Find Eigenvalues & Eigenvectors for Exercise 3 (2), Explained!

In summary, the conversation discusses finding eigenvectors and solving a matrix equation using simple operations. The solution involves finding the determinant of the matrix and solving for possible values of lambda. The conversation also touches on understanding row reduced echelon form and using operations such as subtracting rows and dividing rows to manipulate the matrix. However, it is noted that special treatment is needed in the case where a equals -1.
  • #1
ChiralSuperfields
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Homework Statement
Please see below
Relevant Equations
Please see below
For exercise 3 (2),
1686025959535.png
,
The solution for finding the eigenvector is,
1686025997702.png

However, I am very confused how they got from the first matrix on the left to the one below and what allows them to do that. Can someone please explain in simple terms what happened here?

Many Thanks!
 
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  • #2
You are looking for vectors for which ##Ax=\lambda x##. This is solvable if the determinant of ##A-\lambda I## is zero. Should give you three possible values for ##\lambda## , (different or two or three equal). You then solve ##(A-\lambda I)\vec x=\vec 0## with simple operations like subtracting one row from another.

##\ ##
 
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  • #3
ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

For exercise 3 (2),
View attachment 327486,
The solution for finding the eigenvector is,
View attachment 327487
However, I am very confused how they got from the first matrix on the left to the one below and what allows them to do that. Can someone please explain in simple terms what happened here?

Many Thanks!
When you apply the concept as required; you shall end up with the equation,

##(-1-λ)(λ^2-aλ+λ-a)=0##.

you picked your ##λ=-1## and if you substitute the eigen value back into the matrix, you can confirm that it satisfies the equation.

...

from there, we now have to now deal with the equation, ##(2+2a)x+(1+a)z=0##


 
Last edited:
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  • #4
BvU said:
You are looking for vectors for which ##Ax=\lambda x##. This is solvable if the determinant of ##A-\lambda I## is zero. Should give you three possible values for ##\lambda## , (different or two or three equal). You then solve ##(A-\lambda I)\vec x=\vec 0## with simple operations like subtracting one row from another.

##\ ##
Thank you for your reply @BvU!

Sorry what operations were used to get from,
1686041103340.png

Orange matrix to pink matrix?

Many thanks!
 
  • #5
chwala said:
When you apply the concept as required; you shall end up with the equation,

##(-1-λ)(λ^2-aλ+λ-a)=0##.

you picked your ##λ=-1## and if you substitute the eigen value back into the matrix, you shall end up with,

still posting
Thank you for your reply @chwala!
 
  • #6
ChiralSuperfields said:
Thank you for your reply @BvU!

Sorry what operations were used to get from,
View attachment 327490
Orange matrix to pink matrix?

Many thanks!
...switching of rows combined with understanding of row reduced echelon form...
 
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  • #7
ChiralSuperfields said:
what operations were used to get from orange matrix to pink matrix?
Subtract row 2 from row 3 (gives ##0,0,0##)
Divide row 2 by ##2+2a## (gives ##1,0,{1\over 2}##)
Swap row 1 and 2

:smile:

##\ ##
 
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  • #8
BvU said:
Subtract row 2 from row 3 (gives ##0,0,0##)
Divide row 2 by ##2+2a## (gives ##1,0,{1\over 2}##)
Swap row 1 and 2

:smile:

##\ ##

Note that dividing a row by [itex]2 + 2a[/itex] is only valid if [itex]a \neq -1[/itex]. This case will require separate treatment.
 
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  • #9

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