Eigenvalues and eigenvectors

Jennifer1990

1. The problem statement, all variables and given/known data
Let A and B be similar matrices
a)Prove that A and B have the same eigenvalues


2. Relevant equations
None


3. The attempt at a solution
Firstly, i dont see how this can even be possible unless the matrices are exactly the same :S

matt grime

So you think that eigenvalues uniquely characterize a matrix? What about

10
01

and

11
01

for example?

You've put 'none' for relevant equations. That isn't true - there's a definition of 'similar' and many for 'eigenvalue'. Try it. HINT: polynomials.

Random Variable

EDIT: Changed "equation" to "polynomial"

You have to show that A and B=P^-1AP (for some invertible matrix P) have the same characteristic polynomial.

HallsofIvy

Oh, I think that way is much too complicated!

Do it directly from the equation:
If [itex]Av= \lambda v[/itex] then, for any invertible P, [itex]P^{-1}Av= \lambda P^{-1}v[/itex]. Now define [itex]u= P^{-1}v[/itex].

Random Variable

Your way is too easy.

Random Variable

or let [tex] Av = \lambda v [/tex]

then [tex] AP^{-1}Pv = \lambda v [/tex] and go from there

Jennifer1990

what do u mean by the same characteristic equation?

Random Variable

EDIT: changed "equation" to "polynomial"


The the characteristic polynomial of matrix A is [tex] det(A- \lambda I). [/tex]

The characterisitc polynomial of matrix B is [tex] det(B - \lambda I) = det(PAP^{-1} - \lambda I) [/tex]

so show that [tex] det(A- \lambda I) = det(PAP^{-1} - \lambda I) [/tex]


But there are easier ways as HallsofIvy noted.


Start with [tex] Av = \lambda v[/tex] where v is a nonzero vector

then [tex] AP^{-1}Pv = \lambda v [/tex] since [tex] P^{-1}P = I [/tex]

Do you know what to do next?

jbunniii

A and B are similar if and only if they both represent the same linear map, with respect to two possibly different bases. Eigenvalues are defined independently of what basis, if any, you choose. QED.

Mark44

Make that "The characteristic equation of matrix A is [tex] det(A- \lambda I) = 0. [/tex]"
For it to be an equation, it at least has to have an equals sign.

Random Variable

Sorry. What I should have said is that they have the same characteristic POLYNOMIAL.

Jennifer1990

Av = lambda v
(AP^-1 P)v = lambda v
Bv = lambda v

i think?

Jennifer1990

oh wait....B = P^-1 AP .....so what i said is wrong...

how can i manipulate A P^-1 P to look like P^-1 AP?

Jennifer1990

ohhh i see...

since they can have the same eigenvalues, does this mean that the matrices can also have the same eigenvectors?

Random Variable

They can, but it's not likely.

Jennifer1990

I just tried several similar matrices but they all share the same eigenvector O.o
Can i get an example where two similar matrices have different eigenvectors?

HallsofIvy

I'm surprised you were able to find similar matrices that had the same eigenvectors!

[tex]A= \begin{bmatrix}2 & 0 \\ 0 & 3\end{bmatrix}[/tex]
has, obviously, 2 and 3 as eigenvalues with corresponding eigenvectors <1 0> and <0 1>.

[tex]B= \begin{bmatrix}1 & -1 \\ 2 & 4\end{bmatrix}[/tex]
has the same eigenvalues with corresponding eigenvectors <1, -1> and <1, -2>.

All I did was start with the obvious diagonal matrix, A, choose a simple invertible P:
[tex]P= \begin{bmatrix}2 & 1 \\ 1 & 1\end{bmatrix}[/tex]
and calculate [itex]B= P^{-1}AP[/itex].