 Now yesterday we looked at vector spaces and subspaces and I really want to talk today about let's do let's do this We're going to talk about the subspace subspace of a matrix Now we've already looked at the matrix as far as its Columns are concerned and we saw those as column vectors, but I want to build on to that So imagine that I have some vectors in a space let's call it vector V1 vector V2 vector V3 and they are all elements of this bigger vector space V and so much so that they all part of a subspace and I can even have this V. Let's make this VJ and If these if these are all say V1 V2 V3 If all of them are part of some subspace of V Subspace so listen carefully So they are part of this vector space, but most so they're part of a subspace of These three at least a subspace of V If this is true then the following things hold If I were just to add them V1 to V2 plus V3 imagine that equals this VJ That would mean that this VJ is also in that subspace That guarantees that and if there's some scalars C and I see one and I multiply that by V1 That is also in the subspace And if I have so it's a linear It's this this this scalar times that and if I see one V1 and see to V2 C3 V3 The linear combination of these vectors Remember the each of column major column a linear combination of them so multiplied by scalar will still also be in this subspace So if this is true if this holds then this holds and these two holds, so it's a linear combination Now imagine I have a matrix A Here's my matrix A and it has one one one two one three Let's make it two one two two two three and I have three one three two three three And I have some vector X. It's my vector X and it's X of one X of two and X of three That's it what So imagine that we are used to that we've seen now a Let's do that a X equals B We've seen that a system of linear equations and I'm trying to get that solution That's what linear algebra of most of linear algebra is all about and we can do gas Jordan elimination We can get the inverse of this so that we have X equals a inverse B We can do all of those things, but there's a much more important thing going on here And that is to see the subspace created by this matrix So that's a bit weird because if that is so then this is our matrix of coefficients What do all the coefficients have to do with it? Well, the coefficients have everything to do it. Let's have a look What we're trying to do is to solve for this now. Let's do this Let's do this multiplication that we have here. So I have a one one a One two a one three and I have a 21 a two two a row two column three Three column one three column two three three and I'm multiplying that by X of one X of two and X of three What I'm going to get if I do this multiplication That's a three by three matrix. It's a three by one matrix. It's going to be this three by one matrix another another Vector and that vector better be B Also a three by one matrix a vector in this instance, but what do I get if I do this multiplication? well, I get a Sub one one or let's make it X One X of one plus X of two a one two plus X of three a one three and I get X of two a sub two one plus X sub For this one X of one except two X of two a two two and this one except three a two three And this one's gonna be X of three and then it's a sub three one plus X sub This row in this column one X of two a three two plus X of three a three three Apologies for that horrible writing. What I want you to see is That these are all X of one that's all except two that's all except three So what I could write is something like this except one and that is a one one a two one a three one So here this I'm going to write neatly because this is the important bit and now I have a one two a two two a two three and X of three A sub one three a sub two three a sub three three And that is going to equal. I hope you can still see the B sub one B sub two and B sub three Can you see What's happening here? There's various ways to to look up look at this So we back to what we had here. There's a vector. There's a vector. There's a vector There's a vector. There's a vector. There's a vector and a scalar scalar scalar scalar scalar scalar It's just that this scalar makes up This vector of values that I require and that equals this other vector and That's why we want to talk about the vector space or column space or subspace then more specifically of This matrix because this matrix if I do this column view Gives me the ability to do this linear combination of them to get to a solution And imagine So I need a specific value for that and a specific value for that and a specific value for that to give me this But let's view it just slightly differently imagine I plug in any value here any value here and any value here And that is going to give me a specific if I put another value here another value here another value here I'll get another B And all the B's together make up all the B's together make up this vector space Of this matrix So that matrix has its own vector space and I will only get a solution for this if B exists inside of The vector space of that or the column space then of this matrix and that's a very important thing So, you know, let's get away just from the simplistic view in the linear algebra of just You know the different