In the last video we set out to

find the eigenvalues values of this 3 by 3 matrix, A. And we said, look an eigenvalue

is any value, lambda, that satisfies

this equation if v is a non-zero vector. And that says, any value,

lambda, that satisfies this equation for v is a

non-zero vector. Then we just did a little bit

of I guess we could call it vector algebra up here

to come up with that. You can review that

video if you like. And then we determined, look

the only way that this is going to have a non-zero

solution is if this matrix has a non-trivial null space. And only non-invertible matrices

have a non-trivial null space. Or, only matrices that have

a determinant of 0 have non-trivial null spaces. So you do that, you got your

characteristic polynomial, and we were able to solve it. And we got our eigenvalues where

lambda is equal to 3 and lambda is equal to minus 3. So now, let’s do– what I

consider the more interesting part– is actually find out

the eigenvectors or the eigenspaces. So we can go back to this

equation, for any eigenvalue this must be true. This must be true but this

is easier to work with. And so, this matrix right here

times your eigenvector must be equal 0 for any given

eigenvalue. This matrix right here–

I’ve just copied and pasted from above. I marked it up with the Rule

of Sarrus so you can ignore those lines– is just

this matrix right here for any lambda. Lambda times the identity

matrix minus A ends up being this. So let’s take this matrix for

each of our lambdas and then solve for our eigenvectors

or our eigenspaces. So let me take the case of

lambda is equal to 3 first. So if lambda is equal to 3, this

matrix becomes lambda plus 1 is 4, lambda minus 2 is 1,

lambda minus 2 is 1. And then all of the other terms

stay the same, minus 2, minus 2, minus 2, 1,

minus 2 and 1. And then this times that vector,

v, or our eigenvector v is equal to 0. Or we could say that the

eigenspace for the eigenvalue 3 is the null space

of this matrix. Which is not this matrix. It’s lambda times the

identity minus A. So the null space of this matrix

is the eigenspace. So all of the values that

satisfy this make up the eigenvectors of the eigenspace

of lambda is equal to 3. So let’s just solve for this. So the null space of this guy–

we could just put in reduced row echelon form– the

null space of this guy is the same thing as the null space

of this guy in reduced row echelon form. So let’s put this in reduced

row echelon form. So the first thing I

want to do– let me just do it down here. So let me– I’ll keep my first

row the same for now. 4 minus 2, minus 2. And let me replace my second row

with my second row times 2 plus my first row. So minus 2 times

2 plus 1 is 0. 1 times 2 plus minus 2 is 0. 1 times 2 plus minus 2 is 0. This row is the same

as this row. So I’m going to do

the same thing. Minus 2 times 2 plus 4 is 0. 1 times 2 plus 2 is 0. And then 1 times 2 plus

minus 2 is 0. So the solutions to this

equation are the same as the solutions to this equation. Let me write it like this. Instead of just writing

the vector, v, let me write it out. So v1, v2, v3 are going to

be equal to the 0 vector. 0, 0. Just rewriting it slightly

different. And so these two rows, or these

two equations, give us no information. The only one is this row up

here, which tells us that 4 times v1 minus 2 times v2–

actually this wasn’t complete reduced row echelon form

but close enough. It’s easy for us to work with–

4 times v1 minus 2 times v2 minus 2 times

v3 is equal to 0. Let’s just divide by 4. I could’ve just divided by 4

here, which might have made it skipped a step. But if you divide by 4 you get

v1 minus 1/2 v2 minus 1/2 v3 is equal to 0. Or, v1 is equal to 1/2

v2 plus 1/2 v3. Just added these guys to both

sides of the equation. Or we could say, let’s say that

v2 is equal to– yeah I don’t know, I’m going to just

put some random number– a, and v3 is equal to b, then we

can say– and then v1 would be equal to 1/2 a plus 1/2 b. We can say that the eigenspace

for lambda is equal to 3, is the set of all of vectors, v1,

v2, v3, that are equal to a times times– v2 is a, right? So v2 is equal to a times 1. v3 has no a in it. So it’s a times 0. Plus b times– v2 is just a. v2 has no b in it. So it’s 0. v3 is 1 times– so 0 times

a plus 1 times b. And then v1 is 1/2

a plus 1/2 b. For any a and b, such

that a and b are members of the reals. Just to be a little bit

formal about it. So that’s our– any vector

that satisfies this is an eigenvector. And they’re the eigenvectors

that correspond to eigenvalue lambda is equal to 3. So if you apply the matrix

transformation to any of these vectors, you’re just going

to scale them up by 3. Let me write this way. The eigenspace for lambda is

equal to 3, is equal to the span, all of the potential

linear combinations of this guy and that guy. So 1/2, 1, 0. And 1/2, 0, 1. So that’s only one of

