Linear Algebra: Similar matrices *Prove*

In summary: Isn't that "ah ha!" moment great?In summary, the student is trying to find a way to prove that P^{-1}x is the eigenvector of B. However, they are not able to do it using the equations that they have so far.
  • #1
tua96426
7
0

Homework Statement


if A is similar to B with P^-1 A P = B and x is an eigenvector of A, prove that P^-1x is an eigenvector of B


Homework Equations





The Attempt at a Solution



so far I did this:

A = PBP^-1
Ax = PBP^-1x
P^-1Ax = BP^-1x
B^-1P^-1Ax = P^-1x

is that sufficient to prove that P^-1x is the eigenvector of B? ??

I don't think so .. if yes please explain how.. or else please give a proper "proof" .

I guess the reason I don't think this is right is becaues I am not able to bring it to the format of Ax = tx where t is an eigenvalue and x is the eigenvector. Any help would be appreciated.


Final exam on tuesday ... :(
 
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  • #2
tua96426 said:

Homework Statement


if A is similar to B with P^-1 A P = B and x is an eigenvector of A, prove that P^-1x is an eigenvector of B

so far I did this:

A = PBP^-1
Ax = PBP^-1x
P^-1Ax = BP^-1x
B^-1P^-1Ax = P^-1x

is that sufficient to prove that P^-1x is the eigenvector of B? ??

I don't think so .. if yes please explain how.. or else please give a proper "proof" .

No, but you're on the right track. You got as far as

[tex]P^{-1}Ax = BP^{-1}x[/tex]

Now you need to use the fact that x is an eigenvector of A, so you can replace Ax in this equation with something else, and then you're essentially done.
 
  • #3
jbunniii said:
No, but you're on the right track. You got as far as

[tex]P^{-1}Ax = BP^{-1}x[/tex]

Now you need to use the fact that x is an eigenvector of A, so you can replace Ax in this equation with something else, and then you're essentially done.

so that would mean:
Ax = tx Where t is the eigenvalue:

so [tex]P^{-1}tx = BP^{-1}x[/tex]
[tex] tx = PBP^{-1}x [/tex]
and we know [tex]PBP^{-1} [/tex] is A
so we essentially have [tex]tx = Ax [/tex]

but how does this show that [tex] P^{-1}x [/tex] is the eigenvector of B?
 
  • #4
tua96426 said:
so that would mean:
Ax = tx Where t is the eigenvalue:

so [tex]P^{-1}tx = BP^{-1}x[/tex]
[tex] tx = PBP^{-1}x [/tex]
and we know [tex]PBP^{-1} [/tex] is A
so we essentially have [tex]tx = Ax [/tex]

but how does this show that [tex] P^{-1}x [/tex] is the eigenvector of B?

You can see it from the first equation that you wrote:

[tex]P^{-1}tx = BP^{-1}x[/tex]

Rearrange the left hand slightly, then add some parentheses in appropriate places on both sides, until you see something that looks like an eigenvalue/eigenvector relationship. Notice that this shows something interesting beyond what was asked for: in particular, compare the eigenVALUE of B corresponding to [tex]P^{-1}x[/tex] to the eigenvalue of A corresponding to x.
 
  • #5
jbunniii said:
You can see it from the first equation that you wrote:

[tex]P^{-1}tx = BP^{-1}x[/tex]

Rearrange the left hand slightly, then add some parentheses in appropriate places on both sides, until you see something that looks like an eigenvalue/eigenvector relationship.

I SEE IT!

[tex] t(P^{-1}x) = B(P^{-1}x) [/tex]

cool!
Thanks a lot.
 
  • #6
tua96426 said:
I SEE IT!

[tex] t(P^{-1}x) = B(P^{-1}x) [/tex]

cool!
Thanks a lot.

Isn't that "ah ha!" moment great?

Cheers!
 
  • #7
absolutely!
 

1. What are similar matrices in linear algebra?

In linear algebra, two matrices are said to be similar if they have the same size and shape, and their corresponding entries satisfy a specific relationship. This relationship is that one matrix can be transformed into the other using a combination of elementary row and column operations.

2. How do you prove that two matrices are similar?

To prove that two matrices are similar, we need to show that they have the same size and shape, and their corresponding entries satisfy the relationship mentioned above. This can be done by performing row and column operations on one matrix to transform it into the other.

3. What is the significance of similar matrices in linear algebra?

Similar matrices are important in linear algebra because they represent the same linear transformation in different coordinate systems. This allows us to simplify calculations and make it easier to understand and analyze complex systems.

4. Can two matrices be similar if they have different eigenvalues?

No, two matrices cannot be similar if they have different eigenvalues. This is because the eigenvalues of a matrix are invariant under similarity transformations. In other words, similar matrices have the same eigenvalues.

5. Are similar matrices always invertible?

Yes, similar matrices are always invertible. This is because if two matrices are similar, they have the same determinant and hence, the same rank. A matrix is invertible if and only if its determinant is non-zero, so similar matrices are also invertible.

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