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Suppose I have two Positive Semi-Definite (PSD) matrices, $K_1$ and $K_2 = K_1 + \sigma I$. For $K_1$, I do eigen decomposition, I get the eigenvector matrix $V$ and diagonal eigenvalue matrix $D$, which hold: $K_1 = VDV^T$.

So, for $K_2$, it has the same eigenvector matrix: $V$ and its eigenvalue matrix is: $D + \sigma I$.

My question is how to proof: $K_2$ has such eigenvector and eigenvalue?

Thanks.

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1 Answer 1

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Since $K_1 = V D V^T$, write $K_2 = V D V^T + \sigma I = V D V^T + \sigma V I V^T = V (D + \sigma I) V^T$. Hence $K_2$ has eigenvalues $(D + \sigma I)$.

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