Linear Algebra Spanning Set for a Vector Space

Welcome back to Educator.com and welcome back to linear algebra.0000

Today we are going to talk about something called the span of a set of vectors.0004

It means exactly what you think that it means. If I have a collection of vectors, 2, 5, 10, the number does not actually matter that much.0011

We want to talk about all of the possible linear combinations of those vectors, that are possible... all of the vectors that can be built from that particular set.0021

So, for example, if I take R2... the normal plane, I know that I have the vector in the x direction, I know that I have the vector in the y direction, and if I take any collection of those, multiplied by constants, let us say 5i + 6j, I can represent every single vector in R2.0033

So, those two vectors, we say it actually spans R2.0050

So, that is just sort of the general description that a span is.0055

Unfortunately, in this case, the name actually gives you an idea of what it is that you are talking about, so it is not strange. So, let us start with a couple of definitions and let us see what we can do.0059

Okay. Now, in a vector space, there is an infinite number of elements and then the reason for that is if I have at least one element in that space, and I know there is at least one, I know that I can multiply that element by any number that I want. Any constant.0072

Therefore, since that constant is just a real number, there are an infinite number of elements in that vector space.0091

However, what we want to do is we want to see if we can find a finite number of elements in that vector space that when I take certain combinations of them, linear combinations of them, that we can describe the entire space.0098

That means all infinite vectors based on just that finite set of vectors.0114

So, let us actually write out definitions down.0122

Okay... vectors v1, v2 and so forth onto vk are said to span v, which is our vector space.0134

If every vector in v can be written as a linear combination of the v1, v2, v3.0156

SO, now we have actually written it down. If I have vectors v1 through vk, let us say 6 of them.0188

And... if any linear -- excuse me -- if any combination of those vectors, they do not have to all be included, you know some of the constants can be 0, but if some combination of those vectors can represent very single vector in that vector space, then we say that that set of vectors actually spans the vector space.0194

Okay. Now, let us list the procedure to check to see if a set of vectors actually spans a vector space.0218

Procedure to check if the set of vectors spans a vector space, so vs, for vector space.0233

So, let us see... you know what... let us leave it as blue for right now.0253

Choose an arbitrary vector v in the vector space, so when you are given a vector space, just choose some arbitrary vector.0260

So, if you are given 4-space, R4, then just choose the random vector (a,b,c,d), or you can call it (x,y,z,t), just some random vector and label it... to determine if v is a linear combination of the given vectors.0273

So, this is basically just an application of the definition, which is what definitions are all about.0311

Let me just take a quick second to talk about definitions real quickly.0315

Often in mathematics, we begin with definitions. They are a basic element.0319

We use those definitions to start to create theorems and we sort of build from there, build our way up from the bottom if you will.0325

If you find that you have lost your way in mathematics, more often than not, you want to go back to your definitions, and 90% of the time, the problem is something is either missing from a definition, or there is a definition that the student has not quite wrapped his mind fully around.0333

Again, mathematics is very, very precise. It says exactly what it wants to say, no more and no less.0349

Okay. Determine if v is a linear combination of the given vectors... definition.0357

If so, if it is a linear combination, then yes, the vectors actually span the vector space.0365

If not, if there is no way to form a linear combination, then no. Okay.0373

So, again, when you are forming a linear combination, you are taking a constant, multiplying it by a bunch of vectors and setting it equal to some, in this case an arbitrary constant.0385

So, once again, we are going to investigate the linear system, the linear systems are ubiquitous in linear algebra. Okay.0394

So, let us start with an example here.0403

Let us go to the next page. So, first example. Let us consider R3.0407

So, regular 3 space... (x,y,z), the space we live in.0418

We are going to let v1 = (1,2,1), we will let v2 = (1,0... -- oops, excuse me -- (1,0,2), and we will let our third vector be (1,1,0).0424

I wrote these out in the form of a list, in terms of their coordinates... I could write the vertically, I could write them horizontally without spaces, however you want.0447

Now, the question is these three vectors that I have chosen randomly... do they... so do v1, v2, v3 span R3?0455

Are these vectors enough to represent all of R3.0469

Is a linear combination, any linear combination, any of these three vectors... can I find any vector in R3 and use these three to represent it?0475

Well, let us see. Okay.0484

Well, first thing we do from our procedure, let us choose an arbitrary vector in R3... arbitrary v in R3.0489

So, let us choose, let us say that v is just equal to (a,b,c,d), and again variables do not matter. It is just some random vector.0503

