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

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

### Basis & Dimension

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
- Basis and Dimension 0:23
- Definition
- Example 1
- Example 2: Part A
- Example 2: Part B
- Theorem 1
- Theorem 2
- Procedure for Finding a Subset of S that is a Basis for Span S
- Example 3
- Theorem 3
- Example 4

### Linear Algebra Online Course

### Transcription: Basis & Dimension

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

*Last couple of lessons, we talked about linear independence, and we talked about the span.*0004

*Today we are going to talk about something called basis and dimension, and we are going to use linear independence and span to define those things.*0013

*So, let us get started. Okay. Let us start with a definition here.*0021

*Again, math usually always starts with some definition. Okay.*0031

*Vectors v1, v2, and so on... all the way to vk are said to form a basis for vector space v.*0037

*If 1... v1, v2, all the way to v _{k}, with a span b.*0065

*And 2... v1, v2... v _{k} are independent, linearly independent, but I will just write independent.*0086

*So, again, in the case of a set of vectors that is both independent and happens to span a vector space or some subspace, span something that we are, that we happen to be interested in dealing with. *0103

*We actually give it a special name. It is called a basis. Now, you can have vectors that are independent, but do not necessarily span a space.*0119

*So, for example, if I had 3-space, I could take the i vector and the j vector.*0130

*Well, they certainly are independent, they are orthogonal, they have nothing to do with each other... and yet they do not span the entire space. They only span a part of the space... in other words, the xy plane.*0135

*Or, you can have vectors that span the entire space, but are not necessarily independent.*0145

*So, again, let us take 3-space. I can have i,j,k, and let us say I decided to take also the vector 5k, another vector in the direction of k, but 5 times its length.*0152

*Well, it is a different vector. So, there are four vectors but... and they span the space, you know, every single vector can be written as a linear combination of i,j,k, and 5k, but they are not linearly independent.*0165

*They are dependent, because 5k can be written as a, well, constant × k. It is just they are parallel, they are the same thing,*0179

*So, again, you can have something that spans a space but is not independent, and you can have vectors that are independent but do not necessarily span the space.*0190

*What we want is something that does both. When it does both, it is called a basis for that space... profoundly important.*0199

*Okay. Let us see what we can do. Let us just throw out some basic examples.*0207

*Okay. So, the one we just discussed, e1, e2, and e3, they form a basis for R3.*0218

*Just like e1, e2, e3, e4, e5 would form a basis for R5.*0235

*Let us do a computational example here.*0243

*We are going to take a list of four vectors.*0249

*v1 = (1,0,1,0), v2 = (0,1,-1,2), these are vectors by the way, I better notate them as such.*0254

*v3 = (0,2,2,1) and v4 = (1,0,0,1), again I just wrote them in a list, you can write them as vectors, anything you want.*0273

*Let me see. We want to show that these four vectors are... so, show that these form a basis for R4.*0290

*Well, what do we need to show that they form a basis. Two things, we need to show that the vectors span R4, in this case, and we need to show that they are linearly independent.*0309

*So, let us get started, and see which one we want to do first.*0321

*Let us go ahead and do independence first. So, again, we form the following. *0326

*So, equation 1. Remember, c1v1 + c2v2, I am not going to write out everything, but it is good to write out the equation which is the definition of dependence and independence... c3v3 + c4v4 = the 0 vector.*0336

*When I put the vectors, v1, v2, v3, v4 in here, multiply the c's, get a linear system, convert that to a matrix, I get the following (1,0,0,1,0).*0354

*Again, that final 0 is there... (0,1,2,0,0), (1,-1,2,0,0), again the columns of the matrix are just the vectors through v4 (0,2,1,1,0). Okay.*0368

*When I subject this to reduced row echelon, I get the following. c1 = 0, c2 = 0, c3 = 0, c4 = 0.*0392

*Again, you can confirm this with your mathematical software. This is the non-trivial solution. It implies independence. Good.*0403

