Spectral decomposition of Hermitian positive matrix

Question

Is this always possible to write spectral decomposition of a Hermitian positive definite matrix in terms of unit rank projectors?

Update: I just found this: About the uniqueness of rank-1 decomposition of a positive-definite Hermitian matrix, claiming: "Suppose $T$ is positive-definite Hermitian matrix and I know that it can be expressed by eigen-decomposition as the following sum of rank-1 matrices...", but I am not sure if this claim is correct.

No, I am not interested in finding eigenvalues and eigenvectors for an specific matrix. I just want to now whether it is possible in general to write a Hermitian positive definite matrix in terms of unit rank projectors? — Mah, Jul 12 '18 at 19:20
Yes, every symmetric matrix (not just positive definite ones) can be decomposed in this way, but the decomposition isn't necessarily unique. This is a well-known theorem of linear algebra. — Brian Borchers, Jul 12 '18 at 19:43
@BrianBorchers Thanks. Does this theorem has a name, so I can search for it? — Mah, Jul 12 '18 at 19:46
It's a very simple version of the spectral theorem for self adjoint operators. I've seen it referred to as "the spectral theorem" or "the orthogonal diagonalization theorem" — Brian Borchers, Jul 12 '18 at 20:58

score 2 · Accepted Answer · answered Jul 13 '18 at 01:46

The spectral theorem as it applies to Hermitian matrices can be stated as follows: $T$ is positive definite and Hermitian if and only if there exists a unitary $U$ and real diagonal $D$ such that $T = UDU^*$.

From this version of the spectral theorem, it is easy to obtain the result you're looking for. In particular, let $u_1,\dots,u_n$ denote the columns of $U$. By applying block matrix multiplication, we see that $$ T = UDU^* = \pmatrix{u_1 & \cdots & u_n} \pmatrix{\lambda_1 \\ & \ddots \\ && \lambda_n} \pmatrix{u_1^* \\ \vdots \\ u_n^*} \\ = \lambda_1 u_1u_1^* + \cdots + \lambda_n u_nu_n^* $$

Spectral decomposition of Hermitian positive matrix

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