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Using $\det(A-\lambda{}I)=0$ Find the eigenvalues for the given matrix: $$ A=\begin{bmatrix} 1&-1&0&0\\ 3&5&0&0\\ 0&0&1&5\\ 0&0&-1&1\\ \end{bmatrix} $$

The patterns in this matrix are obvious, so I am assuming there is a way to simplify this problem without expanding by a row/column, which could become messy really fast (although the abundance of zeros should help.) I just need a little direction.

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This matrix is block diagonal. That should point you in the right direction. – Emily Oct 21 '12 at 17:06
Subtract $\lambda$ from all the diagonal elements in the matrix you have written. There should be a simplification for calculating that determinant staring right back at you. – Geoff Robinson Oct 21 '12 at 17:06
up vote 2 down vote accepted


$$(1)\;\;\text{When}\,\,X\,,\,Y\,\,\text{square matrices, we have}\;\;\det\begin{pmatrix} X&0\\0&Y\end{pmatrix}=\det X\cdot \det Y$$

$$(2)\;\;\text{In our particular case}\,\,\,\det(A-\lambda I)=\det\begin{pmatrix}\lambda-1&1&0&0\\-3&\lambda-5&0&0\\0&0&\lambda-1&-5\\0&0&1&\lambda-1\end{pmatrix}$$

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