For more information, please see full course syllabus of Linear Algebra

For more information, please see full course syllabus of Linear Algebra

### Coordinates of a Vector

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Coordinates of a Vector 1:07
- Coordinates of a Vector
- Example 1
- Example 2
- Example 3: Part A
- Example 3: Part B

### Linear Algebra Online Course

### Transcription: Coordinates of a Vector

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

*Today we are going to be talking about something... continue our discussion, of course, about the structure of a vector space.*0005

*We have been talking about bases, linear independence, span, things like that.*0011

*Today we are going to be talking about something called a coordinates and a change of basis.*0016

*So, up to now, we have been talking about a random basis, a set of vectors that actually spans a given space.*0021

*Either the entire space, or a subspace of that space.*0031

*Well, we have not really cared about the order of those vectors -- you know -- we just say v1, v2, v3, the basis will do.*0036

*In this particular lesson, we are going to start talking about the particular order of a basis.*0042

*So, if I put vector 1 in front of vector 2, and if I switch the order, it actually changes the basis and it is going to change something called the coordinates of the particular vector that the basis is actually representing.*0048

*So, let us start with a couple of definitions, and we will jump right on in.*0061

*Now, let us say we have v1... let us make these concepts a little bit more clear so we can see it notationally.*0069

*So, let us say v1 is a set of vectors, v1, v2, and v3.*0080

*So, we have a basis which has 3 vectors, so we are talking about a 3-dimensional vector space.*0088

*If we also have, let us say, a separate basis which is almost the same, it is the same vectors, except now, I am going to put v2 first, and then v3, and then v1.*0096

*Even though these two bases consist of the same vectors, they are not in the same order.*0108

*Turns out, they are not really the same, and you will see why in a minute when we do what we do.*0114

*So, not the same. Now, let us let v... v1, v2, all the way to vN... since there are n vectors, we are talking about an n-dimensional vector space, because again, that is what a basis is.*0120

*The number of vectors in the basis gives you the dimension of that particular space.*0139

*So, let this basis be an ordered basis for an n-dimensional vector space, v.*0145

*Okay. Then of course, as we know, because it is a basis, every vector in v, every v in v, symbol like that, can be written -- excuse me -- as... so v is equal to some constant... c1 × v1, c2 × v2 + all the way + cN × vN.*0171

*Now, some of these constants might be 0, but they cannot all be 0.*0205

*So, again, this is just the definition of a basis. It is a linear combination. A basis allows you to actually write any vector in a vector space as a linear combination of those basis vectors.*0208

*Nothing particularly new here. Now, let us take a look at c1, c2, c3, c4, all the way to cN.*0220

*If we take just the constants and write them as a vector we get this thing.*0230

*I will do the right side of the equation first, and then I will put the symbol on the left hand side.*0260

*So if I take just the constants, so if I have a vector v, and I can express it as some linear combination of the vectors in our basis, and if I just pull these constants out and I write them as a vector.*0265

*So, c1 -- oops, let me do this in red, actually -- c1, c2, all the way to cN>*0275

*So, I am writing it as a -- oops, we do not want these stray lines here.*0287

*I will write it as a column vector. Basically, if I take these, if I have some vector in n-space... well, this vector of the constants that make up the linear combination representing v... it is symbolized as that way.*0298

*We have the vector symbol, and we have a bracket around it, and we put a little b... the b represents the basis, okay?*0320

*We call this... called... this is called the coordinate vector.*0327

*This is called the coordinate vector of v, with respect to basis b.*0342

*This is unique. So, b... so the coordinate vector of any vector with respect to a given basis is unique.*0357

*Let us stop and think about what this means. We know that if we have a given particular vector space, let us say R3, we know that a basis for R3 has to have 3 vectors in it, because that is the definition of dimensions.*0374

*The number of vectors in a basis for that space.*0386

*We also know that there is an infinite number of bases, it does not have to be 1 or the other.*0390

*As it turns out, any vector in a vector space is going to be written as a linear combination of the vectors in that basis.*0394

*Well, the constants that make up that linear combination, I can arrange them as a vector, and I call those the coordinates with respect to that basis of that particular vector that I am dealing with.*0402

