LADR-Ch. 5 #12

Problem Statement: Suppose T\in\mathcal{L}(V) is such that every v\in V is an eigenvector of T. Prove that T is a scalar multiple of the identity operator.

Proof: Let v,w\in V. Then by assumption there exist scalars \alpha,\beta such that T(v)=\alpha v and T(w)=\beta w. Now consider T(v+w). Since V is a vector space it follows that v+w\in V and so there exists a scalar, \gamma, such that T(v+w)=\gamma (v+w). But T is a linear operator and so it follows that T(v+w)=\alpha v+\beta w and so setting these two equations equal we see that \alpha v+\beta w=\gamma(v+w)=\gamma v+\gamma w. Equating like terms we get \alpha=\gamma and \beta=\gamma and so it follows that \alpha=\beta=\gamma. Thus, for any v\in V, T(v)=\beta v for some scalar \beta.

Now we must consider if v,w are already scalar multiples of each other. Consider: w=\eta v and T(v)=\mu v. Then T(w)=T(\eta v)=\eta T(v)=\eta\mu v=\mu\eta v=\mu w. An argument similar to what was done above to show that \alpha=\beta=\gamma will show that all of these scalars are in fact the same scalar.

Thus, T is a scalar multiple of the identity operator.

\Box

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