All right in this video
I'm going to look at the notion of a basis for a vector space and all the basis is it just ties together the
notions of linear Independence
[along] with the notion of a spanning set so we say a collection of vectors v 1 v 2 up to V sub N
We say that those are [a] basis for the vector space if two things happen?
[we] want the vectors to span the [vector] space and we also want these vectors the vectors in the set to be linearly?
Independent and if those two things happen we talk about having a basis of a vector space, so
To me kind of the notion [of] a basis is
it says you have sort of
Just enough vectors
You know to somehow
Write any other vector in that space as a combination of those other vectors, so somehow to me. It's kind of like
You have [a] you have the fundamental tools there to rewrite anything that you want?
They're kind of the basic building blocks to get to get the work done
is to me and there's a
Just enough you know if you maybe if you had some [extra] ones then they wouldn't be linearly independent [anymore]
So it says you have just enough to get the work done
[they're] linearly independent and then they span the entire the entire space
So if that makes any sense probably probably not the best intuitive idea
[so] a couple things here, so
the unit vectors 1 0 0 0 1 0
0 0 1 these are a basis [for] r 3 for example. I'm not going to justify that at all
[I] if you don't believe me [I] will leave that as an exercise
For you dear viewer to think about and to to justify
and hopefully
Shouldn't be too Terribly difficult from the definitions, and if you're not sure well, I am going to do an example here
to determine if these vectors the vector 1 1 1 0 [1] [0] 1 2 3
Are those also a basis [for] R 3?
Well, I'm going to do two things I'm going to show that they span R3 or try to
I'm also going going to show that they're linearly independent so the first thing. I'm going to do is actually show [that] they're
Dependent or not excuse me if they're linearly independent or not
so I guess we could say if they're linearly dependent or not that's also correct, but
Right if you answer one, you answer the other
So again for our vectors to be linearly independent
when we multiply these by
real number scalars a 1 a 2 a 3
We add them up. We set that equal to the [zero] vector. It says that they'll be linearly independent
If the only solution we get to this system is the trivial solution
So I'm really just trying to Determine if I have trivial solutions or non-Trivial solutions
So I'm going to do negative 1 row 1 add that to row 2 to get my new Row 2 and then I'm going to
Do negative 1 [row] 1 add that to Row 3 to get my new?
Row 3
alright, so we'll leave the first Row alone 1 0 1 0
So negative 1 plus 1 is 0 [we'll] get 0 plus 1
We'll get negative 1 plus 2 which is 1 and then we'll get a 0 if we do
negative 1 Plus 1 again that's 0 1 of 0 Plus 0
Then we'll have negative 1 plus 3 which [is] 2 and then again. We'll just have a 0
So the next thing I'm going to do well I can certainly just take a you know one half of
Row 3 to get my New Row 3
And again, you know I want a 1 and the first row first Column zeros everywhere else
I have that I want to want in the second row second column
I have that and I want zeros everywhere else [so] [hey] this problems obviously set up a little conveniently because that immediately happened
And then I want to one in the third Row third column
So I'm just going to divide by 2 so I'm going to instead of rewriting this all one more time
Hey
if we divide everything by 2 the only thing that's going to happen is the 2 will turn into a 1
and now we can just do negative 1 Row 3
add that to row 1 to get our new Row 1 and
negative 1 times Row 3 add that to row 2 [to] get our new Row 2 and again and the
sake of being brief here
The only thing that's going to happen is if we take negative 1 times Row 3 and add it to Row 1 we'll get negative
1 plus 1 if you'll make that into a 0 if we do negative
1 times Row 3 and add it to row 2 it'll make that 1 also into a 0 and
this now tells us that our solution would only be if our constants a 1 a 2 a 3
are all equal to 0 so it says a 1 would have to equal 0 a 2 would have to equal 0 and a 3
Would also have to equal 0 which [tells] us we only have the trivial solution
And since we only have the trivial solution that means yes. They are linearly independent
okay, so they are linearly independent, so
One things down now. Let's check if they span all of our [3]
[ok] so do they span all of our [3] and again what we're looking for [we] want to be able to take linear combinations
so a 1 times 1 1 1
plus a 2 times the vector 0 [1] 0 and a 3 times the vector 1 2 3
and
What we want to be able to do if we can if [we] can use these vectors to span all of our 3 it says?
