# 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 $W$.

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?

• – amd
Commented Apr 14, 2020 at 19:48

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.