# Lower bound on rank of matrix with respect to eigen values

I explored quite a few online resources and math StackExchange questions too. I know that in general, rank is not equal to the number of non-zero eigenvalues. But I was confused by seeing this Solution to Problem 21.1 a. Here they seem to conclude rank based on the theorem "Eigenvectors with distinct eigenvalues are always linearly independent" Also, the column space of the matrix contains all the non-zero eigen vectors.

So, I was wondering if I can make a statement as

Rank >= number of non-zero(counted without repetitions) eigenvalues

or should/can it be

Rank >= number of distinct non-zero eigen values

I could not create any counterexample which disproves this statement. Nor could I analytically prove it is correct. Thank you for the help!

If $$A \in \mathbb{C}^{n \times n}$$, $$\dim R(A) = n - \dim N(A) \geq n - \dim GE(A, 0),$$ where $$GE(A, \lambda) = \{x \in \mathbb{C}^n : (A - \lambda I)^k x = 0 \text{ for some } k \geq 0\}.$$ It is well known that $$\dim GE(A, \lambda)$$ is the multiplicity of $$\lambda$$ in the characteristic polynomial $$K_A(\lambda) = \det(\lambda I - A)$$ of $$A$$ (see chapter 2 of https://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/linalg.pdf).

Thus Rank $$\geq$$ number of nonzero eigenvalues counted with multiplicities as roots of $$K_A$$.

Edit: $$K_A(\lambda) = \det(\lambda I - A)$$ is a complex polynomial of degree $$n$$. By the Fundamental theorem of algebra, $$K_A$$ can be factored as $$K_A = (\lambda - \lambda_1)^{d_1}(\lambda - \lambda_2)^{d_2}\dots(\lambda - \lambda_K)^{d_K}.$$ $$d_j$$ is called the multiplicity of the root $$\lambda_j$$.

• It would be very helpful if you elaborate with a simple example. What do you exactly mean by saying "counted with multiplicities as roots of KA". Sorry if this is too trivial. I am a beginner in LA.
– agr
Oct 9, 2021 at 10:16
• I think I earlier forgot to tag you.
– agr
Oct 9, 2021 at 14:27
• @agr I added what multiplicity of a root of a polynomial means. Oct 9, 2021 at 15:11
• Did you mean algebraic multiplicity or geometric multiplicity. Did you mean that say, eigen values are 1,1,2. Then rank >=2 or did you mean the geometric multiplicity counted per eigen value
– agr
Oct 9, 2021 at 15:36
• @agr I mean the multiplicity of the eigenvalue as a root in $K_A$. This is called the algebraic multiplicity of the eigenvalue. Oct 9, 2021 at 15:42