*So, part of it is set, now let us see about the span. Well, for the span, we need to pick an arbitrary vector in R4, since we are dealing with R4.*0415

*We can just call it -- I do not know -- (a,b,c,d), and, we need to find to set up the following equation.*0425

*I will not use c because we used them before, I will use k... k1v1, constant, k2v2 + k3v3 + k4v4... symbolism in mathematics just gets crazy. Very tedious sometimes.*0432

*And... I will just call it v arbitrary... just some vector v.*0452

*Although, again, we can set up the solution, we can go (1,0,1,0), (0,1,-1,2), (0,2,2,1), (1,0,0,1), and we can do (a,b,c,d).*0458

*You know what, let me go ahead and just write it out, so you see it.*0470

*We have (1,0,0,1), (0,1,2,0), (1,-1,2,0), (0,2,2,1), and of course our vector, this time it is not a (0,0,0,0), it is going to be (a,b,c,d).*0476

*Again, the nice thing about mathematical software is that it actually solves this symbolically. Not necessarily numerically.*0497

*So, it will give you a solution for k1, k2, k3, k4 in terms of a, b, c, and d. Well, there does exist a solution.*0503

*Okay. There does exist a solution. That means that any arbitrary vector can be represented by these 4 vectors.*0513

*So, let us see, so v1, v2, v3, and v4, which are just v1, v2, v3, v4, span R4. *0527

*We found something that spans R4, and we also found that they are linearly independent, so yes, these vectors are a good basis.*0543

*Are they the best basis? Maybe, maybe not, it depends on the problem at hand... but they are a basis and it is a good basis for R4.*0560

*Okay. Let us list a theorem here... theorem... if s, the set of vectors, v1 so on and so forth onto vk, is a basis for v.*0575

*So, if the set is a basis for v, then every vector in v can be written in 1 and only 1 way, as a linear combination of the vectors in s.*0603

*That is not s, we should write the vectors in s. *0655

*So, in other words, if I know that s is a basis for the vector space, any vector in that vector space can only be represented 1 way.*0665

*That means the particular representation, the constants that are chosen is unique. Not multiple ways, it is unique.*0675

*Another theorem... actually, let me write this one in blue because we are possibly going to do something with this one.*0684

*Let s be v1... vk be a set of non-zero vectors in v, and we will let w equal the span of s.*0702

*So, we have this set s, there is a span of it, we will call that w, because it may not span the entire vector space, that is why we are giving it different, but obviously it is going to... I mean it is in v, so it is going to be some subset of it.*0736

*Then, some subset of s is a basis for w. Okay, let us stop and think about what this means.*0752

*I have a vector space v, I have some arbitrary collection of vectors that I have taken from v and I call that set s, just a list of vectors.*0771

*I know that these vectors span some part of v.*0779

*I call that w, if I need to give it a name, or I can just refer to it as the span of s.*0785

*Well, if I take some subset of this, maybe all of it, but so... either k vectors or less than k vectors, some subset of it, it actually forms a basis for the span.*0791

*That makes sense. Again, you have some set of vectors that spans an entire space, well, either all of the vectors together are independent, in which case that is your basis. *0805

*Or, they might be dependent, which means that you should be able to throw out a couple of them and reduce the number.*0820

*But to get something, some set of vectors from here, some collection that actually forms a basis for the span.*0825

*let us see how this works in terms of... a real life example. Okay.*0834

*We are going to list a procedure for finding the subset of s, of any s that is a basis for the span of s.*0841

*Let me actually move forward. Let me write down what this is.*0854

*Procedure for finding a subset of s, that is a basis for this span of s.*0866

*Okay. First thing we are going to do. We want to form c1v1 + c2v2 + so on and so forth... ckvk = 0.*0890

*We want to set up the homogeneous system, okay?*0907

*Now, we want to solve the system by taking it to reduced row echelon form.*0911

*Now, here is the best part. The vectors corresponding to the leading entries form a basis for span s.*0930

