# Homework Help: Linear algebra - Diagonalizable matrix

1. Jan 30, 2012

### BitterX

This question has a lot data but I don't really know how to connect it all together

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

Let A be a 5x5 matrix over R

3 eigenvectors of A are:
$u_1=(1,0,0,1,1)$
$u_2=(1,1,0,0,1)$
$u_3=(-1,0,1,0,0)$

also:
$\rho (2I-A)>\rho (3I-A)$

and: $A(1,2,2,1,3)^t=(0,4,6,2,6)^t$

prove that A is diagonalizable and find a diagonal matrix similiar to it.
2. Relevant equations

3. The attempt at a solution

What I can make of this is:
$(1,2,2,1,3)^t=u_1 + 2u_2 + 2u_3$

(then maybe I can say that P has the eigenvectors as columns and
$AP(1,2,2,0,0)^t=(0,4,6,2,6)^t$

but then what?)

and because
$\rho (2I-A)>\rho (3I-A)$
$\rho (2I-A)\leq 5$
it means that
$5>\rho (3I-A)$
and so
$det(3I-A)=0$
and 3 is an eigenvalue of A.

I would like some hints/suggestions on what to do.

Thanks!

2. Jan 30, 2012

### HallsofIvy

A 5 by 5 matrix is diagonalizable if and only if it has 5 independent eigenvectors. You are given three independent eigenvectors. Can you find two more?

3. Jan 31, 2012

### BitterX

I know how to get the eigenvectors from an existing matrix
but not from an eigenvalue or anything they gave me... :/

4. Jan 31, 2012

### BitterX

I think I found one eigenvector,

if 3 is an eigenvalue of A, then A is similar to a matrix which it's first column is
(3,0,0,0,0)t

so an eigenvector of this matrix is (1,0,0,0,0) and thus it's an eigenvector of A as well

I only need to find one more :D (assuming I'm right).