How to find one corresponding eigenvector?

[Math100]

Homework Statement:

Is lambda=4 an eigenvalue of the given matrix? If so, find one corresponding eigenvector.

Relevant Equations:

None.

The determinant is 0, which means that A-4I has a nullspace, and there is an eigenvector with eigenvalue 4. In the textbook, the answer says "Yes, [1, 1, -1]" for this problem. But I don't know how to find the corresponding eigenvector for this problem. Below is my work.

 

[fresh_42]

Homework Statement:: Is lambda=4 an eigenvalue of the given matrix? If so, find one corresponding eigenvector.
Relevant Equations:: None.

The determinant is 0, which means that A-4I has a nullspace, and there is an eigenvector with eigenvalue 4. In the textbook, the answer says "Yes, [1, 1, -1]" for this problem. But I don't know how to find the corresponding eigenvector for this problem. Below is my work.

You have to find a vector $\vec{v}\neq 0$ such that $(A-4I)\vec{v}=0$. Can you solve this equation system?

 

[Math100]

How should I find the vector then?

 

[fresh_42]

Solve $\begin{bmatrix}-1&0&-1\\2&-1&1\\-3&4&1\end{bmatrix}\cdot \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$ for $x,y,z$. Then you get one or more solutions which are eigenvectors to $4$. I don't know the other eigenvalues though.

 

[DaveE]

I think an easy way to remember these equations is to memorize the concept that the eigenvectors of A are the vectors that "don't change direction" when transformed by the matrix A, they only change in length. That length change is the eigenvalue for that vector. So Aν = λν. All of the equations you need (see @fresh_42) follow from this concept with some matrix algebra.

 

[Math100]

But I didn't get the answer from the book. As a result, I got x=y=z=0, the book says x=1, y=1, z=-1.

 

[fresh_42]

You get $x=-z$ from the first row. This makes $-2z-y+z==-y-z=0$ the second row and thus $y=-z$. Do they solve the third equation, too? And if so, how do we handle $z$?

 

[Infrared]

@Math100 Eigenvectors are by definition nonzero. You are looking for a solution to $(A-4I)x=0$ other than the trivial solution $x=0.$

 

[DaveE]

You will find, when solving for the eigenvectors, that there isn't a single (non-trivial, ν=0) solution. This is because all vectors in the same direction as an eigenvector satisfy the basic requirement that Aν=λν, since these are linear transformations ( A(kν)=kAν for any scalar k). So, we would normally arbitrarily assign a value to one component. Like x=1, for example, and then solve for the others (everyone always chooses x=1, it seems). You could express it as (1,1,-1)^T, (2,2,-2)^T, etc., or as (x,x,-x)^T (which, of course, is x⋅(1,1,-1)^T). These are all solutions.