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Suppose $I$ is an $n \times n$ identity matrix, and $S$ is the $n \times n$ symmetric matrix with rank equals two. I was reading something saying that: $$\det(I-S)=(1-\lambda_1)(1-\lambda_2)$$

where $\lambda_1$ and $\lambda_2$ are the two largest (in absolute values) eigenvalues of $S$. Can anyone provide some clues for proving this? Thanks in advance!

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Notice that $S$ is symmetric real then it's diagonalizable over $\Bbb R$ and since it's rank is $2$ then it's similar to $$\operatorname{diag}(\lambda_1,\lambda_2,0,\ldots,0)$$ where $\lambda_i\ne0$ so $I-S$ is similar to $$\operatorname{diag}(1-\lambda_1,1-\lambda_2,1,\ldots,1)$$ and then the result follows easily.

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  • $\begingroup$ Thank you! That looks great. Just add the following comment that may be helpful for beginners: $D=P^{-1}SP$. Consider $P^{-1}(I-S)P=P^{-1}P-P^{-1}SP=I-D$. So $I-S$ is similar to $I-D$, and therefore they have the same determinant. $\endgroup$ – user175861 Sep 21 '14 at 16:06
  • $\begingroup$ This is a useful comment! You're welcome:-) $\endgroup$ – user63181 Sep 21 '14 at 16:12
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Diagonalizing does not matter much; if you know the full list of eigenvalues and multiplicities, adding, say, $tI$ just adds $t$ to each eigenvalue. The same would work for something with nontrivial Jordan form, you don't need to know full details of the Jordan form, just the diagonal entries.

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