$W=H^{\dagger}H=V diag(\lambda_1,\lambda_2,...,\lambda_N) V^{\dagger}$: Connection between $\textbf{H}$ and $\textbf{V}$

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Let $\textbf{H}\in{C}^{M\times N}$ be a complex-valued matrix. And let $\textbf{W}=\textbf{H}^{\dagger}\textbf{H}\in C^{N\times N}$ with $\{\cdot\}^{\dagger}$ being transpose conjugate operator. By eigendecomposition, we can rewrite $\textbf{W}$ as \begin{align} \textbf{W} = \textbf{V} diag(\lambda_1,\lambda_2,...,\lambda_N) \textbf{V}^{\dagger} \end{align} where $diag(\lambda_1,\lambda_2,...,\lambda_N)$ is the diagonal matrix with entries being eigenvalues of $\textbf{W}$.

Question: How can one find the connection between $\textbf{V}$ and $\textbf{H}$? In other words, given $\textbf{H}$, how do we present a certain column of $\textbf{V}$ mathematically?

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The knowledge of some column of $V$ implies the knowledge of $\sigma^2$, the square of some singular value of $H$. That is, you implicitly know an approximation of some root of the polynomial $\det(H^*H-xI_n)$.

Conclusion. If you want a column of $V$, then the knowledge of $H$ is useless; you must work on $W$.