This is the proof from the book for the Corollary following theorem 2.17.

Theorem 2.17 (Corollary)
If \(T: V \rightarrow W\) is invertible, then \(V\) is finite dimensional if and only if \(W\) is finite dimensional. In this case \(\dim V = \dim W\)


Proof:

\(\Rightarrow:\) Suppose that \(V\) is finite dimensional. If \(T\) is invertible, then \(T\) is onto. By the definition of onto, this means that \(R(T)=W\) (or that for any \(w \in W\), there exists a \(v \in V\) such that \(T(v)=w\)). Since \(V\) is finite dimensional, then let \(\beta\) be a finite basis for \(V\). By Theorem 2.2, \(span(T(\beta)) = R(T)\). But \(R(T) = W\). Therefore, \(W\) must be finite dimensional.

\(\Leftarrow:\) Suppose that \(W\) is finite dimensional. If \(T\) is invertible, then \(T^{-1}\) is linear and invertible. We can now apply the same argument from before. \(T^{-1}\) is invertible and so \(T^{-1}\) is onto. Since \(W\) is finite dimensional, let \(\gamma\) be a basis for \(W\). Therefore, \(R(T^{-1})=V\) and since \(span(T^{-1}(\gamma)) = R(T^{-1}) = V\), then \(V\) is finite dimensional.

So now suppose that \(V\) and \(W\) are finite dimensional. Because \(T\) is one-to-one and onto, then \(nullity(T)=0\) and \(rank(T) = \dim(R(T)) = \dim(W)\). By the dimension theorem we know,

$$ \begin{align*} \dim(V) &= \dim(N(T)) + \dim(R(T)) \\ &= 0 + \dim(W) \\ \end{align*} $$

Therefore, \(\dim(V)=\dim(W)\) as we wanted to show. \(\blacksquare\)

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