00:01
V2 v3 v4 and v5 and the set v2 v3 v4 and v5 are both linearly independent then their spans are equal this is false when you have a linearly independent set and you take its span then the dimension of the span of linearly independent vectors equals how many vectors are there? so in this first one, this span would have a dimension of five, and this second span would have a dimension of four.
01:17
And the same space can't have two different dimensional values.
01:23
So this is false.
01:27
I'm scrolling down to the next one in the true false.
01:30
The set of all diagonal matrices is a vector space.
01:34
All right this is true and the way we're going to show that we need to take a diagonal matrices a and b and i need to have scalers alpha and beta and what i need to show is that alpha a is diagonal plus beta b is diagonal plus beta b is diagonal right because recall we know the set of all matrices is a vector space so i really only need to show the diagonal matrices are a subspace of that and in order to show subspace i really just need closure under vector addition and scalar multiplication which showing that this remains a diagonal matrix will do that so a is a diagonal matrix so i'm going to write a a1 dot dot a n zero zero everything off the diagonal is zero and b as b1 bn zero zero and then alpha a plus beta b b equals alpha a1 a n plus beta b1 bn which is equal to alpha a1 alpha a n and everything off diagonal and beta b1, beta bn, right? because you just multiply the scalar by every entry in your matrix, and anything times zero is zero.
03:37
So the only thing that really gets multiplied is the diagonal entry.
03:41
And now you add up every entry to their corresponding entry, and you get alpha a1 plus beta b1, all the way through alpha an, plus beta bn.
03:54
Everything off diagonal still remains zero, because zero plus zero is zero.
03:58
So this is diagonal.
04:03
Which means we have closure under vector addition and closure under scalar multiplication, which means that, yes, they do form a vector space.
04:15
Moving on to the next one, if both of these lists of vectors span v, then this list is linearly independent.
04:43
Oh, apologies, linearly dependent.
04:50
And this is true.
04:53
And in order to show that, right, recall the fact that i said earlier that if v1, v2, v3, b4, v5 were linearly independent, then their span, which is v, would have a dimension of v, which is their span, would be five.
05:40
However, because v can be spanned by just four vectors, that implies that the dimension of v is no greater than four.
06:02
I don't know that it is four, because i don't know if v1, if b2, v3, b4, v5 is linearly independent, but i know it has to be capped at four because it can be spanned by four vectors.
06:12
So therefore, we have that v1, v2, v3, v4, v5 are veneerly dependent because of this contravention.
06:40
All right.
06:42
On to the next problem...