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I am given question that needs me to prove a lemma, the lemma is as following:

Given a symmetrical matrix $A$ and a matrix $B$ that is constructed by removing the last row and column of $A$, prove that:

$$\lambda_{min}(B) \ge \lambda_{min}(A)$$

I have tried Rayleigh Quotient but it was to no avail. I thought of an eigenvector for $A$ that has its last entry zero and tried to find a relation between the eigenvalues of $A$ and $B$ yet I cannot come up with anything. Can anyone prove this lemma?

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  • $\begingroup$ Replace $A$ by $-A+\lVert A\rVert I$ and then show the corresponding inequality for the maximal eigenvalue. $\endgroup$
    – tomasz
    May 15 at 15:46

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Use the Eigenvalue interlacing theorem. See Theorem 1 of the attached link.

https://people.orie.cornell.edu/dpw/orie6334/Fall2016/lecture4.pdf

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  • $\begingroup$ Thanks very helpful! Though I did not understand the last part of the proof, especially this section: Recall $\lambda_k$ = min_{x∈V} \frac{x^TAx}/{x^Tx} and $\beta_k$ = max _{x∈W}\frac{x^TBx}/{x^Tx}. Then we see that λk ≤ weTAwe weTwe=wTBwwTw≤ βk $\endgroup$ May 15 at 16:04

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