-10-(2)-(b)
Question:
$\qquad$Let $A$ be a Hermitian matrix on $M_{n\times n}(\mathbb{F})$, $\lambda_{i}$ be eigenvalues of $A$ and $E_{\lambda_{i}}$ be the corresponding eigenspace of $\lambda_{i}$.
$\qquad$Let $P_{\lambda_i}$ : $\mathbb{R}^{3}\rightarrow\mathbb{R}^{3}$ be the orthogonal projection of $\mathbb{R}^{3}$ onto $E_{\lambda_i}$.
$\qquad$Show that $A=\sum_{\lambda \in\sigma(A)}\lambda P_{\lambda}$ . Here $\sigma(A)$ is the set of all eigenvalue of $A$ .
Solution:
$\qquad$Because $A$ is a Hermitian matrix , $A$ can be diagonalized like $Q\Lambda Q^{*}$ . $Q$ is unitary matrix and $\Lambda$ is diagonal matrix .
$\qquad$let $Q=\begin{pmatrix}q_1,q_2,q_3,\cdots,q_n\end{pmatrix}$
then $Q^*=Q^{-1}=\begin{pmatrix}q_1^*\\q_2^*\\q_3^*\\ \vdots\\q_n^*\end{pmatrix}$ and $A=\begin{pmatrix}q_1,q_2,q_3,\cdots,q_n\end{pmatrix}\begin{pmatrix}\lambda_1\\&\ddots\\&&\lambda_n\end{pmatrix}\begin{pmatrix}q_1^*\\q_2^*\\q_3^*\\ \vdots\\q_n^*\end{pmatrix}$
$\rightarrow A=\begin{pmatrix}\lambda_1 q_1,\lambda_2 q_2,\lambda _3q_3,\cdots,\lambda_nq_n\end{pmatrix}\begin{pmatrix}q_1^*\\q_2^*\\q_3^*\\ \vdots\\q_n^*\end{pmatrix}=\sum_{i=1}^n\lambda_iq_i q_i^*$
$\qquad$the $P_{\lambda_i}=E_{\lambda_i}(E^*_{\lambda_i}E_{\lambda_i})^{-1}E^*_{\lambda_i}=E_{\lambda_i}E^*_{\lambda_i}=q_i q_i^*$ (the $E_{\lambda_i}$ is a matrix which its column vectors is $q_i$
$\qquad$So $A=\sum_{i=1}^n\lambda_iq_i q_i^*=\sum_{i=1}^n\lambda_iP_{\lambda_i}$ .