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I'm trying to answer this question:

Let $A$ be an $n \times n$ matrix with rank $n-1$. Furthermore, let $Q$ be an orthogonal matrix. Name an eigenvalue of $Q^T(A-1I)Q$.

I know that $Q^T=Q^{-1}$ and that by similarity we can essentially ignore Q leaving us with $(A-1I)$. I also know that if we were to know an eigenvalue of a, call it $\lambda$, that $(A-1I)$ would give us $\lambda - 1$. What I don't know is what the rank of $A$ has to do with determining an eigenvalue. Is it something to do with the number of pivot columns?

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    $\begingroup$ Since $A$ is singular, there is some $v$ such that $Av = 0$. Use this to find an eigenvalue. $\endgroup$
    – copper.hat
    Mar 10, 2015 at 5:24

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$$ Q^T(A-I)Q= Q^{-1}AQ - I $$

Now there exists $v \in \ker A - \{0\}$, so what will happen if we apply the above matrix to $Q^{-1}v$?

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