Can anyone explain to me why eigenvector here is like this

[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?

 

[mfb]

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

 

[sozener1]

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??

 

[Mark44]

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.

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.

 

[CellCoree]

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.