# Proof of finitely dimensional subspaces

Prove: If $V_1, V_2, ...$ are finite dimensional subspaces of a Vector Space $V$, then for $n = 1, 2, ...$ $V_1 +...+ V_n$ is a finite dimensional subspace of V.

• I have the base case
• I assume true in the induction step.

I don't know what to consider about another finite dimensional subspace $V_{n+1}$

I have a gut-feeling that this is one of those "break off pieces" and then put it all back together induction proof.

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If $\beta_1, \ldots, \beta_n$ are bases for $V_1, \ldots, V_n$, then $\mathrm{span}(\beta_1 \cup \ldots \cup \beta_n) = V_1 + \ldots + V_n$, so it must be finite-dimensional. – Christopher A. Wong Sep 26 '12 at 0:06

Hint Why don't you proceed as follows. Suppose you have the base case. Then as the inductive step you assume the thesis to be true for all $k\leq n$, and you want to prove the case $k=n+1$. Now just notice that $$V_1+V_2+\dotso+V_{n+1}=W+V_{n+1}.$$ By inductive hypothesis $W$ is finite dimensional, and again using the case $k=2$ $$W+V_{n+1}$$ is finite dimensional.

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Clearly the case $n=1$ is true, since then you just have $V_1$, which is a subspace. Labeling $$U_n=V_1+V_2+V_3...+V_n$$

The inductive step is that $U_n+V_{n+1}=U_{n+1}$ is a vector space. This is implied by the statement

$$\text{"the sum of any two subspaces is another subspace"}$$

Use the definition of a subspace and of a sum to prove this, then simply use it to look at $V_1+V_2+V_3...V_n$ and $V_{n+1}$.

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