the eigenspaces. That’s the one that

corresponds to lambda is equal to 3. Let’s do the one that

corresponds to lambda is equal to minus 3. So if lambda is equal to minus

3– I’ll do it up here, I think I have enough space–

lambda is equal to minus 3. This matrix becomes– I’ll do

the diagonals– minus 3 plus 1 is minus 2. Minus 3 minus 2 is minus 5. Minus 3 minus 2 is minus 5. And all the other things

don’t change. Minus 2, minus 2, 1. Minus 2, minus 2 and 1. And then that times vectors

in the eigenspace that corresponds to lambda is equal

to minus 3, is going to be equal to 0. I’m just applying this equation

right here which we just derived from that

one over there. So, the eigenspace that

corresponds to lambda is equal to minus 3, is the null space,

this matrix right here, are all the vectors that satisfy

this equation. So what is– the null space of

this is the same thing as the null space of this in reduced

row echelon form So let’s put it in reduced row

echelon form. So the first thing I want to do,

I’m going to keep my first row the same. I’m going to write a little bit

smaller than I normally do because I think I’m going

to run out of space. So minus 2, minus 2, minus 2. Actually let me just

do it this way. I will skip some steps. Let’s just divide the first

row by minus 2. So we get 1, 1, 1. And then let’s replace this

second row with the second row plus this version of

the first row. So this guy plus that guy is 0

minus 5 plus minus– or let me say this way. Let me replace it with

the first row minus the second row. So minus 2 minus minus 2 is 0. Minus 2 minus minus

5 is plus 3. And then minus 2 minus

1 is minus 3. And then let me do

the last row in a different color for fun. And I’ll do the same thing. I’ll do this row

minus this row. So minus 2 minus

minus 2 is a 0. Minus 2 plus 2. Minus 2 minus 1 is minus 3. And then we have minus

2 minus minus 5. So it’s minus 2 plus 5. So that is 3. Now let me replace– and I’ll

do it in two steps. So this is 1, 1, 1. I’ll just keep it like that. And actually, well let me

just keep it like that. And then let me replace my third

row with my third row plus my second row. It’ll just zero out. If you add these terms, these

all just become 0. That guy got zeroed out. And let me take my second

row and divide it by 3. So this becomes 0, 1, minus 1. And I’m almost there. I’ll do it in orange. So let me replace my first row

with my first row minus my second row. So this becomes 1, 0, and then

1 minus minus 1 is 2. 1 minus minus 1 is 2. And then in the second

row is 0, 1, minus 1. And then the last

row is 0, 0, 0. So any v that satisfies this

equation will also satisfy this guy. This guy’s null space is going

to be the null space of that guy in reduced row

echelon form. So v1, v2, v3 is equal

to 0, 0, 0. Let me move this. Because I’ve officially

run out of space. So let me move this lower

down where I have some free real estate. Let me move it down here. This corresponds to lambda

is equal to minus 3. This was lambda is equal to

minus 3, just to make us– it’s not related to this

stuff right here. So what are all of the v1s, v2s

and v3s that satisfy this? So if we say that v3

is equal to t. If v3 is equal to t, then

what do we have here? We have– this tells us that

v2 minus v3 is equal to 0. So that tells us that v2 minus

v3– 0 times v1 plus v2 minus v3 is equal to 0. Or that v2 is equal to v3,

which is equal to t. That’s what that second

equation tells us. And then the third equation

tells us, or the top equation tells us, v1 times 1– so v1

plus 0 times v2 plus 2 times v3 is equal to 0. Or v1 is equal to minus 2v3 is

equal to minus 2 times t. So the eigenspace that

corresponds to lambda is equal to minus 3 is equal to the set

of all the vectors, v1, v2 and v3, where– well, it’s equal

to t times– v3 is just t. v3 was just t. v2 also just ends up being t. So 1 times t. And v1 is minus 2 times t. For t is any real number. Or another way to say it is that

the eigenspace for lambda is equal to minus 3 is equal

to the span– I wrote this really messy– where lambda is

equal to minus 3 is equal to the span of the vector

minus 2, 1, and 1. Just like that. It looks interesting. Because if you take this guy

and dot it with either of these guys, I think you get 0. Is that definitely the case? Take minus 2 times 1/2, you

get a minus 1 there. Then you have a plus 1. That’s 0. And then minus 2 times 1/2. Yeah. You dot it with either of

