Dirac notation - Outer product representation of Normal Matrix

Question

I'm studying some linear algebra applications in quantum mechanics, and I was told that a normal matrix can be written as: $$ M=\sum_{i=1}^{n}\theta_i |\theta_i\rangle \langle\theta_i| $$ where $|\theta_i\rangle$ is the eigenvector associated with it's eigenvalue $\theta_i$.

The problem is that I can't properly visualize that summation as a normal matrix representation.

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Here's my attempt to visualize why that's true.

I know, by spectrum theorem, that I can diagonalize that matrix M by some unitary matrices:

$$ D = U^{\dagger}MU \Rightarrow U^{\dagger}\big(\sum_{i=1}^{n}\theta_i |\theta_i\rangle \langle\theta_i|\big)U $$

So if I manage to calculate the right relation, I'll get why the matrix $M$ can be written as it was said, but how can I do that? How can I include $U$ and $U^{\dagger}$ into that summation to calculate it? Can someone please show me what's really happening in that summation?

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What I've been able to get is: $$ \theta_i|\theta_i\rangle $$ Is a scalar times a "column" vector. $$ \langle\theta_i| $$ Is a bra, or a conjugate transpose ket. $$ \theta_i |\theta_i\rangle \langle\theta_i| $$ Is a matrix, and the summation is actually adding multiple matrices with previous outer product computation.

Can someone please help me out? Thanks!

Answer 1

Note that $\langle \theta_i \mid \theta_j \rangle = \delta_{ij}$. So $M \lvert\theta_j \rangle = \sum_i \theta_i \lvert \theta_i \rangle \langle \theta_i \mid \theta_j \rangle = \theta_j \lvert \theta_j \rangle$ for all $j$. Geometrically $\lvert \theta_i \rangle \langle \theta_i \rvert$ is the orthogonal projection onto the line spanned by $\lvert \theta_i \rangle$.

Answer 2

First let $(\lvert e_i \rangle)_i$ be a basis of your Hilbert space. Note that $\lvert e_i \rangle^\dagger = \langle e_i \rvert$ and the matrix $\lvert e_i \rangle \langle e_i \rvert$ is actually the diagonal matrix filled with zeros except with a $1$ on the $i$-th column and $i$-th row.

So any diagonal matrix $D$ can be written in the form $D = \sum_{i=1}^n \alpha_i \lvert e_i \rangle \langle e_i \rvert$ where $\alpha_i$ are the diagonal entries of $D$. Let $M$ be an arbitrary normal matrix. Then by the spectral theorem, there exists a unitary matrix $U$ such that $$ M = U D U^\dagger = \sum_{i=1}^n \alpha_i U \lvert e_i \rangle \langle e_i \rvert U^\dagger = \sum_{i=1}^n \alpha_i U \lvert e_i \rangle \big(U\lvert e_i \rangle \big)^\dagger. $$ Now we can simply rename $\theta_i = \alpha_i$ and $U\lvert e_i \rangle = \lvert \theta_i \rangle$ to obtain the required form.