Suppose that $\mathcal{B}$ is a basis for $V$. Prove that $T$ is an isomorphism if and only if $T(\mathcal{B})$ is a basis for $W$.

Question

Let $V$ and $W$ be finite-dimensional vector spaces, and let $T: V \rightarrow W$ be a linear transformation. Suppose that $\mathcal{B}$ is a basis for $V$. Prove that $T$ is an isomorphism if and only if $T(\mathcal{B})$ is a basis for $W$.

MY ATTEMPT

We shall prove $(\Rightarrow)$ first.

If $T$ is an isomorphism, then $T$ is injective. Once injective linear mappings take LI sets onto LI sets, we conclude that $T(\mathcal{B})$ is LI. Moreover, once $V$ and $W$ are isomorphic, $\dim V = \dim W$. Since $|T(\mathcal{B})| = \dim V = \dim W$, we conclude that $T(\mathcal{B})$ is a basis for $W$ indeed.

We may now approach the implication $(\Leftarrow)$

Let $\mathcal{B} = \{v_{1},v_{2},\ldots,v_{n}\}$ be a basis for $V$. Consequently, based on the given assumption, $T$ is injective. Indeed, \begin{align*} v = a_{1}v_{1} + a_{2}v_{2} + \ldots + a_{n}v_{n} \in T^{-1}(\{0\}) & \Longrightarrow T(v) = T(a_{1}v_{1} + a_{2}v_{2} + \ldots + a_{n}v_{n}) = 0\\\\ & \Longrightarrow a_{1}T(v_{1}) + a_{2}T(v_{2}) + \ldots + a_{n}T(v_{n}) = 0\\\\ & \Longrightarrow a_{1} = a_{2} = \ldots = a_{n} = 0\\\\ & \Longrightarrow v = 0 \end{align*}

Moreover, it is also surjective. This is because $T(\mathcal{B}) = \{T(v_{1}),T(v_{2}),\ldots,T(v_{n})\}$ spans $T(V)$ as well as $W$. Thus $T(V) = W$.

Hence $T$ is an isomorphism.

Can someone check my reasoning?

Answer 1

Everything is fine, also, I will give you an alternative solution for the $(\Leftarrow)$ part:

Let $\mathcal{B} = \{v_1,v_2,\dots,v_n\}$. If we put $w_i := T(v_i)$ for $i=1,\dots,n$, then $\{w_1,w_2,\dots,w_n\}$ is basis for $W$ by assumption. Therefore, there exists a unique linear transformation $S : W \to V$ such that $g(w_i) = v_i$ for all $i$. Now, it is very straightforward to check that $S$ is the inverse of $T$, and so, $T$ is an isomorphism.