Basis for the eigenspace corresponding

  • MHB
  • Thread starter saqifriends
  • Start date
  • Tags
    Basis
In summary, to solve this problem, you must find the eigenspace corresponding to the eigenvalue of 3 and then find a basis for that eigenspace. This can be done by finding the null space of the given matrix and using the rank-nullity theorem to determine the number of basis vectors needed.
  • #1
saqifriends
2
0

Attachments

  • untitled.bmp
    5.6 KB · Views: 93
Physics news on Phys.org
  • #2
saqifriends said:
View attachment 213

Hi saqifriends, :)

I have outlined the method to do this kind of problems http://www.mathhelpboards.com/threads/1270-basis-for-each-eigenspace?p=6086&viewfull=1#post6086 Since you have been given a particular eigenvalue, find the eigenspace corresponding to that eigenvalue. Then find a basis for that eigenspace.

Kind Regards,
Sudharaka.
 
  • #3
as a slight nudge towards the answer, solve the system:

(A - λI)v = 0. in this case, λ = 3, so you must find the null space of the matrix:

$\begin{bmatrix}1&2&3\\-1&-2&-3\\2&4&6 \end{bmatrix}$

the rank of this matrix should be obvious upon inspection, and the rank-nullity theorem then tells you how many basis vectors you should have for the null space.
 

FAQ: Basis for the eigenspace corresponding

What is the basis for the eigenspace corresponding to a given eigenvalue?

The basis for the eigenspace corresponding to a given eigenvalue is a set of linearly independent vectors that span the subspace of the given eigenvalue.

How do you find the basis for the eigenspace corresponding to a specific eigenvalue?

To find the basis for the eigenspace corresponding to a specific eigenvalue, you must first find the null space of the matrix A-λI, where A is the original matrix and λ is the given eigenvalue. The basis for the null space will be the basis for the eigenspace.

Can a matrix have more than one basis for the eigenspace corresponding to a single eigenvalue?

Yes, a matrix can have an infinite number of bases for the eigenspace corresponding to a single eigenvalue. This is because the eigenspace is a subspace, and any set of linearly independent vectors that span the subspace can be considered a basis.

Is the basis for the eigenspace always unique?

No, the basis for the eigenspace is not always unique. Different sets of linearly independent vectors can span the same subspace, so there can be multiple bases for the eigenspace corresponding to a given eigenvalue.

How does the dimension of the eigenspace relate to the multiplicity of the eigenvalue?

The dimension of the eigenspace will be equal to the multiplicity of the eigenvalue. This is because the multiplicity of an eigenvalue represents the number of linearly independent eigenvectors associated with that eigenvalue, and the number of linearly independent vectors in a basis for a subspace is equal to the dimension of that subspace.

Similar threads

I Standard basis and other bases...
Replies
9
Views
2K
I Non orthogonal basis and the lines of its coordinate grid
Replies
9
Views
1K
I Proof of Column Extraction Theorem for Finding a Basis for Col(A)
Replies
3
Views
2K
I Understanding the concepts of isometric basis and musical isomorphism
Replies
3
Views
2K
I Proof that if T is Hermitian, eigenvectors form an orthonormal basis
Replies
3
Views
3K
MHB 051 how they got the eigenspaces ?
Replies
2
Views
854
I Vector subspace and basis vectors in the context of data science
Replies
6
Views
2K
I Change of Basis Matrix vs Transformation matrix in the same basis....
Replies
12
Views
4K
I Orthonormal basis expression for ordinary contraction of a tensor
Replies
2
Views
1K
I Consistent matrix index notation when dealing with change of basis
Replies
12
Views
2K

Hot Threads

Recent Insights

Back
Top