these guys you get 0. So this line is orthogonal

to that plane. Very interesting. So let’s just graph it just so

we have a good visualization of what we’re doing. So we had that 3

by 3 matrix, A. It represents some

transformation in R3. And it has two eigenvalues. And each of those have a

corresponding eigenspace. So the eigenspace that

corresponds to the eigenvalue 3 is a plane in R3. So this is the eigenspace for

lambda is equal to 3. And it’s the span of these

two vectors right there. So if I draw them, maybe

they’re like that. Just like that. And then the eigenspace

for lambda is equal to minus 3 is a line. It’s a line that’s perpendicular

to this plane. It’s a line like that. It’s the span of this guy. Maybe if I draw that vector,

that vector might look something like this. And it’s the span of that guy. So what this tells us, this is

the eigenspace for lambda is equal to minus 3. So what that tells us– just

to make sure we are interpreting our eigenvalues and

eigenspaces correctly– is look, you give me any

eigenvector, you give me any vector in this, you give me any

vector right here, let’s say that is vector x. If I apply the transformation,

if I multiply it it by a, I’m going to have 3 times that. Because it’s in the eigenspace

where lambda is equal to 3. So if I were to apply a times

x, a times x would be just 3 times that. So that would be a times x. That’s what it tells me. This would be true for

any of these guys. If this was x, and you took a

times x, it’s going to be 3 times as long. Now these guys over here, if you

have some vector in this eigenspace that corresponds to

lambda is equal to 3, and you apply the transformation. Let’s say that this

is x right there. If you took the transformation

of x, it’s going to make it 3 times longer in the opposite

direction. It’s still going to

be on this line. So it’s going to go

down like this. And that would be a times x. It would be the same, it’d be

3 times this length, but in the opposite direction. Because it corresponds to lambda

is equal to minus 3. So anyway, we’ve, I think,

made a great achievement. We’ve not only figured out the

eigenvalues for a 3 by 3 matrix, we now have figured out

all of the eigenvectors. Which are– there’s an infinite

number– but they represent 2 eigenspaces that

correspond to those two eigenvalues, or minus 3 and 3. See you in the next video.

Hallelujah! PRAISE THE LORD!

the gods have answered…

you my respected mare r an absolute legend..!SAVIOUR

hi can i ask if it is necessary to reduce the matrix?

Um I don't think you got this right. An eigenvector is not a basis of a subspace. It is a collection of eigenvalues that are spread out from eachother. For example, if the eigvenvalues for a matrix A are 1 and 3, then the eigenspace is 3+1 = 4.

The same is true for complex eigenvalues and their corresponding eigenspaces.

@ashk0n also eigenmatrices have many applications to number theories aka that if the dominant singular valueso f a matrix P is greater than the dimension of any other matrix then the supremem of P times Q is always equal to the eigenvalues of something

@MartinRyleOShea

if the det(A – lambda*identity) = 0 then lambda is an eigenvalue of A.

I came from precalc, listened to the first minute, and barfed

THANK GOD FOR KHAN ACADEMY

His explanations are pretty clear though he's a little disorded . Very good overall!!

<3 u loads

Thank you this clear many pictures for me 🙂

yyyeeeyyy! I UNDERSTAND! FINALLY!!!! 😀 😀

Thank you – You pushed my Math AND English skill through the roof – Funny that the German word: Eigenvector became a "special" word (It could have been just be translated to "own – vector") =)

Thanks to you I'm going to be able to pass my class…. Thank you soooooo much 😉

totally saving my ass for my exam tomorrow.

I love you <3

YOU ARE THE BEST!!! 😀 You just cleared all the questions I sent to my professor 3 hours ago in 30 minutes ahah!!!.. YOU ARE THE BEST 😀

I LOVE METH!!!!!…..i mean MATH!!!!

YES

Isn't it |A – (lambda)(I)| -> [determinant of {A minus (lambda x Identity matrix)}]?

thank u vry mch…………nw i feel so gud for the eigen vectors….although i watchd ur video jst before a dy of my EXAM 🙂

@unkown1414

totally agree

must be tablets man, he's too precise

All respect to your effort man ….wish that all the world is like you 🙂

you saved me on my final last spring.

I'll have an exam this morning and you ARE a lot of help. Thank you veeeery much!

if i had a nickel for every lab this guys helped me with id have 2 nickels

Thank you! Very clear and comprehensible.

GREAT!!! Really clear and helpful!!!!!!

eigenkosommak

The real superman!

I should had come here earlier, so many tutorials, they avoided taking a 3×3 matrix or explain in detail what's happening, like it is a big deal to work on 2×2 matrix. Thanks a lot!