Okay. Now, we want to see the second thing that we are going to check is are there constants c1, c2, c3 such that, well, c1v1 + c2v2 + c3v3, a linear combination equals v.0510

That is what we are doing. That is all we are doing. We are taking an arbitrary vector, we are setting it equal to the constant times the vectors that we have in our set, and we are going to solve this linear system.0538

So, this is the vector representation, the simplest representation... now we are going to break it down a bit.0548

This is, of course, equal to c1 × the first vector, which is (1,2,1)... I am going to go ahead and write them vertically.0554

That is just a personal choice of mine. You are welcome to write them any way you please. I like to do them vertically because it keeps the coefficient of the ultimate matrix that we are going to do... systematic.0561

These becomes the columns of the particular matrix.0571

Plus c2 × (1,0,2)... I hope that 2 is clear... + c3 × third vector, our third vector is (1,1,0), and we are setting it equal to (a,b,c).0576

We want to find out if this linear system actually has a... so, let us write this system... when we actually multiply these c1's out, we get this... I would like you to at least see one of them.0596

We get c1 + c2 + c3 = a.0607

We get 2c1, c20, I will just leave that over here, + c3 = b, and I get c1 + 2c2, and there is nothing over here = c.0615

This is, of course, equivalent to the following augmented matrix... (1,1,1,a)... (1,1,1,a).0632

Again, I am taking the coefficients of c1, c2, c3... c1, c2, c3 is what I am looking for.0640

Can I find a solution to this? If I can, then yes... (2,0,1,b), (1,2,0)c, so this is the system that we are going to solve.0646

We are going to subject it to reduced row echelon form. I will not go ahead and show you the reduced row echelon form... I of course did this with my computer, with my Maple software.0663

Fast and beautiful. As it turns out, this does have a solution. In other words, it has a non-trivial solution, and here is what it looks like.0672

You end up with c3 = (4a - b - 2c)/3 -- oops, these strange lines that show up on here.0683

Let us see... c2 = (a - b + c)/3... let me go ahead and put parentheses around so that you know the numerator is that way, because I am writing my fractions not two dimensional, but in a line.0703

c1 = (-2a + 2b + c)... and again, all divided by 3.0729

So, for any choice of a, b, or c, I can just put them in here and the constants that I get end up being a solution to this system.0741

So, because there is a solution to this system... this system... that means there is a solution to this because these are all equivalent.0750

That means that any vector that I choose, and again, I just chose it (a,b,c)... is... I can find constants for them and these are the explicit values of those constants no matter what vector I choose.0760

So, yes... let us try this again... so here the answer is yes... v1, v2, v3 do span R3.0771

Turn this and make this red... that vector, that vector and that vector are a perfectly good span for R3.0794

Now, you know, of course, that R3, or if you do not, I am telling you right now... that R3, the standard 3 vectors that we use as the set which spans the space is of course the i vector, the j vector in the y direction, and the k vector in the z direction.0806

Those are mutually perpendicular. Well, as it turns out, you can have any collection of vectors that may span a vector space.0825

There is no particular reason for choosing one over the other, so there are an infinite number of them, but in certain circumstances... it makes sense to choose one over the other.0833

In the case of the i,j,k, we choose it because they have the property that they are mutually orthogonal -- excuse me.0845

Which actually -- excuse me -- makes it easier to deal with certain things.0855

Okay. Let us consider another example. Let us consider... we will do this example in red... p2.0863

If you recall p2 is the vector space of all polynomials... all polynomials of degree 2 or less.0877

So, we will let our set s, this time we will actually do it in set notation... p1t, p2t, oh this p2 right here has nothing to do with this p2.0889

This is a general symbol for p2, the space of polynomials of degree 2. This just happens to be number 2 in the list.0903

p1... let us define that one as t² + 2t + 1, and we will say that p2 of t is t² + 2.0912

Okay. We want to know does s... do these 2 polynomials, are they enough to span all of p2.0928

In other words, can I take two constants c1 and c2, and multiply them by this... by these 2... c1 × this one, c2 × this one.0942

Can I always find constants such that every single polynomial, every second degree polynomial, or first degree polynomial, remember it is degree 2 or less... can be represented by just these 2 vectors... well, let us find out.0953

So, again, the first thing that we do is we choose an arbitrary vector in p2... and an arbitrary vector in the space of polynomials looks as follows... at² + bt + c, right? because it is t², right?0968

And now... we want to show the following... we want c... c1 × p1t + c2 × p2t = our arbitrary vector that we picked.0985

