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There is a matrix H and it has $n \times q_n$ dimension. $$\begin{bmatrix} h_{11}&\cdots& h_{1j}&\cdots& h_{1q_n}\\ h_{21}&\cdots& h_{2j}&\cdots& h_{2q_n}\\ \vdots& &\vdots& \ddots& \vdots \\ h_{n1}& \cdots& h_{nj}& \cdots& h_{nq_n}\end{bmatrix}$$

The singular values $\sigma_i$ of matrix H are $\sqrt{\lambda_i}$, where the $\lambda_i$ are the eigenvalues of $H^TH$. Is there any relationships between ${\|h_{j}\|_2}^2$ and the singular values $\sigma_i$??? where $h_j$ is one column of matrix H. Such as ${\|h_{j}\|_2}^2 \le \sigma_i^2$ or ${\|h_{j}\|_2}^2 \le \lambda_i$

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  • $\begingroup$ You have $\sum_j\|h_j\|_2^2 = \sum_i\sigma_i^2$, which is the square of the Frobenius norm of $H$. $\endgroup$
    – Friedrich Philipp
    Mar 2, 2016 at 21:11
  • $\begingroup$ Is there any inequality between ${\|h_{j}\|_2}^2$ and the maximum singular value? $\endgroup$
    – Vivian Zhang
    Mar 2, 2016 at 21:50
  • $\begingroup$ You can get (a bad) one from the equality: $\|h_k\|_2^2\le\sum_j\|h_j\|_2^2 = \sum_i\sigma_i^2\le\min\{n,q_n\}\sigma_\max^2$. You can replace the min by the rank of H (which is a little better). $\endgroup$
    – Friedrich Philipp
    Mar 3, 2016 at 0:59

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Counterexample: Take for example $$ H=\left( \begin{array}{cc} 0 & 1 \\ 2 & 1 \\ \end{array} \right). $$ The singular values are $\sqrt{3+\sqrt{5}}$ and $\sqrt{3-\sqrt{5}}$, while the columns of $H$ have $2$-norms $2$ and $\sqrt2$. Notice that $\sqrt2>\sqrt{3-\sqrt{5}}\approx0.87$.

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  • $\begingroup$ Can you teach me how to calculate the singular values of a square matrix? $\endgroup$
    – Vivian Zhang
    Mar 2, 2016 at 22:04
  • $\begingroup$ You have given the definition in your post, that's it. $\endgroup$
    – John B
    Mar 2, 2016 at 22:05
  • $\begingroup$ As your example, the eigenvalues are 2 and -1. How can I get the singular values $\sqrt{3+\sqrt{5}}$ and $\sqrt{3-\sqrt{5}}$ ? $\endgroup$
    – Vivian Zhang
    Mar 2, 2016 at 22:24
  • $\begingroup$ 2 and -1 are the eigenvalues of $H$. They are totally unrelated to the singular values as long as the matrix isn't symmetric. Have a look into your own post where you have defined the singular values. $\endgroup$
    – Friedrich Philipp
    Mar 3, 2016 at 1:05

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