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I have the following problem in linear algebra:

Let $A$ be a vector space and $V$, $W$ subspaces of $A$ such that $V \cap W = \{0\}$:

Prove that $\operatorname{span}\{V, W\} := \{λ_1v + λ_2w : v \in V, w \in W, λ_1, λ_2 \in F\}$ is isomorphic to $V \oplus W$:

[Hint: Show that the function $T(v,w) = v + w$ is a linear isomorphism.]

My problem, oddly enough, is with the hint itself. I'm not sure how $T(v,w) = v + w$ can be an invertible linear operator. Without knowing either $v$ or $w$, there should be no way of retrieving them from the result of the operator. I also don't see how it can be used to map between $\operatorname{span}\{V,W\}$ and $V\oplus W$ (or vice versa). In effect, I don't really know where to get started with the problem.

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The map in the hint should be from $V\oplus W$ to $\mathrm{span}\{V,W\}$. Hence it takes an ordered pair $(v,w)$, with $v\in V$ and $w\in W$, and sends it to $v+w$, which is an element of $\mathrm{span}\{V,W\}$. You want to verify that if $T$ is defined in this way, then $T$ is (a) linear, (b) surjective, and (c) injective.

(a) For linearity, you can just directly check that it satisfies the definition.

(b) For surjectiveness, it would help to note that in your definition of the span, $\lambda_1$ and $\lambda_2$ are not needed, due to the fact that $V$ and $W$ are closed under scalar multiplication.

(c) It is for injectiveness that you need to use the condition that $V\cap W=\{0\}$. I recommend using the fact that a linear map is injective if and only if the kernel contains only the zero vector. So if you suppose that $T(v,w)=0$, can you show that $v=w=0$?

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Consider the linear maps $$ V\cap W\ \overset{d}{\to}\ V\oplus W\ \overset{a}{\to}\ V+W, $$ where $d$ sends $x$ to $(x,-x)$, and $a$ is the addition.

Note that $d$ induces an isomorphism of $V\cap W$ onto the kernel of $a$, and that $a$ is surjective.

This shows that $a$ is bijective if and only if $V\cap W=0$.

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A linear map is 1-1 if the kernel $T^{-1}(0)$ is $\{0\}$.
That the map $T$ from $V\oplus W$ to the span of $V$ and $W$ is linear and onto should be easy.

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