This one up here equals at² + bt + c, so again, we are just setting up a basic equation... a linear combination of the vectors that we are given... does it equal our arbitrary vector?1008

Well, let us go ahead and expand this out based on what they are... so, we have c1 × p1t which is t² + 2t + 1 + c2 × t² + 2... okay.1020

We want that to equal at² + bt + c... let us multiply this out.1045

Now, when you multiply this out, this is going to be c1t², 2c1t + c1, and then 2c2² + 2c2.1053

I am going to skip that step... and just... imagine that I just multiplied it, basic distribution, you know, something from algebra 1.1063

Then I am going to combine terms in t², in t, and in t to the 0 power.1068

So, it is going to look like this when I actually expand it. It is just one line that I am skipping.1075

It is going to be c1 + c2 × + 2c1... sorry... t²... 2c1 × t + c1 + 2c2 = at² + bt + c.1080

Well, we have an equality sign here. We have some -- change this to blue -- this is the coefficient of t² here.1103

This is the coefficient of t² in the equality, so that equals that.1112

That is our first equation... c1 + c2 = a.1117

Well, 2c1 is the coefficient of t, here b is the coefficient of t. They are equal on both sides, so this one becomes 2c1 = b.1125

Then, we do this one over here. We have c1 + 2c2 = c... c1 + 2c2 = c.1139

This is going to be equivalent to the following... (1,1,a)... I am just taking coefficients... (2,0,b), and (1,2,c).1150

Now, when I subject that to reduced row echelon form, this one I do want you to see...1164

So, we subject that to Gauss Jordan elimination for reduced row echelon... we get (1,0,2a-c).1173

(0,1,c-a), and we get (0,0,b-4a+2c). We get this as the reduced row echelon of the system that we just took care of.1185

Now, this is possible if and only if this thing right there... so you see this 0 right here, this... that means this thing right here... the b - 4a + 2c has to equal 0 for this thing to be consistent.1201

The only way that that b - 4a + 2c is equal to 0 is if b = 0, a = 0, c = 0. That is just the trivial solution.1223

So, there is no solution to this system. So, the answer to this one is no.1234

p1 and p2, t, those two polynomials do not span p2, which is the vector space of polynomials of degree 2 or less.1239

So, it is not enough. I need some other vector, I do not know, maybe 2 or 3 of them.1254

Let us try... let us go back to red here for example number 3.1268

These will just list, you know these already... i and j span R2.1278

We know that i,j, and k, the three unit vectors in the (x,y,z) direction span R3, and so on, onto R4... R5.1287

So, e1, we do not give the letters anymore, after 4... we just call them e.1298

e1, e2, and so on all the way to eN... They span n-space, which is RN.1305

Now, you have probably already noticed this, but notice, I have 2 vectors that span R2, three vectors that span R3, N vectors needed to span RN, that is a general truth.1318

We will talk more about that in a minute... well, actually the next lesson.1330

Okay. Now we are going to get to a profoundly, profoundly important example.1336

This one, we want to do very, very carefully... therefore, so, let us consider the following homogeneous equation... ax = 0, such that a, and we are actually going to explicitly list this vector... this matrix, excuse me, (1,1,0,2), (-2,-2,1,-5), (1,1,-1,3), (4,4,-1,9). Okay.1343

So, we have this homogeneous system, matrix × some vector x is equal to the 0 vector. Here is our matrix a.1386

In other words, what are my x values that will actually make this true such that when I multiply this by some vector x, which I will put in blue...1397

Let us just say this is going to be x1, x2, x3, x4... what are the values of x that will actually make it true so that when I multiply these, I end up with 0.1407

It might be one vector, it might be an infinite number of vectors, it might be no vectors.1417

So, if you remember what we called the set of vectors that actually make this true, we called it the null space.1423

The solution space of the homogeneous system... in other words, the vectors that make this space be true. We called it the null space.1430

So, let me just write that down again... the null space is a very, very important space. Okay.1437

Now, one thing that we also know about the null space is the null space is a subspace... of... in this case, R4, remember?1445

So, if we happen to be dealing with a null space in R4, you have these 1, 2, 3, 4... 4 by 4, we are looking for a vector which has 4 entries in it, so it is from 4-space.1465

And... remember that we proved that it is actually a null space, it is a subspace of R4.1476

Not just a subset... it is a very special kind of subset... it is an actual subspace.1480

In other words, it is a vector space in its own right. It is as if I can ignore the rest of R4, just look at that, and I can treat it the same way I would the rest of that vector space, it has special properties... all of the properties of a vector space.1486

