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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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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 $Ax=0$. –  user35959 Dec 14 '12 at 19:54
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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. –  Daniel Rust Dec 14 '12 at 19:59

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up vote 1 down vote accepted

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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OK. Looks like I was overthinking things. Thanks! –  zaz Dec 18 '12 at 19:42

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