Equivalence of the nullspace and eigenvectors corresponding to zero eigenvalue

[onako]

Suppose a square matrix A is given. Is it true that the null space of A corresponds to eigenvectors of A being associated with its zero eigenvalue? I'm a bit confused with the terms 'algebraic and geometric multiplicity' of eigenvalues related to the previous statement? How does this affect the statement?

 

[mathwonk]

yes, at least a non zero vector in the nullspace of T is just a non zero vector v such that Tv = 0.v, hence is an eigenvector for the eigenvalue zero.

the geometric multiplicity of this eigenvalue is the dimension of the null space. and the algebraic multiplicity is the power of x that divides the characteristic polynomial. they can be different, as in the case where the matrix has all zeroes on and below the diagonal but all 1's above it. then the null space has dimension one, the ch poly is X^n and the alg mult is n.

 

[onako]

@mathwonk

Thanks. What is the exact meaning of "at least" in your statement? Could the set of vectors of null vectors be a superset of the set of eigenvectors corresponding to zero eigenvalue (of course, any linear comb of such eigenvectors will also be a vector of null space)?

 

[ImaLooser]

@mathwonk

Thanks. What is the exact meaning of "at least" in your statement? Could the set of vectors of null vectors be a superset of the set of eigenvectors corresponding to zero eigenvalue (of course, any linear comb of such eigenvectors will also be a vector of null space)?

Aren't the eigenvectors a basis of the null space? That is, if you have a 2D null space then you have two basis vectors, but infinitely many vectors could be in the null space.

 

[HallsofIvy]

Some texts require that an "eigenvector" of A with eigenvalue \lambda is a nonzero vector, v, such that Av= \lambda v which then requires that we say that "the set of all eigenvectors of A with eigenvalue \lambda together with the 0 vector form a vector space". Other texts will say that "\lambda is an eigenvalue of A if there exists a non-zero vector v such that Av= \lambda v but then say that an eigenvector is any vector v such that Av= \lambda v, even the 0 vector.

If you require that an "eigenvector" not be 0, then, yes, any "eigen space" is a "superset" of the set of a "eigenvectors" specifically because it includes 0. However, if you include the 0 vector as an "eigenvector", the two sets, the null space and the set of eigenvectors with 0 eigenvalue are exactly the same.