*This is actually kind of extraordinary. I love this, and I do not know why, but it is amazing.*0963

*I have this collection of vectors that spans a particular space.*0969

*I set up the homogeneous system and I subject it to Gauss Jordan elimination, bring it down to reduced row echelon form, and as you know, not every column needs to have a leading entry.*0974

*Well, the columns that do have a leading entry, that means I throw out all of the others.*0986

*The original vectors that correspond to those columns that have leading entries, they actually form a basis for my span of s.*0992

*So, let us just do an example and see what happens. Let us take the following vectors.*0998

*let me do this in red, actually... so v1 = 1... this is not going to work... (1,2,-2,1).*1007

*v2 = (-3,0,-4,3). v3, and of course these are vectors, so let me notate them as such... (2,1... this is also 1,-1).*1029

*v4 = (-3,3,-9,6).*1054

*v5 = (9,3,7,-6).*1065

*So, we have 5 vectors... we want to find a subset of these vectors, it might be all 5, it might be 2, it might be 3, it might be 4... that form a basis for the span of s.*1073

*Okay? Okay. So, we form for step... we do this thing right here.*1091

*So, we set up this equation and we put these vectors in for this equation, and we end up with the following system.*1105

*Columns... these vectors just going down... (1,2,-2,1)... or you can do them across... either way.*1115

*(1,-3,2,-3,9), 1, 2, 3, 4, 5 because we have 5 vectors... 5 columns, and of course the augmented is going to be 0.*1126

*(2,0,1,3,3,0), (-2,-4,1,-9,7,0), (3,-1,6,-6,0)... good.*1140

*We are going to subject that to reduced row echelon form.*1161

*When we do that, let me just put that there and write that there. Let me see...*1165

*Let me move on to the next page, that is not a problem.*1172

*So, we have subjected that matrix to reduced row echelon and we end up with the following... (1,0,0,0), (0,1,0,0), (1/2,3/2,3/2,0), (-1/2,3/2,5/2,0), and 0's everywhere else.*1176

*So our reduced row echelon looks like this.*1206

*Well, leading entry, leading entry, no leading entries anywhere else.*1210

*So, vector number 1, vector number 2, v1 and v2 form a basis.*1217

*So, it is not these, it is not (1,0,0,0), (0,1,0,0).*1231

*This is the reduced row echelon from the matrix, the columns of which are the vectors that we are talking about.*1236

*So, those vectors, the actual columns from the original matrix, those 2 vectors, so we started off with 5 vectors, and we found two of them that actually span the entire space.*1242

*We threw out the other three, they are not that important. We can describe the entire span with just these 2 vectors.*1254

*Form a basis for the span of s.*1260

*Again, this is really, really extraordinary.*1266

*Okay. Let us... another theorem... if s, v1... so on and so forth all the way to vk and t, which is, let us say w1 all the way to wk... okay?*1270

*wN, so if s is the set of vectors v1 to vk, k could be 13 so we might have 13 vectors in this one... and t is equal to w1 all the way to wN.*1308

*So, k and n do not have to necessarily be the same, but here is what the theorem says.*1320

*If these 2 sets are bases for v, then k = n.*1328

*In other words, if I have a given vector space, and if I have the bases for them, the bases have the same number of vectors.*1340

*So, the basis set has the same number of vectors. In other words, I cannot have a vector space that has one basis that is 3 vectors and another that is 5 vectors.*1349

*That is not what basis is. Basis expands the set, and it is linearly independent.*1358

*Therefore, if I have 2 bases, they have to have the same number of elements in them.*1364

*It makes sense. Okay. Now, because of this, once again, every basis of a given vector space has the same number of vectors in it.*1368

*There are an infinite number of bases for a vector space... but of that infinite number, they all have the same number of vectors in them.*1382

