MHB Proof of vector dimensions using inequalities

[TheFallen018]

Hello all!

I've got this problem I'm trying to do, but I'm not sure what the best way to approach it is.

View attachment 8713

It's obvious that there can only be 2 dimensions, because there's only two linearly independent vectors in the span. However, what would be a good way of using the inequalities to prove it? I can't think of a good way to do that.

Any ideas would be great!

Thanks :)

 

[I like Serena]

Hey Fallen18!

It is not given that there are 2 linearly independent vectors in the span.
That's why we only get $\le 2$ from the span.

 

[TheFallen018]

Hey Fallen18!

It is not given that there are 2 linearly independent vectors in the span.
That's why we only get $\le 2$ from the span.

Good point. w1 and w2 could technically be multiples of each other, making W a one dimensional set. However, how would you do a $\ge$ 2 proof? Couldn't w2 be a multiple of w1, and w3 also be a multiple of w1? In that case I would be unsure on how to continue. Thanks

 

[I like Serena]

Good point. w1 and w2 could technically be multiples of each other, making W a one dimensional set. However, how would you do a $\ge$ 2 proof? Couldn't w2 be a multiple of w1, and w3 also be a multiple of w1? In that case I would be unsure on how to continue. Thanks

W could even be 0-dimensional, since the $w_i$ could be zero-vectors.
Hpwever, it is also given that the $v_i$ are linearly independent and also in W...

 

[TheFallen018]

W could even be 0-dimensional, since the $w_i$ could be zero-vectors.
Hpwever, it is also given that the $v_i$ are linearly independent and also in W...

Oh, yes, I should have made that clear in my first post. My reasoning for it having to be two dimensional, is since v1 and v2 are in the set and linearly independent, then they must be part of the span, in that for example w1=v1 and w2=v2, or something like that. However, now that I think about it, it could mean that v1 and v2 are only independent to each other, which makes them far less useful. I'm hoping my first assumption was right though. What do you think?

Edit:
Oh, I think I see the significance of that now. If v1 and v2 are linearly independent, then w1, w2 at least must be unique vectors. If they were the zero vectors, or multiples of each other, v1 and v2 couldn't be linearly independent. Therefore w1 and w2 are linearly independent, and since w3 is a combination of w1 and w2 then this must be a two dimensional set. However, that's still not solving it using inequalities, so I'm not sure that will do the trick.

 

[I like Serena]

Oh, yes, I should have made that clear in my first post. My reasoning for it having to be two dimensional, is since v1 and v2 are in the set and linearly independent, then they must be part of the span, in that for example w1=v1 and w2=v2, or something like that. However, now that I think about it, it could mean that v1 and v2 are only independent to each other, which makes them far less useful. I'm hoping my first assumption was right though. What do you think?

Edit:
Oh, I think I see the significance of that now. If v1 and v2 are linearly independent, then w1, w2 at least must be unique vectors. If they were the zero vectors, or multiples of each other, v1 and v2 couldn't be linearly independent. Therefore w1 and w2 are linearly independent, and since w3 is a combination of w1 and w2 then this must be a two dimensional set. However, that's still not solving it using inequalities, so I'm not sure that will do the trick.

Indeed, since $\mathbf v_1$ and $\mathbf v_2$ are linearly independent, $W$ must have at least dimension $2$, which is what we were still looking for.
Effectively you are using that fact to conclude that $\mathbf w_1$ and $\mathbf w_2$ must be linear independent (and not zero) as well.

That is:

1. $\mathbf v_1$ and $\mathbf v_2$ are linearly independent and in $W$, therefore $\text{dim}\ W \ge 2$.
2. $\mathbf w_1$, $\mathbf w_2$, $\mathbf w_3$ span $W$ and $\mathbf w_3$ is a linear combination of $\mathbf w_1$ and $\mathbf w_2$, therefore $\text{dim}\ W \le 2$.
3. Since $\text{dim}\ W \ge2$, $\mathbf w_1$ and $\mathbf w_2$ must be linearly independent.