# How to find one corresponding eigenvector?

• Math100
In summary, the determinant being 0 means that A-4I has a nullspace and there is an eigenvector with eigenvalue 4. The answer in the textbook for this problem is [1, 1, -1], but the process for finding the corresponding eigenvector is not explained. The eigenvector can be found by solving the equation (A-4I)x=0, which will result in multiple solutions due to the nature of linear transformations. One way to find a specific eigenvector is to arbitrarily assign a value to one component and solve for the others.
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.

Math100 said:
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?

DaveE
How should I find the vector then?

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.

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.

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##?

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

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.

## 1. What is an eigenvector?

An eigenvector is a vector that, when multiplied by a matrix, results in a scalar multiple of itself. In other words, the direction of the eigenvector remains unchanged after the matrix transformation.

## 2. Why is it important to find corresponding eigenvectors?

Corresponding eigenvectors are important because they provide insight into the behavior of a matrix transformation. They can help determine the direction and magnitude of the transformation, and are useful in solving systems of linear equations.

## 3. How do you find corresponding eigenvectors?

To find corresponding eigenvectors, you must first find the eigenvalues of the matrix. Then, for each eigenvalue, you can solve the characteristic equation to find the corresponding eigenvector.

## 4. Can a matrix have more than one corresponding eigenvector?

Yes, a matrix can have multiple corresponding eigenvectors for a single eigenvalue. This is because there are infinitely many vectors that can be scaled by the eigenvalue and still remain unchanged after the matrix transformation.

## 5. Are corresponding eigenvectors unique?

No, corresponding eigenvectors are not unique. As mentioned before, there are infinitely many vectors that can be scaled by the eigenvalue and still remain unchanged after the matrix transformation. However, they must all be in the same direction.

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