# Eigenvalues of Inverse Transformations

1. Feb 24, 2012

### TranscendArcu

1. The problem statement, all variables and given/known data

3. The attempt at a solution
So I observed:

T(B) = λB
T-1(B) = λ'B

Also,

T-1(T(B)) = λ'λB = B

This implies,

λ'λ = 1

And so, there should be a relation

$λ = \frac{1}{λ'}$.

Is that right?

2. Feb 24, 2012

### M Quack

Looks OK to me. Note that this is quite obvious if T is diagonal, which can be achieved by a coordinate transformation such that the new coordinates are along the Eigenvectors.

3. Feb 24, 2012

### TranscendArcu

So this is a similar sort of problem, so I'll ask about it here:

T(1,0,0) = (1,1,0)
T(0,1,0) = (2,2,0)
T(0,0,1) = (0,0,1)

Thus, the matrix relative to the standard basis is:

$\left| \begin{array}{ccc} 1 &2&0 \\ 1&2&0 \\ 0&0&1 \end{array} \right|$

$Δ_T (t) = det(\left| \begin{array}{ccc} 1 &2&0 \\ 1&2&0 \\ 0&0&1 \end{array} \right| - tI)$

I have this as equaling,

$(1-t)[(1-t)(2-t)-2] = (1-t)(-3t+2t^2)$. First of all, I don't know if this is right. If it is right, I don't know how to interpret it.

Of course, I know I also have to do the other parts of the problem, but I'll get there eventually :)

Last edited: Feb 24, 2012
4. Feb 24, 2012

### Dick

You are computing the characteristic polynomial. Once you find that the roots are the eigenvalues. $(1-t)[(1-t)(2-t)-2]$ is right, the other side isn't. You want to factor it.

5. Feb 24, 2012

### kai_sikorski

Your conclusion is correct but I'm not sure I agree with your proof. In particular this
T-1(B) = λ'B
is only true if B is an eigenvector of T, and thats not really clear a priori. Just say that B is an eigenvector of T so

T(B) = λB

Apply T-1 to both sides
T-1(T(B)) = T-1(λB)

T is a linear operator (I assume) ... I'll let you finish the argument. It's pretty trivial.

Also as to the diagonalization comment, while I agree that's a good way to think about it intuitively it's not a good technique for a proof. Obviously there are issues with some finite dimensional space maps being non-diagonalizible; but even more fundamentally for a Hilbert space I think you would need the operator to be compact to have any hope of such an approach working.

Last edited: Feb 24, 2012
6. Feb 26, 2012

### TranscendArcu

Should I write: $(1-t)[(1-t)(2-t)-2 = -(t-3)(t-1)(t)$. This is the characteristic polynomial. Thus, the roots are 3,1,0. These are the eigenvalues. If I have equations,

(1-t)x + 2y = 0
1x + (2-t)y = 0
(1-t)z = 0,

and I plug in for t=0,1,3, I find for t=3 that eigenvectors are multiples of (1,1,0). For t=1, eigenvectors are multiples of (0,0,1). For t=0, eigenvectors are multiples of (-2,1,0). The matrix is diagonalizable because T has three linearly indep. eigenvectors.

Because these vectors are linearly independent, and because the number of vectors = dim(R3), these vectors span R3. Thus, R3 is the eigenspace of T (???)

How does that look?

7. Feb 26, 2012

### TranscendArcu

Hmm. What's the problem with writing:

T-1(T(B)) = T-1(λB) = λ'λB = B

And then arriving at an identical conclusion?

8. Feb 26, 2012

### kai_sikorski

T-1(λB) = λ'λB

This is only true if B is an eigenvector of T-1, (sorry guess I forgot the exponent in my first post) and I don't think that's clear ahead of time.

9. Feb 26, 2012

### TranscendArcu

T-1(T(B)) = T-1(λB) = B. This implies,

T-1λ = 1. And thus,
T-1 = 1/λ,

which confirms the relation,

T-1(T(B)) = (1/λ)λ B = 1B =B

It occurs to me that this isn't very good notation. But is the idea here right? Or am I still off?

10. Feb 26, 2012

### kai_sikorski

you have a matrix on one side of the equation and a scalar on the other

11. Feb 26, 2012

### TranscendArcu

T-1(T(B)) = T-1(λB) = B. This implies,

T-1λ(B) = 1(B). And thus,
T-1(B) = 1/λ (B),

which confirms the relation,

T-1(T(B)) = (1/λ)λ B = 1B =B

12. Feb 26, 2012

### kai_sikorski

You're done here. This line means exactly that B is in fact an eigenvector of T-1 and the eigenvalue is 1/λ.

This is self evident from the definition of the inverse, and does not need to be confirmed.

13. Feb 26, 2012

### TranscendArcu

This is a problem nearly identical to the 2nd problem in this thread. Since I'm still kind of sure about the material, I'd like my work checked here as well.

T(1,0,0) = (3,-1,0)
T(0,1,0) = (0,1,0)
T(0,0,1) = (-1,2,4)

Thus, we have the matrix,

$\left| \begin{array}{ccc} 3 &0&-1 \\ -1&1&2 \\ 0&0&4 \end{array} \right|$

$Δ_T (t) = det( \left| \begin{array}{ccc} 3 &0&-1 \\ -1&1&2 \\ 0&0&4 \end{array} \right| - tI)$

I have this equaling: -(t-4)(t-3)(t-1), which is the characteristic polynomial. The roots are the eigenvalues, which are 4,3,1.

To compute the eigenvectors:

When t=4, we have,
-x-z=0
-x-3y+2z=0
0z=0

Which implies that eigenvectors are multiples of (-1,1,1).

When t=3, we have,
-z=0
-x-2y=0
z=0

Which implies that eigenvectors are multiples of (-1,2,0)

When t=1, we have,
2x-z=0
-x+2z=0
3z=0

Which implies that eigenvectors are multiples of (0,1,0).

T is diagonizable because (-1,1,1),(-1,2,0),(0,1,0) are lin. indep.

I'm still unsure if the eigenspace is the span of the eigenvectors. If so, I guess R3, otherwise, I'm not sure.