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In physics, a Hermitian matrix represents an observable and can be constructed using its eigenvalues and eigenvectors in the following way: $$ A = \sum_i \lambda_i v_iv_i^\dagger \qquad \qquad (1)$$ where $\lambda_i$ and $v_i$ are the $i^{th}$ eigenvalue and eigenvector and $v_i^\dagger$ is the transpose conjugate of $v_i$.

The proof is the following:

If the eigenvectors form an orthonormal basis, $\{v_i\}$, then we have:

$$ \sum_i v_iv_i^\dagger =1$$

This must be true becuse we have can write a vector $u$ in the $\{v_i\}$ basis by:

$$ u = \sum_i v_i v_i^\dagger u $$

Therefore, we can apply this identity twice to $A$ and get:

$$ A = \sum_i \sum_j v_iv_i^\dagger A v_jv_j^\dagger = \sum_i \sum_j v_iv_i^\dagger \lambda_j v_jv_j^\dagger= \sum_i \sum_j \lambda_j v_iv_i^\dagger v_jv_j^\dagger = \sum_i \sum_j \lambda_j v_i \delta_{ij} v_j^\dagger= \sum_i \lambda_j v_i v_j^\dagger$$

Is equation (1) only valid for matrices having eigenvectors that form a basis? Or all matrices can be constructed in this way?

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If the eigenvalues of $A$ are real, but $A\neq A^\dagger$, then the right hand side of (1) is Hermitian, but the left hand side is not, so (1) fails. An example is $$\left(\begin{array}{cc} 1&1\\0&2\end{array}\right)$$.

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  • $\begingroup$ Will the eigenvectors of a Hermitian matrix form an orthonormal basis? $\endgroup$
    – Ivan
    May 31, 2020 at 0:51
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    $\begingroup$ @IvanMartinez yes. This is a standard exercise: On each eigenspace you can pick an orthonormal basis. If ${u},{v}$ are eigenvectors with distinct eigenvalues $\lambda,\mu$ then $$\overline{{u}}^TA{v}$$ is both $\lambda(\overline{{u}}^T{v})$ and $\mu (\overline{{u}}^T{v})$, so $\overline{{u}}^T{v}=0$. Also the eigen values are real as $\overline{{v}}^TA{v}=\lambda\overline{{v}}^T{v}$. Finally this means that $B=A-\lambda I$ is also Hermitian, so if $B^2v=0$, then $$0=\overline{{v}}^TB^2{v}=\overline{{v}}^TB^\dagger B{v},$$ so $Bv=0$. Thus the eigenspaces span the whole space. $\endgroup$
    – tkf
    May 31, 2020 at 2:40

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