I am sad to say but once again is proven that internet is full of bad quality job (tutorials)!

thanks god..!! you are great!!!

Because elementary row operations change the value of the determinant, so you'd have to "undo" them again anyway; might as well only do them once.

Jazak allah

( 1 0 1 )

( 0 1 0 )

( 0 0 0 )

What happens when the reduced row echelon form of a 3 x 3 is

Excellent, bad explanation at college, thank you so much for your video!

LOL

Could you possibly do a video of why I am hearing this terminology in my Differential Equations Class?

thank you! my teacher aint got nothin on you

Is this all the same guy? He teaches the Org Chem too. Is this guy just a professor by hobby?

It'd be cool if I had a professor who who any of this stuff.

he's a teacher. he has like 6 degrees, just look him up on wikipedia

thank you

v1+v3=0

v2=0

thank you very much

I love you.

where can I find electricity and magnetism videos which would explain everything just like this.

this guy is definitely jesus. i mean, his voice doesn't sound exactly like what you'd expect it to, but still, he must be jesus. he has come back to help us with maths!

Thank you, my lecturer sucks. You made something he made complicated easy again.

Lets just change colours for fun 😀

What happens if when you row reduce your matrix you get a zero column, how can you find the eigenvectors.

@khanacademy

I'm looking at my book now, shouldn't the eigenvalue solutions be derived from the equation: det ( A – [lambda] I) = 0 ? @1:50, I can see the equation from which the eigenvalues are derived from as: ([lambda] I – A) V =0 , which is the reverse. The book says to "find the null space of the matrix A – [lambda]I. This is the eigenspace E_lambda, the nonzero vectors of which are the eigenvectors of A…" The book is: "Linear Algebra: A Modern Introduction", 3rd E, Poole, p303

much better than my books! thanks a lot

Oh wow, was stressing about the last step in finding the Eigenvalues but this made it incredibly clear, thanks a lot 🙂

Is it because we have free variables, we don't need to normalize it? Thank you

I thought for every NxN matrix you have a character polynomial to the Nth degree with N number of eigenvalues that correspond with the same N number of eigenvectors. So wouldnt you need 3 eigenvalues that have 3 eigenvectors each for this example?

can you explain why it is (lambda I – A) V = A instead of (A – lambda I) v = 0

This is wrong, it is actually (A-λI)v = 0

It could be either (A – lambda*I)v=0 or (lambda*I – A)v=0 . The two are the same, just differing by a multiple of (-1). Because (-1) is a constant, it can multiply into the parentheses and flip the expression inside, leaving the equation unchanged.

I have a matrix A = {{7,-5,0},{-5,7,0},{0,0,-6}}

I have found the Eigenvalues, 2,12,-6 but I'm only getting one Eigenvector, (0,0,1)..

Can someone please help?

Real-life superhero!

wow thanks im from university of cape town,i had a problem in reducing ..now im mastering this! you're the real hero!

Dear Khan,

You da real MVP.

When finding the eigenvectors, do we really have to do gaussian elimination and reduce one of the rows to all 0's? Because I sometimes I have different results than the book provides.

How do i find the eigenvector if when I reduce the nullspace I get the vector [100, 010, 001] instead of [100,010,000]?

This is a strange method for solving for the nullspace. It looks like you're arbitrarily picking either v1, v2, or v3 to be equal to 1t. You should specify that v3=1 because it is a pivot variable.

Thanks for your useful videos. But can you please get a new microphone the noise sometimes makes it hard to follow the video all the way

you choose v3= t out of free choice! but if i choose v2=t my vector will be completely different. or can the "t" adjust the vector?

does it even make a difference? this is the only thing stopping me from understand this subject i math! i understand how to work with it, but i dont understand the outcome!!

10:32 "Free real estate"

Awesome video btw!!

Should have put emphasis on v3 being the free variable (row not containing a leading 1) which is why you chose v3=t. other than that very clear explanation!

is E-3 perpendicular to E3. both span of E-3 is perpendicular to each other, but E3 is not perpendicular to both. this is my thinking. please explain me.

Thank You

don’t understand why you use row reduction when it really isn’t necessary, the eigenvectors are obvious just from looking at A-lamda x identity

also why have you overcomplicated the eigenvector for eigenvalue=3? what’s wrong with (1,1,1)

Thanks Sal!

10:32 It's free real estate.

so eigen vectors and eigen space is the same thing?

didnt understand

The real estate part really helped me out!

Sal I'd really enjoy it if the example you made wasn't of nullity 2, as a full matrix probably would've helped me more.