# singular values of $A-\alpha I_n$ while $A\in\mathbb{C}^{n\times n}$

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.

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 $$\sigma_i + \alpha$$.
• what can $\Lambda$ be? I didn't understand the last part. How did you conclude the singular values are those? Apr 18, 2021 at 14:05
• 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. Apr 18, 2021 at 14:13
• Thanks, would you explain why $P$ is orthogonal? Apr 18, 2021 at 14:40
• 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? Apr 18, 2021 at 16:51