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I have to prove that singular values of $A-\alpha I_n$ are $\sigma_i+\alpha$, while $\sigma_i$s are singular values of $A$ and $A$ is hermitian and positive definite matrix. Also we know that: $$\sigma_1 \ge \sigma_2 \dots \ge \sigma_n $$ and $\alpha \gt -\sigma_n$. I know for hermitian matrices, eigenvalues and singular values are the same, but I don't know what else to do.

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Since $A$ is symmetric it has spectral decomposition $P \Lambda P^T$, where $\Lambda$ is a diagonal matrix with the eigenvalues of $A$ on the main diagonal, and with orthogonal $P$, i.e., $P^TP=I$, hence, $$ A+\alpha I = P \Lambda P^T+\alpha P P^T=P(\Lambda+\alpha I)P^T $$ hence the singular values of $A+\alpha I $ are $\sigma_i + \alpha$.

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  • $\begingroup$ what can $\Lambda$ be? I didn't understand the last part. How did you conclude the singular values are those? $\endgroup$ Apr 18, 2021 at 14:05
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    $\begingroup$ 1. Added explaination. $\Lambda = diag(\lambda_1,...,\lambda_n)$. 2. For symmetric matrix, the SVD equals the eigen-decomposition which is unique, hence by showing that $A+\alpha I$ can be decomposed into the form $UDV=PDP^T$, thus $D$ is the diagonal matrix with the eigenvalues\singularvalues. $\endgroup$
    – V. Vancak
    Apr 18, 2021 at 14:13
  • $\begingroup$ Thanks, would you explain why $P$ is orthogonal? $\endgroup$ Apr 18, 2021 at 14:40
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    $\begingroup$ This is a well known Theorem. You can read here: en.wikipedia.org/wiki/Spectral_theorem $\endgroup$
    – V. Vancak
    Apr 18, 2021 at 16:07
  • $\begingroup$ thanks a lot, I know it's not a part of my question, but would you tell me how can I show that the $A+\alpha I$ is positive definite? $\endgroup$ Apr 18, 2021 at 16:51

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