Proofs where V is a finite-dimensional vector space

These are probably easier than I think they are.

Suppose $\displaystyle V$ and $\displaystyle W$ are vector spaces over a field $\displaystyle F$.

Now suppose $\displaystyle W_1,...,W_k$ are subspaces of $\displaystyle V$ and that $\displaystyle V$ is finite-dimensional.

**a)** Prove that $\displaystyle \dim(W_1+...+W_k)\leq\dim(W_1)+...+\dim(W_k)$.

The question recommends proving this via induction (although proving via bases is possible, albeit more difficult). I think what I have below works, but if it needs correction, I'd like to know.

For $\displaystyle k=1, \dim(W_1)\leq\dim(W_1)$ is true. (base case)

Assume the result is true for $\displaystyle k=n$,

i.e. $\displaystyle \dim(W_1,...,W_n)\leq\dim(W_1)+...+\dim(W_n)$ (hypothesis)

For $\displaystyle k=n+1$, (inductive case)

$\displaystyle \dim(W_1,...,W_n,W_{n+1})\leq\dim(W_1)+...+\dim(W_ n)+\dim(W_{n+1})$

$\displaystyle =\dim(W_1,...,W_{k-1},W_k)\leq\dim(W_1)+...+\dim(W_{k-1})+\dim(W_k)$

Therefore, $\displaystyle \dim(W_1,...,W_k)\leq\dim(W_1)+...+\dim(W_k)$

**b)** Prove that $\displaystyle V=W_1\oplus...\oplus W_k$ if and only if every vector $\displaystyle v\in V$ is equal to $\displaystyle w_1+...+w_k$ for a *unique* choice of $\displaystyle w_j\in W_j, 1\leq j\leq k$.

I don't have an answer yet for this half, but I'm working on it.

Any help on these would be appreciated.