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Is there a way to relate Eigenvalues to the column space and nullspace of a matrix?

I believe a matrices with different eigenvalues would have a different column spaces and/or nullspace. Is this correct?

I am wondering if you can prove that the Eigenvalues of $A$ and $A^T$ are equal using properties of column spaces and nullspaces.

My thinking is:
If you transform a matrix $A$ into $B$, if the row space of $B$ is orthogonal to the nullspace of $A$, and the column space of $B$ is orthogonal to the left nullspace of $A$, then matrices $A$ and $B$ have the same eigenvalues.

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    $\begingroup$ Well, for one, you know that if $\lambda = 0$ is an eigenvalue, then the matrix has a non-trivial null space (and, conversely, if the matrix has a non-trivial null space, then $0$ is an eigenvalue). This follows because $\lambda = 0$ is an eigenvalue implies there exists a non-zero vector $x$ such that $Ax=0$. $\endgroup$ Commented Dec 14, 2012 at 19:54
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    $\begingroup$ The dimension of the nullspace corresponds to the multiplicity of the eigenvalue 0. In particular, A has all non-zero eigenvalues if and only if the nullspace of A is trivial ($null(A)=\{0\})$. You may then use the fact that $dim(Null(A))+dim(Col(A))=dim(A)$ to deduce that the dimension of the column space of A is the sum of the multiplicities of the non-zero eigenvalues. $\endgroup$ Commented Dec 14, 2012 at 19:59
  • $\begingroup$ @DanRust sorry for the delayed comment, but now I came up with the same question. So, what about matrix $A = [2 -2;2 -2]$, it has $\mathcal{N}(A)=\{c[1, 1]^T | c\in \mathbb{R} \}$ , $\mathbb{dim}\mathcal{N}(A)=1$ (a line), but the multiplicity of zero-eigenvalues is two ($\lambda_{1,2}=0$). Could you please give a comment on that? Am I right? $\endgroup$ Commented Sep 25, 2016 at 9:53
  • $\begingroup$ @Thoth Geometric multiplicity, not algebraic multiplicity. $\endgroup$ Commented Sep 25, 2016 at 10:58
  • $\begingroup$ @DanRust in order to complete my understanding, "the dimension of the nullspace corresponds to the (*geometric) multiplicity ( not the multiplicity of the eigenvalue 0 but the multiplicity of the eigen-vectors that correrspond to zero eigen-values.)". The geometric multiplicity equals to $\mathbb{dim}\mathcal{N}(A)$. So the $\mathbb{dim}\mathcal{N}(A)$ does not offer enough information for the multiplicity of eigen-values. Am I right? You comments and discussion valuable in understanding. $\endgroup$ Commented Sep 25, 2016 at 13:53

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Matrices with different eigenvalues can have the same column space and nullspace. For a simple example, consider the real 2x2 identity matrix and a 2x2 diagonal matrix with diagonals 2,3. The identity has eigenvalue 1 and the other matrix has eigenvalues 2 and 3, but they both have rank 2 and nullity 0 so their column space is all of $\mathbb{R}^2$ and their nullspace is $\{0\}$.

This is also probably a negative answer to your question about the transpose - the column space and nullspace don't contain enough information about the eigenvalues.

On the other hand, eigenvalues are certainly related to the nullspace of $A-\lambda I$, where $\lambda$ is an eigenvalue of $A$. Namely, every eigenvector must lie in the nullspace of this matrix.

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  • $\begingroup$ OK. Looks like I was overthinking things. Thanks! $\endgroup$ Commented Dec 18, 2012 at 19:42

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