methods that you have to learn and write that on paper to solve this look at this This has this this specific matrix has a vector column space And it's only if this vector is inside of the all the possible ones by varying these That the final solution would be possible B has to be in the vector space of this this subspace at least because it's linear combinations of these That will give me that There's always the trivial solution. Remember the vector zero zero zero So the zero vector zero vector is always in this This column space of this matrix because I can make that one zero that one zero that one zero That's the trivial solution that will always be there and that's why we always say that this is zero vector Must be in the subspace of a larger vector space. Can you see why that is so? So let's expand our minds and just see any matrix that we have here actually as having its own column space And we do that by noticing that it is a linear combination If I were to just you know, it's x1 times all of those and that's exactly what happens when we do that Exactly what happens when we do this multiplication And it'll only be so you know, we'll only get a solution to these if I swap it around now Only get a solution for these are specific solutions for this if this be that I'm given because you remember I'm giving something equals something something equals something something equals something I'm giving a linear system And that is b1 and b2 and b and b3 So it's only that I will get values of x if and only if this side of the equation Exists in the column space of my matrix of coefficients and that's very profound So let's have a quick look at this notebook that I have prepared and we're looking at the column space of a matrix Once again, we're going to cheat a bit if you look at the top here I've got three equations and three unknowns. I'm going to let My x, y and z but let's make them x sub 1, x sub 2 and x sub 3 I'm going to solve it immediately by getting x sub 1 equals 3 x sub 2 equals 1 x sub 3 equals 2 And I put these coefficients in here so that I can work out what the solution is So I can have my three equations and three unknowns as you would see it in a textbook So I have 3 times x sub 1 plus 2 times x sub 2 plus x sub 3 is 13 And my other two equations and I have to solve for x sub 1 x sub 2 x sub 3 So let's just remind ourselves how to do this in math America I have my matrix of my augmented matrix. I should say and I'm calling it a a Upper case a lower case a augmented matrix Let's execute that I've put a semicolon. So there's not going to be any Expression to the screen here. Let's just view this in matrix form and I see my whole augmented matrix there Very easy for me to do Row reduction there and to reduce row echelon form I'm going to express that in matrix form and we see our solution x sub 1 is 3 x up 2 is 1 x up 3 is 2 312 for x of 1 x up 2 x up 3 very easy to do The other way that you would obviously learn about let's have this our matrix of coefficients And our solution matrix which is 13 5 and 3 And I can do the inverse of a times b and express that in matrix form and I get the solution 312 as we did here with the elementary row operations. I get 312 But let's do this column view. So I'm going to create three columns And each of those are going to be a vector. So the 312 is this first column here 312 the second column 2 negative 2 1 the third column 1 2 and negative 2 So that's all I'm going to create and I claim as This has a column space because Depending on what the values for x of 1 x up 2 and x up 3 are and these are now just Taking that vector and decomposing it as three scalars And it's a scalar multiple of these three column vectors And that you know gives me another vector b And because we have three one and two as those solutions If I remember I have a little space there in between to indicate multiplication So three times a1 plus one times a2 plus two times a3 and I express that in matrix form I get the vector b which was 13 5 and 3. So it's a linear combination I imagine I had some other value. So that's just specifically to this problem So let's move away from this problem and all I have is these three column Vectors the columns Of my matrix a the same matrix depending on the values that I put in for the scalars I'm going to get a different b and it's all those together forms this column space of this very specific matrix So this matrix has a column space And I will only get a solution if b in this instance It was 13 5 and 3 is one of the possibilities We know it's one because we're plugging the scalars 3 1 and 2 we are going to get a solution So please see it from these multiple Points of view. It's very important that you see the column space of any specific matrix And that you see that the zero vector zero zero zero because this is three space You know would also be a solution because if I put in and it's I mean, let's just do that If I have zero times If I have zero times If I have zero and then we get the solution zero zero zero So that is the zero vector is in this column space is in the column space of every matrix that we can construct Such as the one that we have there. So have please have look at it from these Multiple points of view and understand this column space of a matrix