- #1

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## 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

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- Thread starter Jennifer1990
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- #1

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Let A and B be similar matrices

a)Prove that A and B have the same eigenvalues

None

Firstly, i dont see how this can even be possible unless the matrices are exactly the same :S

- #2

matt grime

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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.

- #3

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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.

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

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- #4

HallsofIvy

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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].

- #5

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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.

- #6

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

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

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

- #7

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what do u mean by the same characteristic equation?

- #8

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what do u mean by the same characteristic equation?

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?

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- #9

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- #10

Mark44

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Make that "The characteristic equation of matrix A is [tex] det(A- \lambda I) = 0. [/tex]"The the characteristic equation of matrix A is [tex] det(A- \lambda I). [/tex]

For it to be an equation, it at least has to have an equals sign.

- #11

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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.

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

- #12

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Av = lambda v

(AP^-1 P)v = lambda v

Bv = lambda v

i think?

(AP^-1 P)v = lambda v

Bv = lambda v

i think?

- #13

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

- #14

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

- #15

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They can, but it's not likely.

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

- #16

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Can i get an example where two similar matrices have different eigenvectors?

- #17

HallsofIvy

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[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].

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