*Therefore, we define... again, very, very important definition... the dimension of a non-zero vector space dimension -- fancy word -- is the number of vectors in a basis for the vector space.*1393

*So, read this again, the dimension of a non-zero vector, of a non-zero vector space is the number of vectors in the basis for that space.*1440

*So, dimension is kind of a fancy word that a lot of people throw around.*1452

*So, we talk about 3-dimensional space, the space that we live in. Well, 3-dimensional space, there are a couple of ways to think about it.*1458

*Yes, it means 3-dimensional space because it will require 3 numbers to actually describe a point, (x,y,z)... three coordinates.*1467

*However, the actual mathematical definition is 3 space is 3 dimensional because any basis for 3-space has to be made up of 3 vectors... 5 dimensional space.*1475

*Any basis for 5-dimensional space has to be made up of 5 vectors. I cannot have 4 vectors describing it for 5 dimensional space. It is not going to happen.*1489

*Can I have 6 vectors that actually describe it? Yes, I can have 6 vectors that span the 5-dimensional space, but that span is not linearly independent.*1501

*So, because... and that is the whole idea. The dimension of a space is the number of vectors that form the basis and a basis is expansive and it is linearly independent.*1513

*Okay. Let us see here. p2, which we have used a lot.*1526

*That is the vector space of all polynomials of degree < or = 2.*1534

*The dimension is 3, and here is why. The basis, we will list a basis, and that should tell you.*1548

*This is the best part, if you just want to list a basis you can just count the number of vectors... that is how many dimensions that space is.*1555

*t ^{2}, t, and 1. Any linear combination of t^{2}, t, and 1 will give you every single possible polynomial of degree < or = 2.*1562

*For example, 3t ^{2} + 6t - 10. Well, it is 3 × t^{2}, 6 × t, -10 × 1.*1580

*3t + 2... 3 × t... 2... it is of degree 2, this is degree 1. degree less than or equal to 2.*1592

*So, this one has to be in there. So, p2 has a dimension 3.*1602

*pn has dimension n + 1.*1615

*Okay. Now, here is where it gets really, really interesting and sort of just a sideline discussion, something sort of think about a little bit of mathematical culture. A little bit of abstraction... *1621

*Notice that this p2 has a dimension of 3. Well, our 3-space, our normal 3-space that we live in also has a dimension of 3.*1636

*As it turns out, all vector spaces of a given dimension, the only different between the vector spaces is the identity of their elements.*1649

*In one vector space, R3, we are talking about points, or vectors, arrows.*1658

*In this vector space, where this is a basis, it is a dimension of 3... the elements are actual polynomials.*1664

*As it turns out, the identity of the elements is the only thing that is different about those 2 spaces. These two spaces have the exact same algebraic properties.*1675

*They behave exactly the same way. In fact, I do not even need to think about it... if I can find myself 15 other vector spaces that have a dimension of 3, the identity of those elements completely does not matter.*1683

*In fact, it does not even matter, I can treat it completely symbolically. I can call them whatever I want. I can label them whatever I want.*1697

*What is important is the underlying algebraic property, and it is the same for every single vector space of a given dimension. That is what is extraordinary, that is what gives mathematics its power.*1704

*Once I understand, let us say R3, and we understand R3 pretty well... we live in this space, we enjoy the world around us, look at what we have done with the world around us.*1715

*If I find any other vector space with strange objects in it, if it has a dimension of 3, I know everything about it. I know absolutely everything about it because it behaves the same way that R3 does.*1724

*Again, that is really, really extraordinary... the last thing that I want to leave you with in this particular lesson is that what we have dealt with are finite dimensional vector spaces.*1737

*In other words, we know that R3 has an infinite number of vectors in them, but the basis, the dimension is finite... 3.*1748

*That means I only need 3 vectors in order to describe the entire space.*1757

*Now, that is not always true. There are infinite dimensional vector spaces that require an infinite number of vectors to actually describe them.*1761