*So, needless to say, if I choose one basis, the coordinates are going to be one thing. If I choose another thing, the coordinates are going to be entirely different.*0414

*It is kind of interesting when you think about this. If I pick some random point in a vector space, as it turns out, its identity, its intrinsic identity is actually... It has nothing to do with its coordinates.*0423

*The coordinates are something that we attach to it so that we can actually deal with it. It all depends on the basis that we choose.*0438

*That is kind of extraordinary. You know, were you still thinking of a point in space like (5,6,7) as if it is specifically (5,6,7).*0445

*In a minute you will see that those numbers 5, 6, 7, with respect to a given basis.*0454

*In R3, it is the i,j,k, vectors. It is a very convenient basis because they happen to be unit length, each of the vectors, they happen to be mutually orthogonal, which we will talk about more in a minute.*0462

*But any basis will do, actually. As it turns out, that number (5,6,7), it is specific only to the natural basis.*0473

*It does not really tell me something about the point itself. That is actually kind of interesting to think that it is only so that we can handle that point mathematically as an object, that we have to assign some sort of a value to it.*0482

*We assign a value with respect to an arbitrary choice of basis. Arbitrary in a sense that no one basis is better than another.*0494

*You will have bases that are more convenient than others, but if you... if the problem might call for a basis that is completely different than the one you are used to. Again, that is kind of extraordinary.*0504

*Let us do an example.*0514

*Okay. Let us see -- let us go back to blue.*0520

*We will let s equal the set (1,1,0,0), that is the first vector in the set... then we have (2,0,1,0), that is the second vector in the set, (0,1,2,-1), third vector, and we have (0,1,-1,0). Okay.*0528

*Let s be a basis, an ordered basis for R4.*0555

*Again, we have 4 numbers, we have 4 vectors, so it is R4. Now, let us choose a random vector v, let v in R4 be the vector (-1,2,-6,5).*0564

*Okay. We want to find the coordinate vector of this vector with respect to this basis.*0589

*So, let us stop and think about this for a second. I have some vector, you know, that I just represented as (-1,2,-6,5).*0605

*But, I have a different basis than I am normally accustomed to. So, I want to find the coordinates of this vector with respect to this basis. Okay.*0615

*Okay. Let us see what we are going to do. Well, here is what we want.*0625

*We want constants, c1, c2, c3, c4, such that c1 × the first vector (1,1,0,0) + c2 × the second vector (2,0,1,0), + c3 × (0,1,2,-1), + c4 × (0,1,-1,0) is equal to our vector v, which is (-1,2,-6,5).*0630

*This is what we want. The idea is we take these basis vectors, we write the vector that we are looking for, it is a linear combination of these things.*0667

*Now, we have to solve this. Well, this is just a linear system, so we set it up as a linear system, as a 4 by 5 augmented matrix.*0676

*It is going to be... (1,1,0,0), we just take these as columns, (2,0,1,0), (0,1,2,-1), then we take (0,1,-1,0), (0,1,-1,0).*0686

*And... we augment this with (-1,2,-6,5). We are just solving a × x = b.*0703

*In this particular case, x are the constants. That is what we are looking for. We subject this to reduced row echelon... well, subject it to Gauss Jordan elimination to get the reduced row echelon form.*0712

*We end up with the following. (1,0,0,0), (0,1,0,0), (0,0,1,0) -- that is not a 6, that is a 0 -- and (0,0,0,1), and we end up with... (23,-12,-5,-16).*0724

*Therefore, our coordinate vector for v with respect to the basis that we were given is equal to -- nope, cannot have that -- let us make sure these are clear.*0752

*We have (23,-12,-5,-16). That is our answer, with respect to this basis, the vector is (23,-12,-5,-16).*0774

*These numbers up here, (-1,2,-6,5), this vector was given to us because that is the standard basis.*0797

*In R4, it is the... imagine i,j,k, with one extra vector... basically it is something in the x direction, something in the y direction, something in the z direction, and something in the L direction.*0805

*Again, we are talking about a 4-dimensional space. We cannot see it, but we can still treat it mathematically.*0817

*Mutually orthogonal vectors. That is why this and this are different. We are talking about the same point, but, in order for us to identify that point, to give it a label, to give it a name, we need to choose a basis.*0823

