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TranscendArcu
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Homework Statement
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 thisTranscendArcu said:Homework Statement
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?
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.
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.
TranscendArcu said:T-1λ = 1. And thus,
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: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),
TranscendArcu said:which confirms the relation,
T-1(T(B)) = (1/λ)λ B = 1B =B
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.
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.
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.
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.
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.