Why can we decompose Hermitian Matrices?

Question

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

Answer 1

In general, if you have any basis $\mathbf v_1, \dots, \mathbf v_n$ of $\mathbb R^n$, you can define a linear transformation $f : \mathbb R^n \to \mathbb R^n$ simply by specifying $f(\mathbf v_1), \dots, f(\mathbf v_n)$.

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If $\mathbf v_1, \dots, \mathbf v_n$ is an orthonormal basis, then $f$ has the matrix representation $f(\mathbf x) = M\mathbf x$ where $$ M = f(\mathbf v_1)\mathbf v_1^{\mathsf T} + \dots + f(\mathbf v_n) \mathbf v_n^{\mathsf T}. $$ To check that this works, we can just take $$M \mathbf v_i = f(\mathbf v_1)\mathbf v_1^{\mathsf T}\mathbf v_i + \dots + f(\mathbf v_i)\mathbf v_i^{\mathsf T}\mathbf v_i + \dots f(\mathbf v_n) \mathbf v_n^{\mathsf T}\mathbf v_i$$ and simplify: the term $f(\mathbf v_i)\mathbf v_i^{\mathsf T}\mathbf v_i$ simplifies to $f(\mathbf v_i)$, and all other terms simplify to $0$.

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The key property of a symmetric matrix $M$ is that it has an orthonormal basis of eigenvectors: there is an orthonormal basis $\mathbf v_1, \dots, \mathbf v_n$ of $\mathbb R^n$ such that $M \mathbf v_i = \lambda_i \mathbf v_i$ for $i=1, \dots, n$.

So if we define $f(\mathbf x) = M\mathbf x$, then $f(\mathbf v_i) = \lambda_i\mathbf v_i$ for $i=1, \dots, n$, and therefore we can write $M$ as $$ M = \sum_{i=1}^n f(\mathbf v_i) \mathbf v_i^{\mathsf T} = \sum_{i=1}^n \lambda_i \mathbf v_i \mathbf v_i^{\mathsf T} $$ which is the decomposition you want.

If you replace $\mathbb R^n$ by $\mathbb C^n$, $\mathbf x^{\mathsf T}$ by the conjugate transpose, and "symmetric" by "Hermitian", all of this still holds.

Answer 2

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 $$ M=\sum_{i=1}^n \lambda_i|i\rangle\langle i|. $$