# Doubt about exercise with eigenvalues

1. Apr 30, 2013

### Felafel

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

Given the endomorphism ϕ in $\mathbb{E}^4$ such that:
ϕ(x,y,z,t)=(4x-3z+3t, 4y-3x-3t,-z+t,z-t) find:

A)ker(ϕ)
B)Im(ϕ)
C)eigenvalues and multiplicities
D)eigenspaces
E)is ϕ self-adjoint or not? explain

3. The attempt at a solution

I get the associated matrix:

(4 0 -3 3)
(0 4 -3 -3)
(0 0 -1 1)
(0 0 1 -1)

but i can remove the last row, because it equals the third multiplied by -1

-solving AX=0 i have ker= L((0,0,1,1),(3, -3, 4, 4))

- reducing the columns i get Im= L((1, 0, 0),(0,1,0))

I'm not really sure this results are right, but what I wanted to ask is:
how do I compute the eigenvalues if the matrix is not square? is the rest of the exercise unsolvable?

thank you :)

2. Apr 30, 2013

### micromass

Staff Emeritus
I'm not sure why you think you can do this... If you're solving a system of equation Ax=b, then this is a valid technique. But you're not doing that here.

3. Apr 30, 2013

### Felafel

Okay, so I'll have to keep alle the rows all through the exercise, right?
I've taken it again, with the following results:

$ker(\phi)=L(0, \frac{3}{2}, 1, 1)$

$Im(\phi)=L((-3,-3,-1,-1),(4,0,0,0),(0,4,0,0))$

eigenvalues and multiplicity: $T_1=0^1 ; T_2=-2^1; T_3=4^2$

eigenspaces:

(6 0 -3 3)
(0 6 0 0)
(0 0 -3 3) = T(-2)
(0 6 -5 1)

$V_{-2}= L((2,0,-1,1),(0,1,0,0),(0,0,-1,1),(0,0,5,-1))$

(0 0 -3 -3)
(0 0 0 0)
(3 -3 1 -5) = T(4)
(0 -6 -4 -4)

$V_4=L((3,-3,1,5),(0,3,2,2),(0,0,-1,1))$

$V_0=ker$

which is not selfadjoint because its algebric multiplicity is different from its geometrical multiplicity.

is it okay now?
thanks again