# Relationship between null space and range

### Null space and column space basis (video) | Khan Academy

The dimension of the nullspace of A is called the nullity of A. So if 6 $$\times$$ 3 dimensional matrix B has a 1 dimensional range, then $$nullity(A) = 1$$. The range . MODULE 8. Topics: Null space, range, column space, row space and rank of a matrix. Definition: Let L: V1 → V2 be a linear operator. The null space N(L) of L is . THE RANGE AND THE NULL SPACE OF A MATRIX. Suppose that A is an m × n matrix with real entries. There are two important subspaces associated to the.

But I want to minimize my number of negative numbers. So let me take this third row, minus 3 times this first row.

Mod-04 Lec-15 The Null Space and the Range Space of a Linear Transformation

So I'm going take minus 3 times that first row and add it to this third row. So 3 minus 3 times 1 is 0. These are just going to be a bunch of 3's. And 2 minus 3 times 1 is minus 1.

Now if we want to get this into reduced row echelon form we need to target that one there and that one there. And what can we do? So let's keep my middle row the same. My middle row is not going to change. And to get rid of this one up here I can just replace my first row with my first row minus my second row.

Because then this won't change. I'll have 1 minus 0 is 1. That's what we wanted. That's 1 plus 2. That's 1 plus 1. Now let me do my third row. Let me replace my third row with my third row subtracted from my first row.

### M.6 Range, Nullspace and Projections | STAT ONLINE

They are obviously the same thing. So if I subtract the third row from the second row I'm just going to get a bunch of 0's. Minus 2 minus minus 2 is 0. And minus 1 minus minus 1.

That's minus 1 plus 1. That's equal to 0. And just like that we have it now in reduced row echelon form. So this right here is the reduced row echelon form of A. Now the whole the reason why we even went through this exercise is we wanted to figure out the null space of A. And we already know that the null space of A is equal to the null space of the reduced row echelon form of A. So if this is the reduce row echelon form of A, let's figure out its null space.

So the null space is the set of all of vectors in R4, because we have 4 columns here. The null space is the set of all of vectors that satisfy this equation, where we're going to have three 0's right here.

That's the 0 vector in R3, because we have three rows right there, and you can figure it out. This times this has to equal that 0. That dotted with that essentially is going to equal that 0. That dotted with that is equal to that 0. I say essentially because I didn't define a row vector dot a column vector.

I've only defined column vectors dotted with other column vectors. But we've been over that in a previous video, where you can say this is a transpose of a column vector.

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.

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• Row and column spaces
• Null space 3: Relation to linear independence

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. 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.

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.

The span of the columns of a matrix is called the range or the column space of the matrix. The row space and the column space always have the same dimension. If M is an m x n matrix then the null space and the row space of M are subspaces of and the range of M is a subspace of. If u is in the row space of a matrix M and v is in the null space of M then the vectors are orthogonal.

### Null space 3: Relation to linear independence (video) | Khan Academy

The dimension of the null space of a matrix is the nullity of the matrix. Any basis for the row space together with any basis for the null space gives a basis for.

If M is a square matrix, is a scalar, and x is a vector satisfying then x is an eigenvector of M with corresponding eigenvalue. For example, the vector is an eigenvector of the matrix with eigenvalue.

The eigenvalues of a symmetric matrix are always real. A nonsymmetric matrix may have complex eigenvalues.

## Null space and column space basis

So, why do the, why does x have to be a member of R n? Well just for the matrix multiplication to work, for this to be, if this is m by n, let me write this down, if this is m by n, well in order just to make the matrix multiplication work or you could say the matrix vector multiplication, this has to be an n by one, an n by one, vector, and so it's gonna have n components, so it's gonna be a member of R n.

If this was m by A, well, or, let me use a different letter, if this was m by, I don't know, seven, then this would be R seven, that we would be dealing with. So that is the null space. So, another way of thinking about it is, well if I take my matrix A, and I multiply it by sum vector x, that's a member of this null space, I'm going to get the zero vector.

So if I take my matrix A, which I've expressed here in terms of its column vectors, multiply it by sum vector x, so sum vector x, and, actually let me make it clear that, it doesn't have to have the same, so, sum vector x right over here, we draw the other bracket, so this is the vector x, and so, it's going to have, it's a member of R n, so it's going to have n components, you're gonna have x one, as the first component, x two, and go all the way, to x n.

If you multiply, so if we say that this x is a member of the null space of A, then, this whole thing is going to be equal to the zero vector, is going to be equal to the zero vector, and once again the zero vector, this is gonna be an m by one vector, so it's gonna look, actually let me write it like this, it's gonna have the same number of rows as A, so, I'll try to make it, the brackets roughly the same length, so, and there we go, try and draw my brackets neatly, so you're gonna have m of these, one, two, and then go all the way to the mth zero.

So, let's actually just multiply this out, using what we know of matrix multiplication. And by the definition of matrix multiplication, one way to view this, if you were to multiply our matrix A times our vector x here, you're going to get the first column vector, V one, V one, times the first component here, x one, x one, plus, the second component times the second column vector, x two times V two, V two, and we're gonna do that n times, so plus dot dot dot x sub n times V sub n, V sub n, and these all when you add them together are going to be equal to the zero vector.

Now this should be, this, so it's gonna be equal to the zero vector, and now this should start ringing a bell to you, when we looked at, when we looked at linear independence we saw something like this, in fact we saw that these vectors V, V sub one, V sub two, these n vectors, are linearly independent if and only if, any linear, if and only if the solution to this, or I guess you could say the weights on these vectors, the only way to get this to be true is if x one, x two, x n are all equal zero.