Is the spectral decompsoition of semidefinite matrix unique?

Question

Suppose $A\in M_n(\mathbb{C})$ is a semidefinite matrix, is the spectral decompsoition of $A$ unique?If not,can anyone show me some examples?Thanks

Answer 1

It depends on what you mean by spectral decomposition. The eigenspaces are of course unique, a base of eigenvectors is not. Take for instance the identity matrix $I$. Any base is a base of eigenvectors.

EDIT

I address two sentences appearing in my previous answer.

The eigenspaces are of course unique

Given $A\in M_n(\mathcal C)$, the eigenspace corresponding to $\lambda\in\mathcal C$ is defined as $$ V_\lambda = \ker(A-\lambda I) = \{v\in\mathbb C^n : Av=\lambda v\}. $$ The spectral theorem states that there exist $\lambda_1,\dots,\lambda_k\in\mathbb C$ such that $$ \mathbb C^n = V_{\lambda_1} \oplus \dots \oplus V_{\lambda_k}. $$ Therefore, the decomposition in eigenspaces is well defined an unique. Moreover, any two different eigenspaces are orthogonal, so we can write $$ A = \lambda_1 P_{\lambda_1} + \dots + \lambda_k P_{\lambda_k} $$ where $P_\lambda\in M_n(\mathbb C)$ is the orthogonal projection onto $V_\lambda$. This is the spectral decomposition of $A$.

a base of eigenvectors is not

If instead you mean this eigendecomposition of a matrix, then it is not unique, and non uniqueness doesn't come just from permutations of the eigenvectors or changes in sign. In fact, from every $V_{\lambda_i}$ of the decomposition above, you have to select a basis, and this is not unique as soon as one of them has dimension at least $2$. The identity matrix is an extreme example: any base gives a matrix $Q$ and the eigendecomposition $I=QIQ^{-1}$.