Eigenvalues and eigenvectors

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.

 

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.

 

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].

 

Your way is too easy.

 

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

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

 

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?

 

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.

 

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

 

They can, but it's not likely.

 

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].