$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$

Sum of Ket-Bra

^{First created in July 2018; Updated in Feb 2019}

Let $\Ket i$ and $\Ket j$ be the standard basis. $\Ket i\Bra j =\begin{bmatrix} 0&0&\ldots&\ldots&0\\ \vdots&\ldots&\ldots&\ldots&\vdots\\ 0&\ldots&1&\ldots&\vdots\\ \vdots&\ldots&\ldots&\ldots&\vdots\\ 0&0&\ldots&\ldots&0\\ \end{bmatrix} ,$ where the $1$ is at the $i^\mathrm{th}$ row $j^\mathrm{th}$ column, and therefore $A=[a_{ij}]=\sum_{i,j}a_{ij}\Ket i\Bra j.$

If we restrict the ket and bra to be dual, $A=\sum_k a_{kk}\Ket k\Bra k$ becomes a diagonal matrix.

If $\Ket k$ were replaced by $\Ket{\psi_k}$, and at least one of them is not a standard basis vector, $A$ would not be diagonal.

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Let $A=[a_{ij}]$ and $B=[b_{ij}]$. Also, $A=\big[\Ket{a_1}~\Ket{a_2}~\ldots~\Ket{a_n}\big]$ and $B=\begin{bmatrix}\Bra{b_1}\\\Bra{b_2}\\\vdots\\\Bra{b_n}\end{bmatrix}.~~ \displaystyle\large AB=\sum_{i=1}^n\Ket{a_i}\Bra{b_i} .$

Illustration 1: $~~ \Ket{a_k}\Bra{b_k} =[a_{ik}b_{kj}],~ \sum_k\Ket{a_k}\Bra{b_k} =\sum_k[a_{ik}b_{kj}] =\big[\sum_k a_{ik}b_{kj}\big] .$

Illustration 2: $~~ I=\big[\ldots\Ket i\ldots\big]=\begin{bmatrix}\vdots\\\Bra j\\\vdots\end{bmatrix},~~ A=\sum_j\Ket{a_j}\Bra{j}$ and $B=\sum_i\Ket{i}\Bra{b_i},~~ AB=\sum_{i,j}\Ket{a_j}\Bra{j}\Ket{i}\Bra{b_i} =\sum_k\Ket{a_k}\Bra{b_k} .$

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If $A=\sum_k \lambda_k\Ket{\psi_k}\Bra{\psi_k},$ where $\{\Ket{\psi_k}\}$ is an orthonormal basis, then $\lambda_i$ is an eigenvalue of $A$ with eigenvector $\Ket{\psi_i}$.

Proof: $A\Ket{\psi_i} =\sum_k \lambda_k\Ket{\psi_k}\Braket{\psi_k\Verti\psi_i} =\lambda_i\Braket{\psi_i\Verti\psi_i}\Ket{\psi_i} =\lambda_i\Ket{\psi_i} .$

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$A^\dagger=\sum_k\lambda^*_k\Ket{\psi_k}\Bra{\psi_k}.$ So $A$ is Hermitian ($A^\dagger=A$) if and only if all its eigenvalues are real ($\lambda^*=\lambda$).

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As analysed in Mixed States, this is the unique eigendecomposition.

Therefore, we can conclude that any density matrices can be decomposed into a linear combination of its orthonormal eigenbasis.