Showing that the candidate basis does span C(A) Showing that just the columns of A associated with the pivot columns of rref(A) do indeed span C(A).
Showing that the candidate basis does span C(A)
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- Two videos ago we asked ourselves if we could find the
- basis for the columns space of A.
- And I showed you a method of how to do it.
- You literally put A in reduced row echelon form, so this
- matrix R is just a reduced row echelon form of A.
- And you look at its pivot columns, so
- this is a pivot column.
- It has a 1 and all 0's, this is a pivot column, 1 and all
- 0's, and the 1 is the leading non-zero term in its row.
- And this is a pivot column, let me circle them, these guys
- are pivot columns, and this guy's a pivot
- column right there.
- You look at those in the reduced row echelon form of
- the matrix, and the corresponding columns in the
- original matrix will be your basis.
- So this guy, this guy, so the first,
- second, and forth columns.
- So if we call this a1, this is a2, and let's call this a4,
- this would be a3, and this is a5.
- So we could say that a1, a2, and a4 are a basis for the
- column span of A.
- And I didn't show you why two videos ago.
- I just said this is how you do it.
- You have to take it as a bit of an article of faith.
- Now in order for these to be a basis, two
- things have to be true.
- They have to be linearly independent, and I showed you
- in the very last video, the second in our series dealing
- with this vector.
- I showed you that by the fact that this guy is r1, this guy
- is r2, and this guy is r4, it's clear that these guys are
- linearly independent.
- They each have a 1 in a unique entry, and the rest of their
- entries are 0.
- We're looking at three pivot columns right now, but it's
- true if we had n pivot columns.
- That each pivot column would have a 1 in a unique place,
- and all the other pivot columns would
- have 0 in that entry.
- So there's no way that the other pivot columns, any
- linear combination of the other ones, could never add up
- to each of them.
- So these are definitely linearly independent.
- And I showed you in the last video that if we know that
- these are linearly independent, we do know that
- they are, given that R has the same null space as A, we know
- that these guys have to be linearly independant, I did
- that in the very last video.
- Now the next requirement for a basis, we checked this one
- off, is to show that a1 a2 and an, that their span equals the
- column space of A.
- Now the column space of A is a span of all five of these
- vectors, so I had to throw a3 in there and a5.
- So to show that just these three vectors by themselves
- span our column space, we just have to show that I can
- represent a3 and a5 as linear combinations
- of a1, a2, and a4.
- If I can do that, then I can say then
- these guys are redundant.
- Then the span of a1, a2, a3, a4, and a5 doesn't need the a3
- and the a5 terms, that we can just reduce it to this.
- Because these guys can be represented as linear
- combinations of the other three.
- These guys are redundant.
- And if we can get rid of them we can show that these guys
- can be represented as linear combinations of the other,
- then we can get rid of them.
- And then the span of these three guys would be the same
- as the span of these five guys, which is of course the
- definition of the column space of A.
- So let's see if we can do that.
- Let me fill in each of these column vectors a1 through a5,
- and then each of these column vectors let me label them r1,
- r2, r3, r4, and r5.
- Now let's explore the null spaces again.
- Or not even the null spaces, let's just explore the
- equations Ax is equal to -- let me write it this way --
- instead of x let me write x1, x2, x3, x4, x5 is equal to 0.
- This is how we define the solution set of this.
- All the potential x1's through x5's or all the potential
- vectors X right here, that represents our null space.
- And then also let's explore all of the R times x1, x2, x3,
- x4, x5's is equal to 0.
- This is the 0 vector, in which case you would have four
- entries in this particular case.
- It would be a member of Rm.
- So these equations can be rewritten.
- I can rewrite this as -- what were the column vectors of A?
- They were a1, a2 through a5.
- So I can rewrite this as x1 times a1 plus x2 times a2 plus
- x3 times a3 plus x4 times a4 plus x5 times
- a5 is equal to 0.
- That was from our definition of matrix vector
- multiplication, this is just a bunch of column vectors a1
- through a5, I drew it up here.
- I can just rewrite this equation like this.
- Similiarly, I can rewrite this equation as the vector r1
- times x1 or x1 times r1 plus x2 times r2 plus x3 times r3
- plus x4 times r4 plus x5 times r5 is equal to 0.
- Now we know that when we put this into reduced row echelon
- form the x variables that are associated with the pivot
- columns are -- so what are the x variables associated with
- the pivot columns?
- Well, the pivot columns are r1, r2, and r4.
- The x variables associated with them, we can call them
- pivot variables, and the ones that are not associated with
- our pivot columns are free variables.
