Originally Posted by

**HappyJoe** The last part in particular sounds strange.

What does it mean that the bases are inverse to each other, and why does it imply that $\displaystyle \{\psi(v_1),\ldots,\psi(v_n)\}$ is a basis for $\displaystyle W$?

I would do like this: We need to prove that $\displaystyle \{\psi(v_1),\ldots,\psi(v_n)\}$ is a basis for $\displaystyle W$. So we need to check two things, namely that the vectors in the proposed basis are linearly independent and that they span all of $\displaystyle W$.

Let $\displaystyle x_1\psi(v_1)+\ldots+x_n\psi(v_n)=0$ for some scalars $\displaystyle x_i$. Apply linearity of $\displaystyle \psi$ to get

$\displaystyle \psi(x_1v_1+\ldots+x_nv_n)=0.$

Remember that $\displaystyle \psi$ is an isomorphism, so what do you know about vectors that $\displaystyle \psi$ map to 0?

To see that the basis spans all of $\displaystyle W$, take some vector $\displaystyle w\in W$, and choose a vector $\displaystyle v\in V$, such that $\displaystyle \psi(v)=w$ (why does this vector $\displaystyle v$ exist?).

You know that $\displaystyle \{v_1,\ldots,v_n\}$ is a basis for $\displaystyle V$, so you can write $\displaystyle v$ as a linear combination of these basis vectors. Try applying $\displaystyle \psi$ to this relation and see what happens.