Matrices, eigenvalues, invertibility

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Homework Help Overview

The discussion revolves around the conditions under which a 2x2 matrix \( A \) can be transformed into a diagonal matrix \( D \) with eigenvalues 2 and 3 through the use of an invertible matrix \( S \). Participants explore the relationship between the eigenvalues of \( A \) and the existence of \( S \).

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of \( A \) having eigenvalues 2 and 3, and whether these eigenvalues guarantee the linear independence of the corresponding eigenvectors. There is an exploration of the conditions under which \( S \) exists and how this relates to the properties of matrix \( A \).

Discussion Status

Some participants express confidence in the original poster's reasoning regarding the eigenvalues of \( A \). However, there is an ongoing inquiry into the conditions that ensure the existence of \( S \) and the implications of eigenvalue multiplicity on eigenvector independence.

Contextual Notes

Participants note that the problem requires identifying a specific subset of matrices \( A \) for which \( S \) exists, rather than assuming it exists for all matrices. There is also a reference to the need for further detail regarding the relationship between distinct eigenvalues and the linear independence of eigenvectors.

blue4123
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Homework Statement


For which ##2x2## matrices ##A## does there exist an invertible matrix
##S## such that ##AS=SD##, where
##D=
\begin{bmatrix}
2 & 0\\
0 & 3
\end{bmatrix}
##?
Give your answers in terms of the eigenvalues of ##A##.

Homework Equations


##A\lambda=\lambda\vec{v}##

The Attempt at a Solution


S is a ##2x2## matrix. If x and y are the 1st and second columns of A, then we can write the following:
Ax=2x
Ay=3y

So we can see from these two equations that 2 and 3 must be the corresponding eigenvalues of A and their respective eigenvectors must be linearly independent.I was hoping someone could critique my solution attempt as I am not too confident that I fully answered the question or if there's a better way to express my answer
 
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hi there, welcome to physicsforums :)
I think you meant to say that x and y are the first and second columns of S. I agree with what you're saying. But you haven't shown in what cases does S exist. The problem was to show when S exists. It will not exist for all choices of matrix A, but only a special subset.

Edit: ah, wait, sorry, the problem is to show for what kinds of matrix A, does S exist, such that D is diagonal with values 3,2. OK, I think you pretty much have the answer. You've shown that if S exists, A must have eigenvalues 3 and 2.
 
Last edited:
yeah. sorry for my last post, now I have read your work through again, I think you have answered correctly. The problem is to find the set of all matrices A such that S exists and such that D is diagonal with values 2,3. So to enumerate this set, it is OK to assume S exists, and then find out what this implies for the matrix A. You found that A must have eigenvalues 2,3. So then the last bit of the problem would be to say that all matrices with eigenvalues 2,3 must be able to be diagonalized. And this is what you were implying when you said that the respective eigenvectors of A must be linearly independent.

So maybe you could write a little bit more detail about why if a matrix has eigenvalues 2,3, it must then have linearly independent eigenvectors. Apart from that, I think you have done everything you need for this problem.
 
BruceW said:
hi there, welcome to physicsforums :)
I think you meant to say that x and y are the first and second columns of S. I agree with what you're saying. But you haven't shown in what cases does S exist. The problem was to show when S exists. It will not exist for all choices of matrix A, but only a special subset.

Edit: ah, wait, sorry, the problem is to show for what kinds of matrix A, does S exist, such that D is diagonal with values 3,2. OK, I think you pretty much have the answer. You've shown that if S exists, A must have eigenvalues 3 and 2.
BruceW said:
hi there, welcome to physicsforums :)
I think you meant to say that x and y are the first and second columns of S. I agree with what you're saying. But you haven't shown in what cases does S exist. The problem was to show when S exists. It will not exist for all choices of matrix A, but only a special subset.

Edit: ah, wait, sorry, the problem is to show for what kinds of matrix A, does S exist, such that D is diagonal with values 3,2. OK, I think you pretty much have the answer. You've shown that if S exists, A must have eigenvalues 3 and 2.
Oh sorry, I did mean that x and y are the first and second columns of S, and not A.
 
BruceW said:
yeah. sorry for my last post, now I have read your work through again, I think you have answered correctly. The problem is to find the set of all matrices A such that S exists and such that D is diagonal with values 2,3. So to enumerate this set, it is OK to assume S exists, and then find out what this implies for the matrix A. You found that A must have eigenvalues 2,3. So then the last bit of the problem would be to say that all matrices with eigenvalues 2,3 must be able to be diagonalized. And this is what you were implying when you said that the respective eigenvectors of A must be linearly independent.

So maybe you could write a little bit more detail about why if a matrix has eigenvalues 2,3, it must then have linearly independent eigenvectors. Apart from that, I think you have done everything you need for this problem.
That is the part I was a bit unsure of. I found that A must have eigenvalues 2,3. Are their corresponding eigenvectors always linearly independent?
 
blue4123 said:
That is the part I was a bit unsure of. I found that A must have eigenvalues 2,3. Are their corresponding eigenvectors always linearly independent?

Perhaps a theorem may help:

Suppose ##A \in M_{n \times n}(\mathbb{C})##. If ##A## has ##n## distinct eigenvalues, then any set of ##n## corresponding eigenvectors form a basis for ##\mathbb{C}^n##.

Of course a basis is any nonsingular spanning set.

Think about what this means in terms of scalars. You can create any basis you desire, and all of the vectors inside of it will correspond to the eigenvalues, regardless of being scaled.
 

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