# Thread: Linear Algebra - Find a matrix P that diagonalizes A

1. ## Linear Algebra - Find a matrix P that diagonalizes A

This question is broken up into 4 parts to help you get to the right answer. Below is where I am so far.

Find eigenvalues of A:
A=[2 13
-1 -12]

det(lI - A) = 0
So the equation is l^2-10l - 11=0
So lambda = -11,1

b) Find corresponding eigenvectors
if l=-11
eigenvector is [-1
1]
if l=1
Eigenvector is [-13
1]

c) Find a matrix P that diagonalizes A
P = [-1 -13
1 1]

d) Verify that P^-1 A P is diagonal
This verification does not work for me, so I assume I did something wrong above? Any idea where I went wrong?

2. With a 2x2, you don't have to use the det method.

The characteristic polynomial is simply $\displaystyle \displaystyle \lambda^2-tr(A)\lambda+det(A)\rightarrow \lambda^2+10\lambda-11=(\lambda+11)(\lambda-1)$

Your eigenvalues are correct. Looks correct to me though.

3. All your work is correct. You may have made a clerical error when doing the multiplication.

4. Originally Posted by saxophonemaster
d) Verify that P^-1 A P is diagonal
This verification does not work for me, so I assume I did something wrong above? Any idea where I went wrong?
All is right. Verify:

$\displaystyle AP=\begin{pmatrix}{2}&{13}\\{-1}&{-12}\end{pmatrix}\begin{pmatrix}{-1}&{-13}\\{1}&{1}\end{pmatrix}=\begin{pmatrix}{11}&{-13}\\{-11}&{1}\end{pmatrix}$

$\displaystyle PD=\begin{pmatrix}{-1}&{-13}\\{1}&{1}\end{pmatrix}\begin{pmatrix}{-11}&{0}\\{0}&{1}\end{pmatrix}=\begin{pmatrix}{11}& {-13}\\{-11}&{1}\end{pmatrix}$

So, $\displaystyle P^{-1}AP=D$

Regards.

Fernando Revilla

5. Your eigenvalues and P are correct. What did you get for $\displaystyle P^{-1}$?