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I am asked to relate eigen{value,vectors} with equalizers in the category of matrices. However, in the category of matrices $g\circ f$ means the (matrix) product $fg$. And $\lambda$ is an eigenvalue of $A$ if there exists $v$ such that $Av=\lambda v$.

To relate this with equalizers, I was expecting the definition of eigenvalue to be such that $vA=v\lambda$.

Am thinking right?

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up vote 1 down vote accepted

Yes. The eigenspace associated with the eigenvalue $\lambda$ is usually defined as the kernel of $f - \lambda \operatorname{id}$, but is equally well defined as the equaliser of $f$ and $\lambda \operatorname{id}$. Accordingly, you can define eigenvalue as a scalar $\lambda$ such that the equaliser of $f$ and $\lambda \operatorname{id}$ is non-zero. (Beware: a scalar $\lambda$ such that $f - \lambda \operatorname{id}$ is non-invertible is not necessarily an eigenvalue if your vector spaces are not finite-dimensional!)

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