# Thread: Proofs where V is a finite-dimensional vector space

1. ## Proofs where V is a finite-dimensional vector space

These are probably easier than I think they are.

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

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

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

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 $k=1, \dim(W_1)\leq\dim(W_1)$ is true. (base case)
Assume the result is true for $k=n$,
i.e. $\dim(W_1,...,W_n)\leq\dim(W_1)+...+\dim(W_n)$ (hypothesis)
For $k=n+1$, (inductive case)
$\dim(W_1,...,W_n,W_{n+1})\leq\dim(W_1)+...+\dim(W_ n)+\dim(W_{n+1})$
$=\dim(W_1,...,W_{k-1},W_k)\leq\dim(W_1)+...+\dim(W_{k-1})+\dim(W_k)$
Therefore, $\dim(W_1,...,W_k)\leq\dim(W_1)+...+\dim(W_k)$

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

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

Any help on these would be appreciated.

2. Your inductive step looks weird.

You have typed commas instead of plusses between the W_i. Anyway, I suggest that you prove your base case _also_ for k=2.

Then the inductive case could go like: Suppose the result is true for k=n, and let k=n+1. Then

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

where the first equality is just putting in parentheses, the first inequality is using the base case k=2 (with W_1+...+W_n and W_{n+1} being the two vector spaces), and the next inequality is using the induction hypothesis that the result is true for k=n.

As for b), what is your definition of direct sum?

3. Originally Posted by HappyJoe

You have typed commas instead of plusses between the W_i. Anyway, I suggest that you prove your base case _also_ for k=2.

Then the inductive case could go like: Suppose the result is true for k=n, and let k=n+1. Then

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

where the first equality is just putting in parentheses, the first inequality is using the base case k=2 (with W_1+...+W_n and W_{n+1} being the two vector spaces), and the next inequality is using the induction hypothesis that the result is true for k=n.

As for b), what is your definition of direct sum?
Thanks for clearing up my answer on the first half. I knew I'd done SOMETHING wrong.

As for the second half, I have quite a bit done, but again it could use some proof-reading. I probably have a few things out of proper order, which I feel would be the main issue.

We first have $v=w_1+...+w_k$, as provided by the question.
Then, we suppose that $v=w'_1+...+w'_k$.
This leads to the following: $w_1+...+w_k=w'_1+...+w'_k$
This implies that $0=(w_1-w'_1)+...+(w_k-w'_k)$,
$\Rightarrow w_1-w'_1=0,...,w_k-w'_k=0$
$\Rightarrow w_1=w'_1,...,w_k=w'_k$
Assuming the above is true, since each $w_j\in W_j$ is a unique choice, there could only be one solution to $w_1+...+w_k=0$, which is that $w_1=0,...,w_k=0$ for all $k>0$.

Therefore, $V=W_1\oplus...\oplus W_k$.

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Tell me if I've messed up anything. I'm suspecting right now that my proof just might be a bit out of order.

4. Here's a quickly written answer, I hope you can read my non-LaTeX'ified scribbles.

What's your definition of direct sum? Is it that V is the direct sum of W_1, W_2, ..., W_k, iff W_1+...+W_k = V and the W_i have only the 0-vector in common?

For the first part of your proof, where you show that (V is a direct sum of the W_i) implies (each v is a unique sum of elements w_i), how do you argue from 0 = (w_1-w_1') + ... + (w_k - w_k') to w_1-w_1'=0, ..., w_k-w_k'=0?

I don't quite get the other part. It may be because I'm not sure which definition you are working from, or maybe because I'm busy.