Finding a vector A from given eigen values and eigenvectors

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Homework Help Overview

The problem involves a matrix A with given eigenvectors [2,1] and [1,-1], along with their corresponding eigenvalues 2 and -3. The task is to determine the product of matrix A with a vector b = [1,1].

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss expressing vector b as a linear combination of the eigenvectors. There is uncertainty about the coefficients used in this combination, particularly regarding the factor associated with the second eigenvalue.

Discussion Status

The discussion includes various interpretations of the eigenvalue application and the resulting vector. Some participants express confusion about the correctness of the proposed solution and the factors used in the linear combination. There is no explicit consensus on the final outcome, but participants are exploring different approaches to the problem.

Contextual Notes

There are indications of potential errors in the problem setup or the provided solution, as participants question the eigenvalue associated with the second eigenvector. The conversation reflects a mix of attempts to clarify the problem and verify the calculations involved.

sg001
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Homework Statement



A matrix A has eigenvectors [2,1] [1,-1]
and eigenvalues 2 , -3 respectively.

Determine Ab for the vector b = [1,1].


Homework Equations





The Attempt at a Solution



First I put be as a combination of the two eigenvectors
ie

2/3[2,1] -1/3[1,-1] = b

so A(2/3[2,1] -1/3[1,-1]) = Ab

but not sure what to do from this point as the sltn says it went from this

A(2/3[2,1] -1/3[1,-1]) = (2[2,1] -1/3[1,-1]) but I am not sure how??
 
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hi sg001! :smile:
sg001 said:
A matrix A has eigenvectors [2,1] [1,-1]
and eigenvalues 2 , -3 respectively.

A(2/3[2,1] -1/3[1,-1]) = (2[2,1] -1/3[1,-1]) but I am not sure how??

no, that can't be right :redface:

if the second eigenvalue is -3, that factor must be -3, not -1/3
 
tiny-tim said:
hi sg001! :smile:


no, that can't be right :redface:

if the second eigenvalue is -3, that factor must be -3, not -1/3

hmm

they have the answer of Ab = 1/3[11,1] ?

is this correct?
 
you mean 1/3[11,7] ?

then the question must be wrong, the eigenvalue must be -1/3
 
tiny-tim said:
you mean 1/3[11,7] ?

then the question must be wrong, the eigenvalue must be -1/3

no, that's the sltn and the exact question they give,,,

so how would you approach this type of question then..

rewrite b as a liner combination of the given eigenvalues... then how do you solve for A from here...

2/3[2,1] -1/3[1,-1] = b

just so I know if it comes up in my test.

Thanks for the help.
 
hi sg001! :smile:

you want to find A[1,1].

start with [1,1] = 2/3[2,1] -1/3[1,-1]

so A[1,1] = 2/3 A[2,1] -1/3 A[1,-1]

= 2/3 2[2,1] -1/3 -3[1,-1]


= [11/3,1/3] …

oh that is right!

(i shouldn't have tried doing it in my head :redface:)

the important step is the bit in bold :wink:
 
tiny-tim said:
hi sg001! :smile:

you want to find A[1,1].

start with [1,1] = 2/3[2,1] -1/3[1,-1]

so A[1,1] = 2/3 A[2,1] -1/3 A[1,-1]

= 2/3 2[2,1] -1/3 -3[1,-1]


= [11/3,1/3] …

oh that is right!

(i shouldn't have tried doing it in my head :redface:)

the important step is the bit in bold :wink:

cool thanks
 

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