Okay. Now, can we find a set of vectors... here is our problem here... can we find a set of vectors that stands the null space?1499

Let me write that down... can we find a set of vectors... set which spans... singular because it is set... this null space.1511

So, again, we are not looking to expand the entire R4. We are just looking for something that will span this particular null space. The null space based on this particular matrix.1535

Well, let us solve this system... Let us see what x's we can come up with.1546

So, when we solve the system, we create the augmented matrix. When I take this and I put 0's over on the final column in matrix a, and I end up subjecting that to reduced row echelon form, which I will not do... well, actually you know what, let me go ahead and write it all out. It is not a problem.1555

So, I have 1...1... yes, (1,1,0,2,0), (-2,-2,1,-5,0), (1,1,-1,3,0)... and (4,4,-1,9,0).1581

So, this is our augmented matrix. This is just this thing in matrix form, and then I am going to subject this to Gauss Jordan elimination to convert it into reduced row echelon, so let us write out the reduced row echelon form of this.1613

It is going to be (1,1,0,2,0), (0,0,1,-1,0), (0,0,0,0,0)... 0's everywhere.1626

Okay. So, this is our reduced row echelon, and remember... this represents x1, x2, x3, x4. This is the -- we are looking for a vector... a 4 vector.1646

So, let us take a look here. This one is fine. This one has a leading entry -- let me go to red -- so, this one is good, and this one has a leading entry.1661

This one does not have a leading entry. The x2. So... x2... x3, x4, yes.1672

So, that second does not have a leading entry and the fourth does not have a leading entry.1681

Remember, when we do something, when we convert it to reduced row echelon, the columns that do not have a leading entry actually are free parameters... I can call them s, t, x, y, they can be any number I want.1686

Then I solve for the other two. So, let us go ahead and set x2 = r, and we will set x4 = s, they can be any numbers, they are free parameters.1697

Then, what I get is x1 + x2 + 2x4, that is what this line here says... x1 + x2 +2x4 = 0.1716

Therefore, x... x2 is R, x4 is S, so x1 = -R - 2s.1734

So, I have x2, I have x4, I have x1 that I just calculated, and now I will do x3... x3 here... x3... this right here...x3 - x4 = 0.1745

So, x3 = x4, which is equal to x... s... I am sorry.1768

So, we have x2 is r, x4 is s, x3 is s, excuse me, and x1 is -r - 2s. Let me rewrite that a little bit differently.1774

I am going to write that as the following -- let me go back to blue here.1785

x1 = -r - 2s, x2 = r, and there is a reason why I am writing it this way, you will see in a minute... x3 = s -- I will put the s over there in that column.1791

x4 = s, well take a look at this x1, x2, x3, x4... I have r's here, 0, I have 2, 0, s, s.1811

I can rewrite this as the following in vector form. This is equivalent to (x1, x2, x3, x4) equal to, let me pull out the r from here, and I can just take these coefficients (-1,1,0,0).1823

And... plus, now I can pull out an s from here, and I can take these coefficients, (-2,0,1,1), if that is not clear, just stop and take a look at what it is that I have done.1850

I will just treat this as a column... treat this as a column... Okay.1864

So, notice what I have done. I have taken the solution space which is x1, x2, x3, x4, this is because i have solved the system... these are all the possible solutions.1870

There is an infinite number of them because r and s can be anything. I have written it in vector form this way, as a linear combination of this vector and that vector.1884

These are just arbitrary numbers, right? The r and the s are just arbitrary numbers, therefore I have expressed the solution set, which is this thing of the homogeneous system based on that matrix.1893

I have expressed it as the linear combination of this vector and this vector.1906

Therefore, those 2 vectors (-1,1,0,0) and (-2,0,1,1), they actually span the null space.1912

The null space has an infinite number of solutions, that is what our system tells us here.1930

Well, I know that I can describe all of those solutions by reducing it to two vectors, any linear combination of which will keep me in that null space.1937

It will give me all of the vectors, all of the vectors are represented by a linear combination of this vector and that vector.1947

They span the null space of the homogeneous system.1953

Again, profoundly important example. Go through this example again carefully to understand what it is that I did.1957

I had a homogeneous system, I solved that homogeneous system for the solution space. I represented that solution space... once I have that... I represent it as vectors, and these vectors that I was able to get -- because these are arbitrary constants -- well, that is the whole idea behind a span.1965

I was able to represent the higher solution space but only 2 vectors. That is extraordinary.1983

Okay. Thank you for joining us here at Educator.com for the discussion of span, we will see you next time.1990