Basis for the eigenspace corresponding

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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.
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saqifriends said:
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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.

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