*We need to choose a point of reference, a frame of reference. That is what all of modern science is based on. All of measurement is based on.*0843

*We need something from which to measure something. Our frame of reference, well here it is the standard basis. The basis of mutually orthogonal unit vectors.*0851

*Here, it is a completely different basis. Well, this one basis is not necessarily better than this one.*0864

*We are just accustomed to this one. We think that that is the one, that this vector is actually (-1,2,-6, 5). It is not.*0869

*This (-1,2,-6,5), actually has nothing to do intrinsically with that point. It has to do with our imposing a label on that point so that we can deal with it mathematically.*0877

*This set of coordinates is just as good as this set of coordinates. This basis is just as good as the natural basis.*0892

*That is what you have to... so now, we are getting into the idea of linear algebra we want to sort of disabuse ourselves of the things that we have become accustomed to.*0899

*That, just because we have become accustomed to them, it does not mean that they are necessary, or necessarily better that anything that we might develop for these mathematical objects.*0909

*Okay. Let us actually demonstrate this mathematically. This whole idea of the standard basis. Okay.*0918

*Let s, this time we will let s equal the standard basis, remember? e1, e2, e3, and e4.*0929

*Which is equal to... (1,0,0,0), that is e1. (0,1,0,0), that is e2.*0945

*Again, the e _{i}, the 1, 2, 3, 4, that means all of the entries for that vector are 0, except that number.*0956

*So, for example, e3, all of the entries are going to be 0 except the third entry which is going to be 1... (0,0,1,0).*0965

*You notice all of these vectors have a length of 1, and if you actually took the dot product of this with this, you would get 0.*0974

*So, they are length 1, which is very convenient, and they are also mutually orthogonal, perpendicular... and (0,0,0,1).*0980

*So, this is our set. Now we are using this basis. Well, we are going to let v equal the same thing... (-1,2,-6,5).*0994

*Okay. So, we set up the same system. We want constants c1e1 + c2e2 + c3e3 + c4e4... and these are vectors, I should actually notate them as such -- excuse me.*1007

*Such that they equal v. Well, again, this is just a system. Well, we take these vectors in the basis, set them up as a matrix, augment them with v, and we solve it.*1029

*So, we have (1,0,0,0), (0,1,0,0), e3 is (0,0,1,0), and (0,0,0,1).*1041

*We augment it with (-1,2,-6,5).*1050

*Now, we take a look at this, we want it to subject it to Gauss Jordan elimination to take it to reduced row echelon form.*1055

*Well, it is already in row echelon form. So, as it turns out, with respect to this basis, s, which is the natural basis... it is the vector itself (-1,2,-6,5), which is what we said from before.*1060

*The natural basis is the basis that we use all the time to represent a point. That is why... so... a particular vector does not own this set of numbers.*1080

*So this point that is represented by (-1,2,-6,5). It is only a representation of that point. It is not as if this (-1,2,-6,5) actually belong to that point. It is not an intrinsic property, in other words.*1094

*It is simply based on the basis that we chose for our frame of reference.*1109

*Those of you in engineering and physics, you are going to be changing frames of reference all the time, and you are always going to be choosing a different basis.*1114

*So, your coordinates are going to change. The relationship of the points themselves that you deal with, the vectors that you deal with do not change, but the coordinates are simply representations of those points - they are not intrinsic properties of those points.*1123

*Very curious, isn't it? Okay.*1138

*In other words, any basis will do. Any basis that is convenient.*1143

*Alright. Let us take one more look here. Let us do one more example, this time with the space of polynomials.*1148

*Okay. This time we will let our vector space v equal p1. It is the space of polynomials of degrees < or = 1.*1164

*So, for example, t + 6, 5t - 7, things like that, a degree less than or equal to 1... 8... that works because it is less than one.*1187

*We will let s be one of our basis and consist of t and 1, and we will let t be another basis, and it will be t + 1, and t - 1.*1198

*We will let our v random be 5t - 2, so first thing we want to do is we want to find... or the first issue is find the coordinate vector of v with respect to the basis s.*1213

