Basis for column space of matrix - MATLAB colspace
This tutorial contains many matlab scripts. You, as the Theorem: The row space and the column space of a matrix have the same dimension. Theorem: If a. the column and null space of a matrix, linear dependence, and bases of vector (c) Enter a matrix C whose columns are the vectors from part (b) into MATLAB. . gives us a surprising connection between these two notions. how to print the rowspace and coloumn space of Learn more about row Symbolic Math Toolbox.
So they're all members of Rm. So the column space is defined as all of the possible linear combinations of these columns vectors. So the column space of A, this is my matrix A, the column space of that is all the linear combinations of these column vectors. What's all of the linear combinations of a set of vectors? It's the span of those vectors. So it's the span of vector 1, vector 2, all the way to vector n.
And we've done it before when we first talked about span and subspaces. But it's pretty easy to show that the span of any set of vectors is a legitimate subspace. It definitely contains the 0 vector.
If you multiply all of these guys by 0, which is a valid linear combination added up, you'll see that it contains the 0 vector. If, let's say that I have some vector a that is a member of the column space of a. That means it can be represented as some linear combination. So a is equal to c1 times vector 1, plus c2 times vector 2, all the way to Cn times vector n. Now, the question is, is this closed under multiplication?
If I multiply a times some new-- let me say I multiply it times some scale or s, I'm just picking a random letter-- so s times a, is this in my span?
Well s times a would be equal to s c1 v1 plus s c2 v2, all the way to s Cn Vn Which is once again just a linear combination of these column vectors. So this Sa, would clearly be a member of the column space of a. And then finally, to make sure it's a valid subspace-- and this actually doesn't apply just to column space, so this applies to any span.
This is actually a review of what we've done the past. We just have to make sure it's closed under addition. So let's say a is a member of our column space. Let's say b is also a member of our column space, or our span of all these column vectors.
Then b could be written as b1 times v1, plus b2 times v2, all the way to Bn times Vn. And my question is, is a plus b a member of our span, of our column space, the span of these vectors? Well sure, what's a plus b? I'm just literally adding this term to that term, to get that term.
This term to this term to get this term. And then it goes all the way to Bn and plus Cn times Vn. Which is clearly just another linear combination of these guys. So this guy is definitely within the span. It doesn't have to be unique to a matrix. A matrix is just really just a way of writing a set of column vectors. So this applies to any span. So this is clearly a valid subspace. So the column space of a is clearly a valid subspace. Let's think about other ways we can interpret this notion of a column space.
Let's think about it in terms of the expression-- let me get a good color-- if I were to multiply my-- let's think about this. Let's think about the set of all the values of if I take my m by n matrix a and I multiply it by any vector x, where x is a member of-- remember x has to be a member of Rn.
It has to have n components in order for this multiplication to be well defined. So x has to be a member of Rn.
Let's think about what this means. This says, look, I can take any member, any n component vector and multiply it by a, and I care about all of the possible products that this could equal, all the possible values of Ax, when I can pick and choose any possible x from Rn.
Let's think about what that means. If I write a like that, and if I write x like this-- let me write it a little bit better, let me write x like this-- x1, x2, all the way to Xn.
Well Ax could be rewritten as x and we've seen this before-- Ax is equal to x1 times v1 plus x2 times v2, all the way to plus Xn times Vn. We've seen this multiple times. So let's just take this, and write a system of equations with this. So we get 1 times x1. So this times this is going to be equal to that 0. So one times x1, that is x1. Plus 0 times x2. Let me just write that out. Plus 3 times x3. Plus 2 times x4 is equal to that 0. And then -- I'll do it in yellow right here -- I have 0 times x1.
Plus 1 times x2. Minus 2 times x3. Minus x4 is equal to 0. And then this gives me no information. So it just turns into 0 equals 0. So let's see if we can solve for our pivot entries, or our pivot variables. What are our pivot entries? This is a pivot entry. That's a pivot entry. That's what reduced row echelon form is all about, getting these entries that are 1 and they're the only non-zero term in their respective columns. And that every pivot entry is to the right of a pivot entry above it.
And then the columns that don't have pivot entries? These columns represent the free variables. So this column has no pivot entry. And so when you take the dot product, this column turned into this column in our system of equations.
So we know that x3 is a free variable. We can set it equal to anything. Likewise x4 is a free variable.
Null space and column space basis
X1 and x2 are pivot variables, because their corresponding columns in our reduced row echelon form have pivot entries in them. So let's see if we can simplify this into a form we know. And we've seen this before. So if I solve for x1 -- this 0 I can ignore. That 0 I can ignore -- I could say that x1 is equal to minus 3x3 minus 2x4. I just subtracted these two from both sides of the equation and I can say that x2 is equal to 2x3 plus x4. And if we want to write our solution set now, so if I wanted to find the null space of A, which is the same thing as the null space of the reduced row echelon form of A, is equal to all of the vectors -- let me do a new color.
Maybe I'll do blue -- is equal to all of the vectors x1, x2, x3, x4 that are equal to -- So what are they going to be equal to? X1 has to be equal to minus 3x3 minus 2x4. Just to be clear, these are free variables because I can set these to be anything. And these are pivot variables because I can't just set them to anything.
When I determine what my x3's and my x4's are, they determine what my x1's and my x2's have to be.
