Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
This matrix is block diagonal. That should point you in the right direction. –  Arkamis Oct 21 '12 at 17:06
1  
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
add comment

1 Answer 1

up vote 1 down vote accepted

Hint:

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.