*Okay -- let us go to blue here -- well, we want to solve the following.*1236

*We want to go c1 × t + c2 × 1 = 5t - 2.*1240

*That is what we are doing, it is a linear combination. Constants × the individual members of the basis, and we set it equal to the vector.*1253

*Well, when we see c1t + c2 × 1 = 5t - 2, this is just c1t + c2 = 5t - 2.*1260

*Well, c1t, this is an equality, so what is on the left has to equal what is one the right.*1269

*So, c1t is equal to 5t, that means c1 = 5, and c2 = -2. Well, there you go.*1276

*With respect to this basis, it is equal to 5 - 2. Well, 5 - 2, that is exactly what these numbers are here... 5 and -2.*1289

*So, you see that this basis... t and 1... this is the natural basis for the space of polynomials of degree < or = 1.*1300

*If you were talking about the space of polynomials of degree < or = 2, your natural basis would be t ^{2}, t and 1.*1310

*If you were talking about degree < or = 3, you would have t ^{3}, t^{2}, t, and 1. This is the natural basis, the basis that we have become accustomed to talking about.*1320

*However, we have a different basis that can still represent -- you know -- this particular polynomial. This particular point in the space of polynomials... 5t - 2.*1333

*Let us calculate... now we want to find the coordinates of v with respect to the basis t.*1347

*So, we end up doing the same thing. We are going to go c1 × t + 1 + c2 × t - 1, and it is equal to 5t - 2.*1358

*Linear combination, set it equal to the vector. We go ahead and we solve for this. We get c1t + c1 + c2t - c2 = 5t - 2.*1375

*I collect terms... t × (c1 + c2) + c1 - c2 = 5t - 2.*1387

*Okay. Let me rewrite that on this page. t × c1 + c2 + c1 - c2 = 5t - 2.*1402

*Well, t, t, c1 + c2, so I get c1 + c2 = 5, and I get c1 - c2 = -2, right?*1418

*I can just go ahead and add this directly. So, I end up with 2c1 = 3, c1 = 3/2, and when I put that into any one of the other equations, I end up with c2 = 7/2.*1434

*Therefore, with respect to that vector... 5t - 2, let me write it up here again... this was our original random vector... 5t - 2.*1452

*The coordinates of that vector with respect to the other basis that we chose, is equal to 3/2 and 7/2.*1465

*This is extraordinary. All of our lives from early pre-algebra into algebra, algebra 1, algebra 2, a little bit of geometry, trigonometry, calculus... we think that a polynomial like 5t - 2 actually is 5t - 2.*1479

*Well, 5t - 2 is our way of dealing with that particular polynomial with respect to the standard basis.*1499

*As it turns out, this 5t - 2 can be written in a completely different way with respect to another basis.*1507

*I can write it as 3/2 t + 7/2, with respect to the other basis that I gave for that space of polynomials.*1515

*No one basis is better than another. This 5t - 2 is not... the polynomial itself is something that exists in a space... but in order for us to deal with something that exists, we have to put a label on it.*1523

*We need a frame of reference for it. That is what is going on with linear algebra. This is the difference between mathematics and science.*1540

*Science actually labels things and deals with them in a given frame of reference, but what mathematics tries to do is... these things exists but we need to understand that the labels that we give them are not intrinsic to those objects.*1548

*They are simply our way of dealing with them because at some point we have to deal with them in a certain way, and we deal with things from a frame of reference.*1562

*From a point of reference. For example, measurement does not mean anything.*1570

*If I said something is 5 feet long, well, it is based on a certain standard. It is based on a point of reference, and a certain definition of a distance.*1574

*As it turns out, those things are actually arbitrary. It has nothing to do with the relationship between the two points that I am measuring the distance between. Those are deeper mathematical properties.*1583

*Linear algebra is sort of an introduction to that kind of thinking.*1594

*So, there we go. Today we dealt with coordinates and ordered bases, and notice that we can actually deal with 2 different bases to talk about the same mathematical object.*1599

*Next lesson, we will actually start talking about change of basis and how we go from one to the other.*1613

*Thank you for joining us at Educator.com today, we will see you next time for some more linear algebra.*1620

0 answers

Post by Burhan Akram on November 12, 2013

Fascinating!!