# Can anyone explain to me why eigenvector here is like this

• sozener1
In summary: So, in summary, you need to recheck your calculations and make sure you are using the correct matrix and values to find the eigenvectors for the given eigenvalues.
sozener1
Im supposed to find the eigenvectors and eigenvalues of A

I found that eigenvalues are 2 12 and -6

then I found eigen vectors substituting -6 to lambda

and someone has told me I get 0 0 1 for eigenvector which I cannot understand why??

can anyone pleasezzzzzzzz explain why this is?

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Did you multiply this vector with your matrix? What do you get?
Does this fit to the definition of an eigenvector?

mfb said:
Did you multiply this vector with your matrix? What do you get?
Does this fit to the definition of an eigenvector?

I get 0 0 0

it doesn't fit with the definition of eigenvectors

so does that mean it should be 0 0 0 instead of 0 0 1??

I mean when you try to calculate for v3 = [ v1 v2 v3]

since the last row of the matrix is 0 0 0 should v3 come out as 0?? couldn't v3 be any number??

sozener1 said:
I get 0 0 0
That's not what I get.

The matrix you showed in the second attachment is not A. It is [A - (-6)I]. Of course if you multiply this matrix times your eigenvector, you'll get the zero vector.
sozener1 said:
it doesn't fit with the definition of eigenvectors

so does that mean it should be 0 0 0 instead of 0 0 1??

I mean when you try to calculate for v3 = [ v1 v2 v3]

since the last row of the matrix is 0 0 0 should v3 come out as 0?? couldn't v3 be any number??

Assuming that your eigenvalue is λ and that x is an as-yet unknown eigenvector for λ, what you're doing is solving the equation Ax = λx for x. That's equivalent to solving the equation (A - λI)x = 0. In other words, of finding the kernel of the matrix A - λI. This should be something that you have already learned to do.

The kernel here should not consist of only the zero vector - an eigenvector cannot be the zero vector.

Sure, I would be happy to explain why the eigenvector for the eigenvalue -6 is [0 0 1].

First, let's review the definition of eigenvectors and eigenvalues. Eigenvectors are special vectors that do not change direction when multiplied by a matrix. In other words, when we multiply a matrix A by an eigenvector v, the result is a scalar multiple of v. This scalar multiple is called the eigenvalue, denoted by λ. Mathematically, we can write this as Av = λv.

Now, let's look at your specific example. You have a matrix A and have found the eigenvalues to be 2, 12, and -6. This means that when we multiply A by an eigenvector v, we will get either 2v, 12v, or -6v.

Let's focus on the eigenvalue -6. When we substitute this into the equation Av = λv, we get A(v) = (-6)v. In other words, when we multiply A by the eigenvector v, we will get a vector that is a scalar multiple of v, with the scalar being -6.

Now, let's think about what this means for the eigenvector v. Since A(v) = (-6)v, this means that v is a vector that is not changed in direction when multiplied by A, but is scaled by -6. In other words, v points in the same direction as the vector (-6)v, but is just a different length.

So, to find the eigenvector for the eigenvalue -6, we just need to find any vector that is unchanged in direction when multiplied by A, and then scale it by -6. This can be any vector that is perpendicular to the other two eigenvectors (since it cannot be a scalar multiple of them), and has a magnitude of 1/6. And since there are infinite vectors that fit this criteria, we can choose the simplest one, which is [0 0 1].

I hope this explanation helps to clarify why the eigenvector for the eigenvalue -6 is [0 0 1]. Let me know if you have any further questions or need more clarification.

## 1. What is an eigenvector?

An eigenvector is a vector that does not change its direction when a linear transformation is applied to it. In other words, it is a special vector that is only scaled by the transformation, but its direction remains the same.

## 2. Why is the eigenvector important in mathematics?

Eigenvectors are important because they help us understand the behavior of linear transformations. They also have various applications in fields such as physics, engineering, and computer science.

## 3. How do you find eigenvectors?

To find eigenvectors, we first need to find the eigenvalues of a matrix. Then, we can use these eigenvalues to solve for the corresponding eigenvectors by setting up and solving a system of linear equations.

## 4. Can you explain the relationship between eigenvectors and eigenvalues?

Eigenvectors and eigenvalues are closely related, as eigenvalues represent the scaling factor of an eigenvector. In other words, an eigenvector is an eigenvector of a matrix if it is scaled by its corresponding eigenvalue when the matrix is applied to it.

## 5. How is the concept of eigenvectors used in real-world applications?

Eigenvectors have various real-world applications, such as in image and signal processing, data compression, and machine learning. They are also used in physics to study the behavior of quantum mechanical systems.

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