Finding a vector A from given eigen values and eigenvectors

[sg001]

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??

 

[tiny-tim]

hi sg001!

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 …

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

 

[sg001]

hi sg001!


no, that can't be right …

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?

 

[tiny-tim]

you mean 1/3[11,7] ?

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

 

[sg001]

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.

 

[tiny-tim]

hi sg001!

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 )

the important step is the bit in bold ​

 

[sg001]

hi sg001!

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 )

the important step is the bit in bold ​

cool thanks