# eigenvector proof

Let $V$ be a vector space and $T : V\to V$ a linear transformation with the property that $T(W)\subset W$ for every subspace $W$ of $V$. How can we prove that there is an element $\lambda$ in the field of scalars such that $T(v) = \lambda v$ for all $v \in V .$

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Here are 2 hints.

1. Given $0\neq v\in V$, let $W = \text{span}\{v\}$. The hypothesis on $T$ imply what about $v$?

2. Assume for a contradiction that $v$ and $w$ are nonzero vectors and $Tv = \lambda v$ and $Tw = \mu w$ with $\lambda \neq \mu$. If $W = \text{span}\{v+w\}$, then what happens?

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If $\mathbf{v}\neq\mathbf{0}$, let $\mathbf{W}=\mathrm{span}(\mathbf{v})$. Then $T(\mathbf{v})\in \mathbf{W}$, so $T(\mathbf{v}) = \mu_{\mathbf{v}}\mathbf{v}$ for some scalar $\mu_{\mathbf{v}}$ which, for all we know right now, may depend on $\mathbf{v}$.

If $\dim(\mathbf{V})\leq 1$, this suffices. (Why?)

Now, assume $\mathbf{v}\neq\mathbf{w}$ are two nonzero vectors that are linearly independent Then: $$\mu_{\mathbf{v}+\mathbf{w}}(\mathbf{v}+\mathbf{w}) = T(\mathbf{v}+\mathbf{w}) = T(\mathbf{v}) + T(\mathbf{w}) = \mu_{\mathbf{v}}\mathbf{v} + \mu_{\mathbf{w}}\mathbf{w}.$$ That means, $$(\mu_{\mathbf{v}+\mathbf{w}} - \mu_{\mathbf{v}})\mathbf{v} = (\mu_{\mathbf{w}}-\mu_{\mathbf{v}+\mathbf{w}})\mathbf{w}.$$ Given that we are assuming $\mathbf{v}$ and $\mathbf{w}$ are linearly independent, what can we conclude from this?

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