Eigenspace and basis of eigenvectors

[Locoism]

Homework Statement

Given the matrix
0 1 0
0 0 1
-3 -7 -5

Find the eigenspaces for the various eigenvalues
Prove that there cannot be a basis of R3 consisting entirely of eigenvectors of A

Homework Equations

The Attempt at a Solution

The eigenvectors are not a problem, I end up with (λ+3)(λ+1)² so my eigenvalues are -3 and -1. Substituting in I get [1, -3, 9] and [1, -1, 1]. Now would the eigenspace simply be {[1, -3, 9], [1, -1, 1], [0, 0, 0]} or am I missing some other step?

Also, how can I prove there cannot be a basis of R3 consisting of eigenvectors of A? Could it just be because there must be n vectors in a basis of Rn?

 

[Mark44]

Homework Statement

Given the matrix
0 1 0
0 0 0
-3 -7 -5

Find the eigenspaces for the various eigenvalues
Prove that there cannot be a basis of R3 consisting entirely of eigenvectors of A

Homework Equations

The Attempt at a Solution

The eigenvectors are not a problem, I end up with (λ+3)(λ+1)² so my eigenvalues are -3 and -1. Substituting in I get [1, -3, 9] and [1, -1, 1]. Now would the eigenspace simply be {[1, -3, 9], [1, -1, 1], [0, 0, 0]} or am I missing some other step?

Don't include the zero vector. It can't be a vector in a basis.

Also, how can I prove there cannot be a basis of R3 consisting of eigenvectors of A? Could it just be because there must be n vectors in a basis of Rn?

For R³, any basis must contain 3 linearly independent vectors. Since your basis contains only two vectors, these two vectors could not possibly span R³, and so aren't a basis for R³.

 

[Dick]

Homework Statement

Given the matrix
0 1 0
0 0 0
-3 -7 -5

Find the eigenspaces for the various eigenvalues
Prove that there cannot be a basis of R3 consisting entirely of eigenvectors of A

Homework Equations

The Attempt at a Solution

The eigenvectors are not a problem, I end up with (λ+3)(λ+1)² so my eigenvalues are -3 and -1. Substituting in I get [1, -3, 9] and [1, -1, 1]. Now would the eigenspace simply be {[1, -3, 9], [1, -1, 1], [0, 0, 0]} or am I missing some other step?

Also, how can I prove there cannot be a basis of R3 consisting of eigenvectors of A? Could it just be because there must be n vectors in a basis of Rn?

Your characteristic polynomial is wrong. So the eigenvalues and eigenvectors are too. Did you put the correct matrix in the problem statement?

 

[Mark44]

Dick makes a good point. After putting in all that effort to find eigenvalues and eigenvectors, you should at least check your work. If x is an eigenvector with eigenvalue $\lambda$, then it should be true that Ax = $\lambda$x.

 

[Locoism]

Aaahhh sorry, I missed a 1, it's corrected now

 

[Dick]

Aaahhh sorry, I missed a 1, it's corrected now

That's better. Now listen to Mark44 on the basis part of the question.

 

[Locoism]

Thank you. So I know I can't have the zero vector as part of a basis, but should it be included in the eigenspace? or is the eigenspace juste a basis of the eigenvectors? The terminology confuses me a little.

 

[Mark44]

The eigenspace is the set of all linear combinations of the basis vectors. The eigenspace is a vector space, which like all vector spaces, includes a zero vector.

No one is asking you to list the eigenspace (an impossible task) - just a basis for it.

 

[Dick]

Thank you. So I know I can't have the zero vector as part of a basis, but should it be included in the eigenspace? or is the eigenspace juste a basis of the eigenvectors? The terminology confuses me a little.

Be careful. You have two different eigenspaces here. One corresponding to the eigenvalue -3 and another to the eigenvalue -1. What's a basis for each?

 

[Locoism]

Oh ok so basis for lambda=-3 is span(1, -3, 9) and for -1 it is span(1, -1, 1).
Why is it that there can't be a basis for R3 of only eigenvectors?

 

[Dick]

Oh ok so basis for lambda=-3 is span(1, -3, 9) and for -1 it is span(1, -1, 1).
Why is it that there can't be a basis for R3 of only eigenvectors?

It's basically what you said. To span R3 you need three linearly independent eigenvectors. The eigenspaces only give you two.

 

[Locoism]

Alright thank you so much guys, this was really helpful. Is there some way to +rep or something?

 

[Dick]

Alright thank you so much guys, this was really helpful. Is there some way to +rep or something?

If you mean a ratings boost, no, we don't have that. Thanks is enough. Very welcome.

 

[Mark44]

Same here. Posters don't always say "thank you," but it's appreciated when they do.