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?


1 Answer 1


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 $$\left(\begin{array}{cc} 1&1\\0&2\end{array}\right)$$.

  • $\begingroup$ Will the eigenvectors of a Hermitian matrix form an orthonormal basis? $\endgroup$
    – Ivan
    May 31, 2020 at 0:51
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.