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

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For exercise 3 (2),

,
The solution for finding the eigenvector is,

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!

 

[BvU]

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.

$\$

 

[chwala]

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, $(2+2a)x+(1+a)z=0$

 

[ChiralSuperfields]

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,

Orange matrix to pink matrix?

Many thanks!

 

[ChiralSuperfields]

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!

 

[chwala]

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...

 

[BvU]

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



$\$

 

[pasmith]

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



$\$

Note that dividing a row by $2 + 2a$ is only valid if $a \neq -1$. This case will require separate treatment.

 

[ChiralSuperfields]

Thank you for your replies @chwala, @BvU, and @pasmith!

I understand now :)