*Those of you that go on into higher mathematics, or not even that, those of you who are engineering and physics majors, at some point you will be discussing something called the Fourier series, which is an infinite series of trigonometric polynomials.*1772

*Sin(x), cos(x), sin2(x), cos2(x), sin3(x), cos3(x), and so on. That is an infinite dimensional vector space.*1785

*Okay. So, I will list... let us see... 2 infinite dimensional vector spaces, we of course are not going to deal with it.*1796

*Linear algebra, mostly we stick with finite dimensional vector spaces, but I do want you to be aware of them.*1804

*p, the space of all polynomials... all polynomials, that is an infinite dimensional vector space.*1811

*It requires... it has an infinite number of vectors in its basis. Not like p2 or R3, that only have 3.*1817

*The other one is the space of continuous functions on the real line.*1824

*So, the space of continuous functions, you will see it represented like this... from negative infinity to infinity... that is defined on the entire real line.*1838

*That space has an infinite number of dimensions. I need an infinite number of functions in order to be able to describe all of the other functions, if I need to do so.*1845

*I just wanted to throw that out there. Really, what I wanted you to take away from this is that the identity for a vector space of any given dimension, the identity of the elements is completely irrelevant.*1859

*The underlying behavior is what we are concerned with, and the underlying behavior is exactly the same.*1870

*Thank you for joining us here at Educator.com, linear algebra, we will see you next time.*1876

1 answer

Last reply by: Professor Hovasapian

Fri Jan 20, 2017 3:34 AM

Post by Shih-Kuan Chen on January 19 at 08:32:39 PM

Are orthogonal vectors always linearly independent, and are linearly independent vectors always orthogonal?

2 answers

Last reply by: Professor Hovasapian

Fri Mar 25, 2016 10:40 PM

Post by David LÃƒÂ¶fqvist on March 19, 2016

Maybe wrong place again, but could you explain the Hector triple production? I know hos to calcu?ate it, but I havet no idea och what I'm calculating?

Thanks again for your great Messina!

1 answer

Last reply by: Professor Hovasapian

Fri Mar 14, 2014 7:29 PM

Post by Josh Winfield on February 26, 2014

6:48 "This is the non-trivial solution"..........should be trivial solution

1 answer

Last reply by: Professor Hovasapian

Thu Jan 9, 2014 3:39 AM

Post by Joel Fredin on January 8, 2014

Raffi.

For the first, I totally love your videos. You are a really really great teacher, so keep it up! :)

For the second, are you going through how to use the determinant to calculate the area or the volume of a Geometric shape? I can't seem to find it anywhere :( Maybe a little bit off topic but i hope you will answer my question relative soon. Thank you very much for all of your hard work, I don't think you know how much it helps me.

Joel

2 answers

Last reply by: Christian Fischer

Tue Oct 1, 2013 2:29 AM

Post by Christian Fischer on September 25, 2013

Hi Raffi: Just a question for Theorem 2: is the following true

a) S is a subset of V (not a subspace)

b) w=SpanS is a few of the vectors in S (so a subset of S if not all the vectors in S are required to span S)

Have a great day!!

1 answer

Last reply by: Professor Hovasapian

Tue Oct 16, 2012 4:02 AM

Post by Suhaib Hasan on October 16, 2012

Thanks for the quick explanations in the beginning with unit vectors i, j, and k; it definitely helped me get a better understanding for the basis of a vector space.

1 answer

Last reply by: Professor Hovasapian

Tue Aug 14, 2012 8:29 PM

Post by Shahaz Shajahan on August 14, 2012

if you solved the system and you are left with the trivial solution, is that enough to show that all the vectors in the set are basis? as i've seen a few examples which show the solution involving the inverse of the matrix, im not entirely sure why the inverse relates to the unique solution, would you be able to explain please, thanks

0 answers

Post by Kamal Almarzooq on December 12, 2011

i like this theorem too :)