Eigenvalues and eigenvectors

[Jennifer1990]

Homework Statement

Let A and B be similar matrices
a)Prove that A and B have the same eigenvalues

Homework Equations

None

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 $Av= \lambda v$ then, for any invertible P, $P^{-1}Av= \lambda P^{-1}v$. Now define $u= P^{-1}v$.

 

[Random Variable]

Oh, I think that way is much too complicated!

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

Your way is too easy.

 

[Random Variable]

or let $$Av = \lambda v$$

then $$AP^{-1}Pv = \lambda v$$ and go from there

 

[Jennifer1990]

what do u mean by the same characteristic equation?

 

[Random Variable]

what do u mean by the same characteristic equation?

EDIT: changed "equation" to "polynomial"


The the characteristic polynomial of matrix A is $$det(A- \lambda I).$$

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

so show that $$det(A- \lambda I) = det(PAP^{-1} - \lambda I)$$


But there are easier ways as HallsofIvy noted.


Start with $$Av = \lambda v$$ where v is a nonzero vector

then $$AP^{-1}Pv = \lambda v$$ since $$P^{-1}P = I$$

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]

The the characteristic equation of matrix A is $$det(A- \lambda I).$$

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

 

[Random Variable]

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

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]

ohhh i see...

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

They can, but it's not likely.

 

[Jennifer1990]

I just tried several similar matrices but they all share the same eigenvector
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!

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

$$B= \begin{bmatrix}1 & -1 \\ 2 & 4\end{bmatrix}$$
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:
$$P= \begin{bmatrix}2 & 1 \\ 1 & 1\end{bmatrix}$$
and calculate $B= P^{-1}AP$.