$V$ and $W$ are finite-dimensional isomorphic vector spaces, where we have a basis for $W$

Question

Say I have two finite-dimensional vector spaces $V$ and $W$. I know that the two are isomorphic. I also know a basis for $W$.

I claim a basis for $V$. In order to show that this indeed is a basis for $V$, is it enough to show that there exists a bijection between the claimed basis for $V$ and the basis for $W$ ?

Thanks!

Answer 1

First result:

Given two finite-dimensional vector spaces $V$ and $W$, they are isomorphic iff $\dim V = \dim W$.

Second result:

If the linear transformation $T:W\rightarrow V$ is injective, then it takes LI sets onto LI sets.

Third result:

If the set of vectors $\mathcal{B}_{W} =\{w_{1},w_{2},\ldots,w_{n}\}\subset W$ is LI and $\dim W = n$, then $\mathcal{B}_{W}$ is a basis for $W$.

We may now tackle the given problem.

Since $W$ and $V$ are isomorphic, we conclude that $\dim V = \dim W$. Let us consider a basis $\mathcal{B}_{W} = \{w_{1},w_{2},\ldots,w_{n}\}$ for $W$. Since $T:W\rightarrow V$ is injective, $T(\mathcal{B}_{W}) = \{T(w_{1}),T(w_{2}),\ldots,T(w_{n})\}$ is LI. But we know that $\dim V = n$. Consequently, $\mathcal{B}_{V} = \{T(w_{1}),T(w_{2}),\ldots,T(w_{n})\}$ is a basis for $V$.

Hopefully this helps.