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Suppose $V_1, \dots, V_m$ are vector spaces such that $V_1 \times \cdots \times V_m$ is finite-dimensional. Prove that $V_j$ is finite-dimensional for each $j=1,\dots, m$.

I have written up a proof, but it just doesn't quite sound like the kinds of proofs I see in my book. Here's what I've got:

$0 \times \cdots \times 0 \times V_j \times 0 \times \cdots \times 0$ is clearly a subspace of $V_1 \times \cdots \times V_m$ (I could prove this, but I think it's pretty obvious). Using a previous theorem, we know that if a vector space is finite-dimensional then all of its subspaces are finite-dimensional. This implies that $0 \times \cdots \times 0 \times V_j \times 0 \times \cdots \times 0$ is finite-dimensional. Because the linear map given by $$0 \times \cdots \times 0 \times V_j \times 0 \times \cdots \times 0 \ni (0,\dots, 0,v,0,\dots, 0)\mapsto v \in V_j$$ is an isomorphism, we know that $\dim(0 \times \cdots \times 0 \times V_j \times 0 \times \cdots \times 0) = \dim(V_j)$. Thus $V_j$ is finite dimensional, completing the proof. $\square$

Can you give me some advice on how to write this up so it sounds a bit more formal?

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    Seems formal to me. – 2017-02-13
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    You might use the canonical projection $\pi_j$ to simplify the notation but the proof is good enough. – 2017-02-13
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    @JulioMaldonadoHenríquez (After Googling) By the canonical projection map do you mean what my book calls the quotient map: $\pi: V\to V/ U: \pi(v) = v+U$? If so, could you expand on how that function can be used to simplify the notation? Or did you just mean the $T_j$ that levap defines in his answer? – 2017-02-13
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    neither. I mean $\pi_j :V_1\times V_2\times ... \times V_m\rightarrow V_j $ where $\pi_j(v_1,v_2,...,v_m)=v_j$ which is linear and epijective – 2017-02-13
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    OK. Thanks for the clarification. – 2017-02-13

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Your proof is good and is already quite formal. You can make it more explicit by defining a map $T_j \colon V_j \rightarrow V_1 \times \dots \times V_m$ given by $$ T_j(v) = \underbrace{(0,\dots,v,\dots,0)}_{j\text{th place}} $$

and then arguing that this map is linear and injective. Hence, $T_j$ is an isomorphism from $V_j$ to $T_j(V_j)$. Since $T_j(V_j)$ is the image of a linear map, it is a subspace of $V_1 \times \dots \times V_m$ and since $V_1 \times \dots \times V_m$ is finite dimensional, so is $T_j(V_j)$ (being a subspace of a finite dimensional space).

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    That's a good idea. Thanks! – 2017-02-13