We should be able to figure out constants, so that we get any old vector x Y&Z that we want okay?
So we have to justify that we can do this so I'm going to write it as a just a linear system
So we would have 1 a 1 plus
well 0 a 2 plus 1 a 3 equals x and again
I've got some examples about this when we talk about in the video with spanning sets, so this is just a really a little bit
redundant just tying those [two] those two concepts together independence and
expanding sets so our next equation would be 1 a 1 plus 1 a 2
Plus to a sub 3 we can set that equal to y, and [then] we would have 1 a 1 plus
well, we've got 0 a 2 and then we would have 3 a 3
equal to z
so [all] I'm going to try to do is just simply solve for a 1 a 2 and a 3
in terms of xy and [z]
The way I'm going to do it is just a little bit of substitution. I think
So let's see if we look at our first two
Equations we've got 1 a 1 plus 1 a 3 that equals x we [have] 1 a 1 plus
3 a 3 that equals
z
well if we solve for the first one for a 1 that's as well a 1 would equal x minus a
sub 3 and now I'm going to plug that into
my other equation that I kind of peeled off here, so
[it] says we have x minus
A sub 3 again, I'm just plugging that in for a 1 here, okay?
But then we still have plus 3 times a sub 3 equals z
Well that tells us. Hey that's going to be x plus 2 times a sub 3 equals z and again
I'm trying to solve for my constants a 1 a 2 a 3 well here. We can solve for a sub 3
We can just say well
we can subtract the x over so z minus x and then we could divide by 2
Or equivalently multiply that by 1/2 and that will give us our value for a sub [serie]
Okay, so that's one thing that's going to be useful well now. I'm just going to simply plug that back into our other equation
So we said hey a 1 is going to be x minus a sub 3
well, let's plug that in we've got 1/2 times z minus x
And again, I'm just going to solve for a 1 here. I've already got it sold. I'm just going to simplify
So we've got x minus 1/2 times z I guess we would have what plus
so plus 1/2 x and
Now I'm just going to combine my like terms so 1 [x] plus 1/2 x is going to be 3 halves x minus 1/2
z so now I've got a 1 in terms of my components
X Y&Z we don't need the y obviously in this case just x and z and last but not least
Let's see I think the lat the only equation we haven't really played with yet is our second equation from before right?
[we] haven't done anything with this one yet
We've got a 1 solved for we've got a 3 solved for we just need to solve for a sub 2 now
So it says a sub 1 let's plug that in well. We know that that's 3 over [2x] minus 1 over 2
z it says then we have plus a sub 2 that's the one we don't know and then 2 times a
sub 3 and again a sub 3 we said is 1/2 times z minus x
So let's see 2 times
1/2 Times z Minus x
And we said [all] [of] that is going to be equal to y, so let's set it equal to y
And again now. It's just the same thing as before we're going to combine [like] terms and then solve for a sub 2
well, let's see the 2 and a half would cancel I guess we have 3 halves x it looks like we would have minus an
x so that would leave us with 1/2 x
It looks like we have negative 1/2 z plus 1 z so it looks like [a] positive 1/2
z to me
[we] still have our plus a sub 2 hanging out?
Then that's equal to y and again now. We can just simply solve for [a] sub 2 it says a sub 2 would be
if we subtract negative 1/2 x
plus
Y
minus 1/2
z and again now we have
We've solved for a sub 2 as well, so the idea is again by doing this you know we've got values
It says if you want to you know write any vector x Y&Z you know pick your favorite numbers x
[Y&Z] and now we've got a formula [that] will determine what a 1 should equal what a 2 should equal and what three should equal?
We know a formulas to figure out these real number coefficients
So that after we do the arithmetic
we'll get our desired vector and again these are what the
Coefficients would have to equal you know in terms of xy and z in terms of xy and z and?
We've also got a sub 3 in terms of xy and z
Again obviously two of them don't use y but that's all there is to it
so now we've [shown] that our vectors are linearly independent they also form a spanning set and
therefore
these
three vectors also form a basis for R
Sub 3 so the Moral of the story is [you] don't just have to use the unit vectors for a basis you can certainly use
Other vectors as well