Let $v_{ij}$ be a $n\times n$ real positive definite matrix with $\omega_{\alpha}^2\neq0$ and $a_{\alpha}$ being its $\alpha^{th}$ eigenvalue and eigenvector, respectively. So we have
$$ \Sigma_{j}^nv_{ij}a_{j\alpha}=\omega^2_{\alpha}a_{i\alpha}\tag{1}, $$ where $a_{j\alpha}$ is the $j^{th}$ element in the eigenvector $a_{\alpha}$. Now I want to prove that the matrix,
$$\tau_{jk}=\Sigma_{\alpha}^n\frac{a_{j\alpha}a_{k\alpha}}{\omega_{\alpha}}\tag{2},\quad(\omega_{\alpha}>0)$$
is the reciprocal square root of the matrix $v_{ij}$ under the condition of normalized and orthogonal eigenvectors, i.e.,
$$\Sigma_{i}^na_{i\alpha}a_{i\beta}=\delta_{\alpha\beta}\tag{3}.$$
My attempt was to prove the following equation, $$ \Sigma_l^n\Sigma_m^n\tau_{jl}\tau_{lm}v_{mk}=\delta_{jk}\tag{4}. $$
If I substitute Eqn. (2) into Eqn. (4) and use Eqn. (3), I get,
$$ \Sigma_l^n\Sigma_m^n\tau_{jl}\tau_{lm}v_{mk}=\Sigma_m^n\Sigma_{\alpha}^n\frac{a_{j\alpha}a_{m\alpha}}{\omega_{\alpha}^2}v_{mk}=\Sigma_{\alpha}^na_{j\alpha}a_{k\alpha}\tag{5}, $$
where Eqn. (1) is used at the last equivalence. But I'm not sure how to get $\delta_{jk}$ from Eqn. (5). Can anyone give me a hint? Is there anything I'm missing? I really appreciate any help you can provide.