# Can we always construct a matrix using its eigenvectors?

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?

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