I'm trying to prove a theorem from "linear Algebra Done right" by Sheldon Axler.
$\lambda$ is an eigenvalue of T if and only if $T-\lambda \mathrm{I}$ is not injective. I'm a bit confused about what we assume to be true. So, I'm not sure what to do.
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Axler's approach is non-traditional. Start by getting the definitions clear. – Ted Shifrin Mar 26 '14 at 03:00
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If $\lambda$ is an eigenvalue of $T$ then there is nonzero vector $v$ s.t. $Tv=\lambda v$. So then $T-\lambda I$ is not injective since it sends both $0$ and $v$ to $0$.
Conversely if $T-\lambda I$ is not injective then it's kernel is not $0$ so there is a nonzero vector $v$ such that $(T -\lambda I)v=0$. Then $v$ is an eigenvector of $V$.
Seth
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To expand some on Seth's answer, the intuition behind an eigenvalue is that it creates a linear dependence in the matrix. By definition, a linear dependence on a set of vectors implies that there are at least two linear combinations of the vectors in the set to form a given resultant. Hence, $(T - \lambda I)$ fails to be injective.
ml0105
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Did you mean: By definition, a linear $dependence$ on a set of vectors implies.. – user180708 Jan 29 '15 at 01:54
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No problem! I don't mean to be super meticulous, I'm just exceptionally new to linear algebra and just wanted to make sure I understood correctly. – user180708 Jan 30 '15 at 04:41