Question

: Linear Algebra: Let A be a nxn matrix and let B = I - 2A + A^2. Show that if x is an eigenvector of A belonging to an eigenvalue L of A, then x is also an eigenvector of B belonging to an eigenvalue U of B. How are L and U related?

Answer 1

With that set up, Ax = Lx and Bx = Ux

Now calculate (I - 2A + A^2)x and set it equal to Bx

Answer 2

oh yeah i forgot that Ax = Lx.. thanks, i'll try that out.

Answer 3

Got it?

Answer 4

Sorry, but I'm getting nowhere..

Answer 5

All we have to start with are the following three facts:
1. Ax = Lx, L is a scalar
2. Bx = Ux, U is a scalar
3. B = (I - 2A + A^2)

So let's look at using number 3 by calculating
Bx = (I - 2A + A^2)x

Note first that A^2x = A(Ax) = A(Lx) = L Ax = L^2x

Hence
(I - 2A + A^2)x = Ix - 2Ax + A^2x
= x - 2Lx + L^2x
= (1 - 2L + L^2)x

Now as Bx = (I - 2A + A^2)x and Bx = Ux, we have that

Ux = (1 - 2L + L^2)x

As x is an eigenvalue and not zero, the two coefficients here must be equal. Hence

U = 1 - 2L + L^2

Answer 6

Make sense?

Answer 7

Wait, "As x is an eigenvalue.." -> was that a typo and did you mean "x is an eigenvector"?

Answer 8

yes, x is an eigenVECTOR with eigenvalue L

Answer 9

Okay thanks, and how exactly in the proof was "x also an eigenvector of B belonging to an eigenvalue U" proven?

Answer 10

We never proved that Bx = Ux; that was a given. But we used it in the proof in order to write Bx = Ux here:

Bx = (I - 2A + A^2)x
=> Ux = (expression in L)x

Answer 11

Sorry, let me rephrase that I was ambiguous haha. The problem said that if x is an eigenvector of A corresponding to eigenvalue L (i.e. Ax=Lx), prove that that same x is also an eigenvector of B corresponding to eigenvalue U (i.e. Bx=Ux). I'm wondering if that was already implied in your proof or what?

Answer 12

Oh, sorry, I misread the question.

Nonetheless, using the same calculation

Bx = (1 - 2L + L^2) x

and hence x is an eigenvector.

Answer 13

Associate with that eigenvector of B the eigenvalue U, then Bx = Ux and hence

Ux = (1 - 2L + L^2)x

and therefre U = 1 - 2L + L^2

Answer 14

Bx = (1 - 2L + L^2) x ->how does this prove that x here is the same x as Ax=Lx though?

Answer 15

Because x is not an arbitrary vector. It is a vector with the property that Ax = Lx

Answer 16

If that were not the case, we could have never shown that
(I - 2A + A^2)x = (1 - 2L + L^2)x

Answer 17

Okay, I'll try to answer the question myself, please tell me if this proof is correct and complete:

We actually start with only these facts:
1. Ax = Lx
2. B = I - 2A + A^2

The problem asks first to prove that if x is an eigenvector of A corresponding to eigenvalue L, then that same x is also an eigenvector of B corresponding to U.

So we try to prove that Bx = Ux is true, where x is the same x in Ax = Lx
Bx = Ux
(I - 2A + A^2)x = Ux since B = I - 2A + A^2
(I - 2A + AA)x = Ux
(I - 2L + L^2)x = Ux since Ax=Lx
(I - 2L + L^2) = U since x is an eigenvector and not zero.

Therefore Bx=Ux is true and the relationship between L and U is that (I - 2L + L^2) = U

Answer 18

Almost. You can't write where you do that Bx = Ux because you don't know yet that x is an eigenvector of B.

Instead just begin the calculation:

Bx = (I - 2A + A^2)x
= (1 - 2L + L^2)x

NOW, because Bx = (scalar)x, x must also be an eigenvector of B. Call the eigenvalue of B associated with this eigenvector, U. Then Bx = Ux. Then

Ux = (1 - 2L + L^2)x and hence ....

Answer 19

THANK YOU!! Got it!! I always get confused with how to logically write proofs haha.

Answer 20

It takes practice.

Answer 21

Note also in your proof above you've used "I" for both the identity matrix and for the number 1. They are of course different. Just be careful with that.

Answer 22

oh yes thank you.

Answer 23

Can you please help me in the next part of the question? "Show that if L = 1 is an eigenvalue of A, then the matrix B will be singular."

Answer 24

What I've done so far:

U = 1 - 2L + L^2
U = 1 - 2(1) + (1)^2
U = -2

det(B) = det(UI)
det(B) = det((-2,0,0),(0,-2,0),(0,0,-2))
det(B) = -8 != 0, which just shows otherwise.

What did I do wrong here?

Answer 25

Note that \( U = 1 - 2(1) + (1)^2 = 1 - 2 + 1 = 0 \)

Answer 26

oh wow i messed up in the arithmetic.. thanks. I'm not thinking right anymore haha.

Answer 27

;-)