- So the free variables in this case, x3 and
- x5, are free variables.
- And that applies to A.
- All of the vectors x that satisfy this equation also
- satisfy this equation, and vice versa.
- They're the exact same null space, the exact
- same solution set.
- We can also call this x3 and this x5 as free variables.
- Now what does that mean?
- We've done multiple examples of this.
- The free variables, you can set them to anything you want.
- So x3 in this case and x5 you can set it to any real number.
- You can set to anything you want.
- And then from this reduced row echelon form we express the
- other pivot variables as functions of these guys.
- Maybe x1 is equal to Ax3 plus Bx5.
- Maybe x2 is equal to Cx3 plus Dx5.
- And maybe x4 is equal to Ex3 plus Fx5.
- That comes directly out of literally multiplying this guy
- times this equals 0, you'd get a system of equations that you
- could solve for your pivot variables in terms of your
- free variables.
- Now given this, I want to show you that you can always
- construct one of your -- in your original matrix.
- So if we go to our original matrix, you can always
- construct one of the vectors that are associated with the
- free columns.
- You can always construct one of the free vectors using the
- linear combination of the ones that were associated with the
- pivot columns before.
- And how do I do that?
- Well, let's say that I want to find some linear combination
- that gets me to this free column, that gets me to a3.
- So how could I do that?
- Let me rearrange this equation up here.
- So what do I get?
- I'm sorry.
- That's x3 a3.
- If I subtract x3 a3 from both sides of the equation, I get
- minus x3 a3 is equal to x1 a1 plus x2 a2 plus -- I don't
- have the 3 there -- plus x4 a4 plus x5 -- sorry, x isn't a
- vector-- x5 a5.
- This, I guess salmon colored statement here, is just
- another rewriting of this equation right here.
- And all I did is I subtracted this term right here, x3 a3,
- from both sides of the equation.
- Now x3 is a free variable.
- We can set it to anything we want, and so is x5.
- So let's set x3 is equal to minus 1.
- Then this term right here becomes a 1, because that was
- a minus x3.
- And let's set x5 equal to 0.
- So if x5 is equal 0, this term disappears, and I did that
- because x5 is a free variable.
- I can set them to anything I want.
- Now I've written a3 as a linear combination of, I guess
- you could call it my potential basis vectors right now, or
- the vectors a1, a2, and a4.
- They're the vectors in the original matrix that were
- associated with the pivot columns.
- Now in order to show that I can always do this, we have to
- show that for this combination there's always some x1, x2,
- and x4 that satisfy this.
- Well, of course there's always some x1, x2 that satisfy this,
- we just have to substitute our free variables, x3 is equal to
- minus 3 and x5 is equal to 0, into these equations that we
- get from our system when we did it with the reduced row
- echelon form.
- In this case you have x1 is equal to minus A plus 0, x2 is
- equal to minus C, so on and so forth.
- So you can always do that.
- You can always express the vectors that are associated
- with the non-pivot columns as linear combinations of the
- vectors that are associated with the pivot columns.
- What I just did for a3, you could just as easily have done
- for a5 by subtracting this term from both
- sides of the equation.
- Setting x5 to negative 1 and setting x3 to 0 so that the 3
- term disappears, and you could run the same exact argument.
- So given that, I've hopefully shown you, or least helped you
- see or made you comfortable with the idea, that the
- vectors -- let me do them in a nice vibrant color -- these
- magenta color vectors here that are associated with the
- free columns or with the free variables, the free variables
- were x3 and x5, those were these columns right here, that
- they can always be expressed as linear combinations of the
- other columns.
- Because you just have to manipulate this equation, set
- the coefficient for whatever you're trying to find a linear
- combination for equal to minus 1, and set all the other free
- variables equal to 0 that you're not solving for.
- And then you can get a linear combination of the vectors
- that are associated with the pivot columns.
- So given that, we've shown you that these free vectors, and
- I'm using my terminology very loosely, that these ones that
- are associated with the non pivot columns can be expressed
- as linear combinations of these guys.
- So they're unnecessary.
- The span of this is equivalent to the span of this, the span
- of this is the column space of A, so the span of this is the
- column space of A.
- So in the last video I showed you that these guys are
- linearly independent, and now I've showed you that the span
- of these guys is the column space of A.
- So now you should be satisfied that these vectors that are
- associated -- let me do it in a blue color -- that that
- column vector, this column vector, and this column
- vector, that are associated with the pivot columns in the
- reduced row echelon form of the matrix, do indeed
- represent a basis for the column space of A.
- Anyway, hopefully you didn't find that too convoluted.
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