1 Attachment(s)

Diagonalisable problem ? (Similar Eigenvectors)

Attachment 23735

I've worked out the first two bit's which are easy enough:

i)Multiply and get eigenvalues

ii)Turns out that the eigenvalue = 2 has multiplicity 2(algebraic !?!) I did this by row reduction at each eigenvalue to check dimension of solution set.

iii) Now this bit seems impossible as there do not seem to be enough eigenvector's. I've plugged the matrix into a few programs and they can't quite solve it.

So does anyone know what's going on ?

Here is some data so that you don't have to work out anything:

**Characteristic polynomial:**

*x*4−3*x*3−2*x*2+12*x*−8

Real eigenvalues:

{-2, 1, 2, 2}

Eigenvector of eigenvalue -2:

(1, 1, 0, 0)

Eigenvector of eigenvalue 1:

(0, 0, -1, 1)

Eigenvector of eigenvalue 2:

(1, 0, -1, 1)

Eigenvector of eigenvalue 2:

(1, 0, -1, 1)

The best I can do is say that diagonal of T is made up of the eigenvalues of A.