Eigenvalues of Inverse Transformations

In summary,The relation between T-1 and λ is 1/λ. T-1 is a linear operator and can be diagonalized to obtain the eigenvectors and eigenvalues of T.
  • #1
TranscendArcu
285
0

Homework Statement



Screen_shot_2012_02_24_at_3_34_33_PM.png


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

[itex]λ = \frac{1}{λ'}[/itex].

Is that right?
 
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  • #2
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
So this is a similar sort of problem, so I'll ask about it here:

Screen_shot_2012_02_24_at_4_28_11_PM.png


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:

[itex]\left| \begin{array}{ccc}
1 &2&0 \\
1&2&0 \\
0&0&1 \end{array} \right|[/itex]

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

I have this as equaling,

[itex](1-t)[(1-t)(2-t)-2] = (1-t)(-3t+2t^2)[/itex]. 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 :)
 
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  • #4
You are computing the characteristic polynomial. Once you find that the roots are the eigenvalues. [itex](1-t)[(1-t)(2-t)-2][/itex] is right, the other side isn't. You want to factor it.
 
  • #5
TranscendArcu said:

Homework Statement



Screen_shot_2012_02_24_at_3_34_33_PM.png


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

[itex]λ = \frac{1}{λ'}[/itex].

Is that right?
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 that's 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:
  • #6
Dick said:
You are computing the characteristic polynomial. Once you find that the roots are the eigenvalues. [itex](1-t)[(1-t)(2-t)-2[/itex] is right, the other side isn't. You want to factor it.

Should I write: [itex](1-t)[(1-t)(2-t)-2 = -(t-3)(t-1)(t)[/itex]. 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
kai_sikorski said:
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 that's 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.

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
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
Okay. How about this:

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
TranscendArcu said:
T-1λ = 1. And thus,

:confused:

you have a matrix on one side of the equation and a scalar on the other
 
  • #11
Okay. How about this:

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
TranscendArcu said:
Okay. How about this:

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

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

TranscendArcu said:
which confirms the relation,

T-1(T(B)) = (1/λ)λ B = 1B =B

This is self evident from the definition of the inverse, and does not need to be confirmed.
 
  • #13
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.
Screen_shot_2012_02_26_at_3_59_09_PM.png


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,

[itex]\left| \begin{array}{ccc}
3 &0&-1 \\
-1&1&2 \\
0&0&4 \end{array} \right|[/itex]

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

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.
 

What are eigenvalues of inverse transformations?

Eigenvalues of inverse transformations refer to the values that when multiplied by a particular vector, result in the same vector after the transformation has been applied. In other words, these are the values that remain unchanged under the transformation.

How are eigenvalues of inverse transformations calculated?

Eigenvalues of inverse transformations can be calculated by finding the roots of the characteristic polynomial of the inverse transformation matrix. This polynomial is obtained by subtracting the identity matrix from the inverse transformation matrix and then taking its determinant.

What is the significance of eigenvalues of inverse transformations?

The eigenvalues of inverse transformations provide important information about the transformation. They can indicate whether the transformation will stretch, shrink, or rotate a given vector, and by how much. They also play a crucial role in determining the stability and behavior of a system represented by the transformation matrix.

Can eigenvalues of inverse transformations be complex numbers?

Yes, eigenvalues of inverse transformations can be complex numbers. This typically occurs when the transformation involves rotation or shearing. In such cases, the eigenvalues may have both real and imaginary components, indicating a combination of rotation and stretching in the transformation.

How do eigenvalues of inverse transformations relate to eigenvectors?

Eigenvalues of inverse transformations are closely related to eigenvectors. The eigenvectors of a transformation are the vectors that are transformed only by a scalar multiple, which is equal to the corresponding eigenvalue. In other words, the eigenvectors of a transformation are the vectors that remain unchanged under the transformation, just like the eigenvalues.

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