Diagonalizable matrix problem with parameters

[ryebread]

Homework Statement

A is a given 3x3 matrix
A=$$\begin{pmatrix}<br /> a & 0 & 1 \\<br /> 0 & a & b \\<br /> 0 & 0 & c <br /> \end{pmatrix}$$
a,b,c are real numbers
1) determine for what values of a,b,c the matrix A is diagnoizable. (advice: distinguish between the cases a=c and a≠c)

Homework Equations

characteristic polynomial of A is det(tI-A)

The Attempt at a Solution

finding the solution for det(tI-A)=0
tI-A=$$\begin{pmatrix}<br /> t-a & 0 & -1 \\<br /> 0 & t-a & -b \\<br /> 0 & 0 & t-c <br /> \end{pmatrix}$$
the characteristic polynomial is (t-c)(t-a)^2 and therefore the eigenvalues are a and c.
for t=c we get the matrix $$\begin{pmatrix}<br /> c-a & 0 & -1 \\<br /> 0 & c-a & -b \\<br /> 0 & 0 & 0 <br /> \end{pmatrix}$$
let it be C, x=(x,y,z), for Cx=0 we get the system of linear equations:
(c-a)x-z=0
(c-a)y -bz=0
likewise, for t=a we get
-1z=0
-bz=0
(a-c)z=0

after that I'm pretty much lost. my guess is that you need to find the dimension of the eigenspace that corresponds to each eigenvalue, but I'm nut sure how.

Help would be very welcomed

 

[jbunniii]

As the problem suggests, consider the a=c case separately. If a=c, then your equations above simplify to
-z = 0
-bz = 0

What constraints does this impose on x, y, and z?

 

[ryebread]

As the problem suggests, consider the a=c case separately. If a=c, then your equations above simplify to
-z = 0
-bz = 0

What constraints does this impose on x, y, and z?

i have no idea.
I finally managed to solve it with algebraic and geometric multiplicities, but i just don't get nullity.