$L:\Bbb{V} \to \Bbb{W}$ is an invertible linear mapping, prove that $\dim\Bbb{V} = \dim\Bbb{W}$?

Question

Suppose $L:\Bbb{V} \to \Bbb{W}$ is an invertible linear mapping, and $\Bbb{V}$, $\Bbb{W}$ are finite dimensional vector spaces. How would you prove that $\dim\Bbb{V} = \dim\Bbb{W}$?

My attempt for this proof relies on proving that $\Bbb{V}$ and $\Bbb{W}$ are isomorphic. I have proved that $L$ is an isomorphism from $\Bbb{V}$ to $\Bbb{W}$. Is this correct?

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Answer 1

The rank-nullity theorem says that $$ \dim\operatorname{im}L+\dim\ker L=\dim V $$ Since $L$ is invertible, you also have $$ \dim\operatorname{im}L^{-1}+\dim\ker L^{-1}=\dim W $$ On the other hand $\ker L=\{0\}$ and $\ker L^{-1}=\{0\}$, so we have $$ \dim V=\dim\operatorname{im}L\le\dim W=\dim\operatorname{im}L^{-1}\le\dim V $$

Answer 2

$dim V = n$

Given that $L$ is invertible, that means that $ \dim N(L) = 0$, and $\dim Im(L) = n$.

$L^{-1}$ is the inverse of $L$. $L^{-1}: W \rightarrow V$. $L^{-1}$ is injective and surjective, given that $L$ is invertible. That means $\dim N(L^{-1}) = 0$.

It also means that for every $x$ in $V$ there exists $y$ in $W$ such that $L^{-1}(y) = x$. Given that $\dim V = n$, it follows that $Im(L^{-1}) = n$, which proves that $\dim V = \dim W$.