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So these are pivoted variables. These are free variables. I can make this guy pi. And I can make this guy minus 2. We can set them to anything. So x1 is equal to -- let's see, let me write it this way -- they're equal to x3 -- let me do it in a different color -- do x3 like this.
So it's equal to x3 times some vector plus x4 times some other vector. So any solution set in my null space is going to be a linear combination of these two vectors. We can figure out what these two vectors are just from these two constraints right here. So -- let me do it in a neutral color -- x1 is equal to minus 3 times x3 minus 2 times x4. What's x3 equal to? Well x3 is equal to itself. Whatever we set x3 equal to, that's going to be x3.
So x3 is going to be 1 times x3 plus 0 times x4. It is not going to have any x4 in it. X3 is going to be kind of an independent variable. It's going to be free.
We can set whatever it is. We set it and then that's going to be x3 in our solution set. It's just going to be 1 times x4. And so our null space is essentially all of the linear combinations of these two vectors. This can be any real number. This is just any real number and x4 is just a member of the real space. So all of these, the set of all of the valid solutions to Ax is equal to 0 -- where did I write that. Did I even write that down? No I haven't even written that anywhere. The set of all Ax is equal to 0, where this is my x, it equals all the linear combinations of this vector and that vector right there.
And we know what all of the linear combinations mean. It means my null space is equal to the span of these two guys, the span of minus 3, 2, 1, 0. And minus 2, 1, 0, 1. Now let me ask you a question. Are the columns in A, are they a linearly independent set?
Are they a linearly independent set? So if we write these vectors right there, these are the column vectors of A. So let me write that down. So are the column vectors of A -- so what were they? No it's 1, 2, 3. So this is just the column vectors of A.
I could just write A is just this much of columns, but my question is, is this a linearly independent set? And here you might immediately start thinking, well when we said that something is linearly independent -- so linearly independence implies that there's only one solution -- we saw this I think two videos ago, that there's only one solution -- one solution to Ax is equal to 0.
And that is the 0 solution, that x is equal to the 0 vector. Or another way to say that is that the null space of my matrix A is equal to just the 0 vector. That's what linear independence implies. And it goes both ways. If my null space is just a 0 vector, then I know it's linearly independent. If my null space includes other vectors, then I am not linearly independent.
Now my null space of A, what does it include? Is it just the 0 vector? Well, no it includes every linear combination of these guys. It includes actually an infinite number of vectors, not just one solution.
Obviously 0 vector is contained here, if you just multiply both of these -- if you pick 0 for that and that. It's contained, but you can get a whole set of vectors. So because the null span of A, the null space, sorry, the null space of A does not just contain the 0 vector.
So it has more than just 0. So what does that mean? Well that means that there's more than one solution to this. And that means that this is a linearly dependent set. And what does that mean?
At the very beginning of the video I said, what's the column space of A. And we said, the column space of A is just the span of the column vectors. I just wrote it out like that. And I said, well it's not clear whether this is a valid basis for the column space of A.
And what's a basis? A basis is a set of vectors that span a subspace, and they are also linearly independent. And we just showed that these guys are not linearly independent. So that means that they are not a basis for the column space of A. They do span the column space of A, by definition really.
But they're not a basis. They need to be linearly independent for them to be a basis. So let's see if we can figure out what a basis for this column space would be. And to do that we just have to get rid of some redundant vectors. If I can show you that this guy can be represented by some combination of these two guys, then I can get rid of that guy.
He's not adding any new information. Same with that guy. So let's see if we can figure this piece of the puzzle out. So we know already that x1, let me write it this way, that x1 times -- Maybe I'll just kind of leave you hanging and continue this in the next video. But we know that x1 times 1, 2, 3. Plus x2 times 1, 1, 4. Plus x3 times 1, 4, 1. Plus x4 times 1, 3, 2.
how to print the rowspace and coloumn space of matrix in matlab - MATLAB Answers - MATLAB Central
We know that this is equal to 0. Now if we are able to solve for x4 in terms of -- let me just think that I can solve for the vectors that are associated with my free variables using the other vectors. Let me see if I can do that. And you'll see it's actually pretty straightforward. So let's say I want to solve for x4. So if I subtract this from both sides of this equation, I get what?
Let me put it this way, let me set x3 equal to 0. It was a free variable. I can do that. So if I set x3 is equal to 0, then what do I get here? If I said x3 equals 0, this guy disappears. And if I subtract this from both sides of this equation, I get x1 times 1, 2, 3. Is equal to -- I'm just setting x3 equal to 0. That was a free variable. So I'm setting x3 equal to 0.
So this whole thing disappears. So that is equal to minus x4 times 1, 3, 2. Now I set x3 equal to 0. Let me set x4 to be equal to minus 1. If x4 is equal to minus 1, what is minus x4? Well then this thing will just be equal to 1. And I'll have x1 times 1, 2, 3. Plus x2 times 1, 1, 4 will equal this fourth vector right here. And can I always find things like this?
Well sure I can actually find the particular ones. If x3 is equal to 0, and x4 is minus 1 -- Let me copy and paste this that I have up here -- Let me scroll down a little bit. This is what we got when we figured out our null space, right there. So if I'm setting -- remember these are the free variables -- if I set x3 equal to 0 and x4 is equal to minus 1, what is x1? Then this will imply that x1 is equal to minus 3 times x3, that's just 0, minus 2 times x4.