What is the relation between rank of a matrix, its eigenvalues and eigenvectors?

Question

I am quite confused about this. I know that zero eigenvalue means that null space has non zero dimension. And that the rank of matrix is not the whole space. But is the number of distinct eigenvalues ( thus independent eigenvectos ) is the rank of matrix?

Answer 1

Let's dissect this with a 3x3 matrix that has rank 2

[1 0 1]

[0 1 1]

[0 0 0]

The first two columns can serve as a basis for the 2-d column space, and [0 0 1]^T is a basis for the 1-d null space. Together they span all of R3.

The characteristic polynomial for the matrix is (1-λ)(1-λ)(0-λ)=0, so λ={1, 1, 0}.

The λ=1 eigenvector equation is

[0 0 1]

[0 0 1]. v = 1 v

[0 0 -1]

which should (hopefully) yield two independent eigenvectors because of the repeated root. In this case we can use [1 0 0] and [0 1 0]. The λ=1 eigenvalue has algebraic multiplicity 2, because it is a repeated root in the characteristic equation. We were able to obtain two independent eigenvectors from the same eigenvalue, so λ=1 has geometric multiplicity 2. When we solve for an eigenvector, we are essentially looking for a basis for the nullspace of (A-λI). Geometric multiplicity corresponds to the dimension of this nullspace. In the best case, we have a full basis and the geometric multiplicity is equal to the algebraic multiplicity. Otherwise, geometric multiplicity is less than algebraic multiplicity.

The λ=0 equation is

[1 0 1]

[0 1 1] v = 0 v

[0 0 0]

and the solution is [1 1 -1]. The third eigenvector is independent of the first two, but it is not necessary for it to be orthogonal.

Different eigenvalues yield independent eigenvectors, so if an NxN matrix gives you N unique values, you should have a set of e-vectors that span the entire space, regardless of the rank of A. When you have repeated e-values, there's a possibility for an eigenspace that does not span N-dimensions (depending on geometric multiplicity of repeated e-vals)