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!
Best Answer
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.