Finding A Matrix, given eigenvalues, and eigenvectors

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 55K views
Laura1321412
Messages
23
Reaction score
0
Find a matrix that has eigenvalues 0,18,-18 with corresponding eigenvectors (0,1,-1), (1,-1,1), (0,1,1).

... I know the diagonlize rule, and the the rule to find a a power of A

A= PDP^-1
D=P^-1AP


... but i am lost as to how to contine... help please?
 
Physics news on Phys.org
Call you eigenvectors u1,u2,u3. Call your eigenvalues l1,l2,l3. Call you matrix A. I guess A is 3x3, so it has 9 coefficients.

You have 3 vector equations

Au1=l1u1
Au2=l2u2
Au3=l3u3

Consider the matrix coefficients a11,a12,a13, etc as unknowns. You have 3x3=9 linear equations for nine unknowns. Not too bad. There is a hope.

Of course there are other, more elegant methods. But this one is pretty straightforward.
 
It's not clear to me what you are asking. You say you have found the eigenvalues to be 0, 18, -18 with corresponding eigenvectors (0,1,-1), (1,-1,1), (0,1,1).

So your matrix P is given by
[tex]P= \begin{bmatrix} 0 & 1 & 0 \\ 1 & -1 & 1 \\ -1 & 1 & 1\end{bmatrix}[/tex]
and D is
[tex]D= \begin{bmatrix}0 & 0 & 0 \\ 0 & 18 & 0 \\ 0 & 0 & -18\end{bmatrix}[/tex]

Yes, it is true that [itex]A= PDP^{-1}[/itex], [itex]D= P^{-1}AP[/itex], and that [itex]A^n= PD^nP^{-1}[/itex].

Can you find [itex]P^{-1}[/itex]?

It is
[tex]P^{-1}= \begin{bmatrix}1 & \frac{1}{2} & -\frac{1}{2} \\ 1 & 0 & 0 \\0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix}[/tex].
 
Last edited by a moderator: