Let’s say I have this matrix

B, here, and I want to know what the null space of B is. And we’ve done this multiple

times but just as a review, the null space of B

is just all of the x’s that are a member. It’s all the vector x’s that

are member of what? 1, 2, 3, 4, 5 that are members

of r to the fifth, where B, my matrix B, times any of these

vector x’s, is equal to 0. That’s the definition

of the null space. I’m just trying to find the

solution set to this equation right here. And we’ve seen before, that the

null set of the reduced row echelon form of B is equal

to the null set of B. So what’s the reduced row

echelon form of B? And this is actually almost

trivially easy. Let me just take a couple of

steps right here– to get a 0 here, let’s just replace row

2 with row 2 minus row 1 So what do we get? Row 2 minus row 1. Row 1 doesn’t change, it’s

just 1, 1, 2, 3, 2. And then row 2 minus row 1. 1 minus 1 is 0. 1 minus 1 is 0. 3 minus 2 is 1. 1 minus 3 is minus 2. 4 minus is 2 is 2. We’re almost there. Let’s see, so this is a free

variable right here. This is a pivot variable

right here. We have a 1. So let me get rid of that

guy right there. And I can get rid of that guy

right there, by replacing row 1 with row 1 minus

2 times row 2. So now row 2 is going

to be the same. 0, 0, 1 minus 2, 2. And let me replace row 1 with

row 1 minus 2 times row 2. So 1 minus 2 times 0 is 1. 1 minus 2 times 0 is 1. 2 minus 2 times 1 is 0. 3 minus 2 times minus 2. So that’s 3 plus

4 is 7, right? 2 times this is minus 4 and

we’re subtracting it. And then 2 minus 2 times

2– that’s 2 minus 4– it’s minus 2. So this is the reduced row

echelon form of B is equal to that right there. And then if I wanted to figure

out its null space, I have x1, x2, x3, x4, and x5 equaling–

I’m going to have two 0’s right here. Now I can just write this

as just a set of or a system of equations. So let me do that. I get x1. I’m going to write my pivot

variables in a green color. x1 plus 1 times x2, so plus

x2, plus 0 times x3. Plus 7 times x4. Minus 2 times x5 is equal

to that 0 right there. And then I get my–

this is x3, right? 0 times x1 plus 0 times

x2 plus 1 times x3. So I get x3 minus 2 times x4

plus 2 times x5 is equal to that 0 right there. And then if we solve for our

pivot variables, right? These are our free variables. We can set them equal

to anything. If we solve for our pivot

variables what do we get? We get x1 is equal to– I

should do that in green. The color coding helps. I get x1 is equal to minus x2

minus 7×4 plus 2×5, just subtracted these from both

sides of the equation. And I get x3 is equal to–

we’ve done this multiple times– 2×4 minus 2×5. And so if I wanted to write

the solution set in vector form, I could write my solution

set or my null space, really, is– or all

the possible x’s. x1, x2, x3, x4, x5. This is my vector

x, that’s in r5. It is equal to a linear

combination of these. So let me write it out. The free variables are x2 times

some vector right there. Plus x– is x3, no x3 is

not a free variable. Plus x4, that’s my next free

variable, times some factor. Plus x5 times some vector. I’ve run out of space. Plus x5 times some vector. And what are those vectors? Let’s see. I don’t want to make this too

dirty, so let me see if I can maybe move– nope that’s not

what I wanted to do. Let me just rewrite this. I haven’t mastered this pen

tool yet, so let me rewrite this here. So x3 is equal to

2×4 minus 2×5. Let me delete this right

over here so I get some extra space. Cross that out. I think that’s good enough. So I can go back to what

I was doing before. x5 times some vector

right here. And now what are

those vectors? We just have to look

at these formulas. x1 is equal to minus

1 times x2. So minus 1 times x2. Minus 7 times x4. Plus 2 times x5. Fair enough. And what is x3 equal to? x3 is equal to 2×4. 2×4, right? It had nothing to do with x2

right here, so it’s equal to 2×4 minus 2×5. And then 0 times x2, right? Because it had no x2

term right here. And then what is x2 equal to? Well x2 is just equal

to 1 times x2. And so all of these terms

are 0 right there. And I want you to pay

attention to that. I’ll write it right here. x2 is a free variable, so it’s

just equal to itself, right? 1 and you write a 0 and a 0. x4 is a free variable. And this is the important

point of this exercise. So it’s just equal to

1 times itself. You don’t have to throw in any

of the other free variables. And x5 is a free variable. So it just equals 1 times itself

and none of the other free variables. So right here we now say that

all of the solutions of our equation Bx equals 0, or the

reduced row echelon form of B times x is equal to 0,

will take this form. Or they are linear combinations

of these vectors. Let’s call this v1,

v2, and v3. These are just random

real numbers. I can pick any combination here

to create this solution set, or to create

our null space. So the null space of A, which is

of course equal to the null space of the reduced row echelon

form of A, is equal to all the possible linear

combinations of these 3 vectors, is equal to the span

of my vector v1, v2, and v3. Just like that. Now, the whole reason I went

through this exercise– because we’ve done this multiple

times already– is to think about whether these guys

form a linear independent set. So my question is are these

guys linearly independent? And the reason why I care is

because if they are linearly independent then they

form a basis for the null space, right? That we know that they span the

null space, but if they’re linearly independent,

then that’s the 2 constraints for a basis. You have to span the subspace,

and you have to be linearly independent. So let’s just inspect these

guys right here. This v1, he has a

1 right here. He has a 1 in the second term

because he corresponds to the free variable x2, which is

the second entry, so we just throw a 1 here. And we have a 0 everywhere

else in all of the other vectors in our spanning set. And that’s because for the other

free variables we always wanted to multiply them

times a 0, right? And this is going to

be true of any null space problem we do. For any free variable, if this

free variable represents a second entry, we’re going

to have a 1 in the second entry here. And then a 0 for the second

entry for all of the other vectors associated with the

other free variables. So can this guy ever be

represented as a linear combination of this

guy and that guy? Well there’s nothing that I can

multiply this 0 by and add to something that I multiply

this 0 by to get a 1 here. It’s just going to get 0’s. So this guy can’t be

representated as a linear combination of these guys. Likewise, this vector

right here has a 1 in the fourth position. Why is it a fourth position? Because the fourth position

corresponds to its corresponding free

variable, x4. So this guy’s a 1 here. These other guys

will definitely always have a 0 here. So you can’t take any linear

combination of them to get this guy. So this guy can’t be represented

as a linear combination of those guys. And last, this x5 guy, right

here, has a 1 here. And these guys have 0’s here. So no linear combination of

these 0’s can equal this 1. So all of these guys are

linearly independent. You can’t construct any of

these vectors with some combination of the other. So they are linearly

independent. So v1, the set v1, v2, and v3

is actually a basis for the null space, for the null space

of– Oh, you know what, I have to be very careful. For the null space of B. Just for variety, I defined my

initial matrix as matrix B, so let me be very careful here. So the null space of B was equal

to the null space of the reduced row echelon form of B. It’s good to switch things up

every once in a while, you start thinking that

every matrix is named A if you don’t. And that’s equal to the

span of these vectors. So these vectors, and we just

said that they’re linearly independent. We just showed that because

there’s no way to get that one from these guys, that one from

these guys, or that one from these guys. These guys form a basis for

the null space of B. Now this raises an interesting

question. In the last video, I defined

what dimensionality is. And maybe you missed

it because that video was kind of proofy. But the dimensionality, the

dimension, of a subspace– I’ll redefine it here– is the

number of elements in a basis for the subspace. And in the last video I took

great pains to show that all bases for any given subspace

will have the same number of elements. So this is well defined. So my question to you now is:

what is the dimension of my null space of B? What is that the dimension

of my null space of B? Well, the dimension is just

the number of vectors in a basis set for B. Well this is a basis set

for B right there. And how many vectors

do I have in it? I have 1, 2 3 vectors. So the dimension of the

null space of B is 3. Or another way to think about

it– or another name for the dimension of the null space

of B– is the nullity, the nullity of B. And that is also equal to 3. And let’s think about

it, you know I went through all this exercise. But what is the nullity of any

matrix going to be equal to? It’s the dimension of

the null space. Well the dimension of the null

space– you’re always going to have as many factors here as

you have free variables. So in general, the nullity of

any matrix of any matrix– let’s say matrix A– is equal

to the number of I guess you could call it free variable

columns or the number free variables in, well, I guess we

call it in the reduced row echelon form, or I guess we

could say the number of non-pivot columns. The number of non-pivot columns

in the reduced row echelon form of A. Because that’s essentially the

number of free variables– all of those free variables have

an associated, linearly independent vector with

each of them, right? So the number of variables is

the number of vectors you’re going to have in your basis

for your null space. And the number of free variables

is essentially the number of non-pivot columns

in your reduced row echelon form, right? This was a non-pivot column,

that’s a non-pivot column, that’s a non-pivot column. And they’re associated

with the free variables x2, x4, and x5. So the nullity of a matrix is

essentially the number of non-pivot columns in the reduced

row echelon form of that matrix. Anyway, hopefully you found

that vaguely useful.

Dude, awesome.

You make it so clear. Thank you

yup

thx mate..u saved me from failing maths…..lolz

A thank you would not be enough . . . I dont know wat to say

God Bless You

Thank you sr!

Hi there, I would like to know if I did exactly like you did in the video, except I put in row echelon form instead of a row reduced echelon form. Would I still be able to get the right answer?

Doesn't this mean that the dimension of A is equal to the number of collumns minus the number of rows of A. Since the reduced row echelon form will allways contain a number of pivots equal to the number of rows?

Thanks guy you just saved me from my mid-term!

absolutely amazing!!! u are a genius, thanks sooooo much

Thanks so much, this is EXACTLY what i needed! ðŸ˜€

Amazing job, trying to get ready for linear algebra exam tomorrow.

I come listen to this guys voice and all my stress about up coming exam goes away.

Sick

Not just vaguely useful, EXTREMELY useful.

I'm pretty sure he uses a USB tablet to do these

Then how does he move the pointer around without writing?

When the pen hovers closely above the tablet, the mouse is responsive. You don't have to be in contact with it. I use one for Photoshop and that's how mine works.

Damn this was really helpful.

so good

i have a test in 20 minutes and his videos saved my life

Teach at my school please???

Hes so smart, not even funny

skipped whole 6 month classes watch all videos . passed

Jeeeez what snoozzzzeefest. #YAAAAWWWWWNNNN

oh my god you go extremely slow thanks for wasting my timeÂ

Why can't every lecturer explain as clearly and simply as this?

This guys is the best. But let's face it if universities and colleges could hire professors like this guy who can "teach" a concept and "explain" thoroughly then people would not even have to resort to watching his videos as last minute life savers before a test or exam. Just because someone has a PHD in a subject does not necessarily mean they can teach it. They might be a complete expert and genius in that subject but getting other people to learn it and teaching it to them is a completely different story. The institutes should look at Khan's videos as an example how a great teacher should explain concepts and questions to someone rather than looking at whether they have got a PHD or doctorate in a course.Â

Thank you so much man!!!! You are awesome!

perfect!!!

your handwriting is just like mine!

Vaguely useful? are you kidding me, thus was great! Thanks for the help Sal!

thank you very much sir , i have final exam in this after one hour

you really explained everything for me better than that one we called prof in our college

Useful… thank you..

thanks it helps a lot

God bless you! The tutorial was amazing!

I wanna kiss you right on the fuckin mouth Mr. Khan Academy

what if the free variable is not linearly independent?

11:40 for Dimension of Null space, you're welcome

how are x2, x4, x5 free variables?

It actually doesn't need to be in RREF btw. REF will work better since it's a time saver.

Row eschelon form is good enough because the numbers will change but the pivots will not. Since the rank-nullity theorem is dealing strictly with the # of pivots col and # of non-pivots col

Let me ask you a question… is number of non pivot columns the nullity of a matrix and number of pivot columns is the rank of the matrix?

Love you Sal. Your videos make me look like some sort of super genius in class!

Jesus christ that ending "Anyway hope you found that, vaguely useful" LOL

God I love you.

Excellent video, though why does the calculation for nullspace start with B but end with A?

i wish i can hug uh for ur help

luv ya m8