Why can we decompose Hermitian Matrices

eigenvalues-eigenvectorshermitian-matriceslinear algebra

In my quantum physics class today, I learned about Hermitian Matrices decomposition. The process can be described as follows:

Decomposing Hermitian Matrices:

  1. Find eigenvalues.
  2. Find eigenvectors.
  3. Find generator matrices using eigenvectors (where a generator matrix is defined as the outer product of an eigenvector with itself).
  4. Find the decomposition using eigenvalues and generator matrices.

I would like to know what relationship between eigenvalues and generator matrices allows such a decomposition. Namely, why does the relationship

$$
[M] = \sum\limits_{i=1}^n \lambda_i\ {| i\rangle} {\langle i|}
$$

hold?

note:
$[M] =$ the Hermitian matrix; $\lambda_i =$ eigenvalue;
${| i\rangle} {\langle i|} =$ generator

Best Answer

The crucial relationship between what you call generator matrices and the eigenvalues is that the generator matrices are orthogonal projections (namely $|i\rangle\langle i|$ projects onto the linear subspace spanned by $|i\rangle$) and $M$ acts as $\lambda_i$ times the identity on this subspace.

Once you know that every Hermitian matrix has at least one eigenvalue, this is enough to show this decomposition as follows. Let $\lambda_1$ be an eigenvalue of $M$ with corresponding normalized eigenvector $|1\rangle$. Knowing that $|1\rangle\langle 1|$ is the orthogonal projection onto the linear subspace spanned by $|1\rangle$, we can write $$ M|=M(|1\rangle\langle 1|+I-|1\rangle\langle 1|)=\lambda_1|1\rangle\langle 1|+M(I-|1\rangle\langle 1|). $$ Now $I-|1\rangle\langle 1|$ is the orthogonal projection onto the orthogonal complement of $|1\rangle$. Since $M$ is hermitian, this space is also invariant under $M$, or in other words, $M(I-|1\rangle\langle 1|)=(I-|1\rangle\langle 1|)M(I-|1\rangle\langle 1|)=:M_1$.

The matrix $M_1$ is again hermitian on the $(n-1)$-dimensional subspace $|1\rangle^\perp$. This means you can play the same game again: Find and eigenvalue $\lambda_2$ with corresponding normalized eigenvector $|2\rangle$ and write $M_1$ as $$ M_1=\lambda_2|2\rangle\langle 2|+M_1(I-|2\rangle\langle 2|). $$ Thus $$ M=\lambda_1|1\rangle\langle 1|+M_1=\lambda_1|1\rangle\langle 1|+\lambda_2|2\rangle\langle 2|+M_1(I-|2\rangle\langle 2|). $$ Note that in every step we reduce the dimension by $1$, so that after $n$ steps (if $M$ is an $n\times n$ matrix) we arrive at $$ M=\sum_{i=1}^n \lambda_i|i\rangle\langle i|. $$

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