# Diagonalize matrix: please check work

1. Apr 15, 2014

### 939

1. The problem statement, all variables and given/known data

2 -3
2 -5

2. Relevant equations

2 -3
2 -5

3. The attempt at a solution

(note: I apologize for poor notation)

eigenvalues = -4, 1
eigenvectors = (1/√5)(1/2), (1/√10)(3/1)

The matrix is not symmetric, thus the diagonal is (Q^-1)(A)(Q) = λ

Last edited: Apr 15, 2014
2. Apr 15, 2014

### HallsofIvy

What you have done looks good but your last line is not at all clear. You have not said what "Q" is. $Q^{-1}AQ$ is a 2 by 2 matrix, not a "diagonal", and you have not said what "$\lambda$" is intended to mean. Your attachment is correct.