# Eigenvectors and eigenspaces for a 3×3 matrix | Linear Algebra | Khan Academy 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.

### 81 thoughts on “Eigenvectors and eigenspaces for a 3×3 matrix | Linear Algebra | Khan Academy”

• December 3, 2009 at 11:54 pm

Hallelujah! PRAISE THE LORD!

• April 16, 2010 at 9:59 am

• April 21, 2010 at 10:40 pm

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

• May 14, 2010 at 7:30 pm

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

• May 23, 2010 at 11:29 pm

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.

• May 23, 2010 at 11:34 pm

@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

• May 23, 2010 at 11:35 pm

@MartinRyleOShea

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

• September 12, 2010 at 2:27 am

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

• September 29, 2010 at 12:33 pm

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

• October 24, 2010 at 7:48 pm

• December 18, 2010 at 10:39 am

Thank you this clear many pictures for me 🙂

• January 27, 2011 at 4:55 pm

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

• February 14, 2011 at 4:36 pm

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") =)

• February 21, 2011 at 2:13 am

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

• April 28, 2011 at 1:59 am

totally saving my ass for my exam tomorrow.

• May 12, 2011 at 12:11 pm

I love you <3

• June 9, 2011 at 5:31 am

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 😀

• June 11, 2011 at 11:59 pm

I LOVE METH!!!!!…..i mean MATH!!!!

• June 12, 2011 at 7:57 pm

YES

• June 12, 2011 at 8:53 pm

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

• June 14, 2011 at 9:29 am

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 🙂

• June 18, 2011 at 9:53 pm

@unkown1414
totally agree
must be tablets man, he's too precise

• August 10, 2011 at 2:43 am

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

• August 13, 2011 at 6:09 am

you saved me on my final last spring.

• October 7, 2011 at 6:07 pm

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

• October 30, 2011 at 9:03 pm

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

• November 3, 2011 at 5:33 am

Thank you! Very clear and comprehensible.

• November 14, 2011 at 3:35 am

• December 2, 2011 at 3:53 am

eigenkosommak

• January 3, 2012 at 11:44 am

The real superman!

• February 5, 2012 at 4:15 pm

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)!

• March 10, 2012 at 5:46 pm

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

• April 3, 2012 at 8:18 pm

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.

• July 12, 2012 at 2:56 pm

Jazak allah

• August 16, 2012 at 8:57 pm

( 1 0 1 )
( 0 1 0 )
( 0 0 0 )

• August 16, 2012 at 8:58 pm

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

• August 18, 2012 at 12:22 pm

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

• October 17, 2012 at 1:36 pm

LOL

• November 6, 2012 at 8:03 am

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

• December 2, 2012 at 12:48 am

thank you! my teacher aint got nothin on you

• December 2, 2012 at 3:35 am

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

• December 9, 2012 at 7:00 am

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

• December 20, 2012 at 12:36 am

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

• December 27, 2012 at 11:23 am

thank you

• January 1, 2013 at 1:11 am

v1+v3=0
v2=0

• January 6, 2013 at 1:26 pm

thank you very much

• March 10, 2013 at 5:51 pm

I love you.

• April 6, 2013 at 12:16 am

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

• April 9, 2013 at 12:50 pm

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!

• April 18, 2013 at 4:02 pm

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

• April 18, 2013 at 11:53 pm

Lets just change colours for fun 😀

• April 29, 2013 at 6:59 pm

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

• May 3, 2013 at 9:35 am

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

• July 21, 2013 at 11:11 pm

much better than my books! thanks a lot

• July 25, 2013 at 11:26 am

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

• October 2, 2013 at 7:52 am

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

• October 4, 2013 at 1:46 pm

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?

• December 3, 2013 at 6:02 pm

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

• December 9, 2013 at 8:45 pm

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

• December 16, 2013 at 6:35 am

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.

• March 28, 2014 at 5:48 am

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)..

• May 14, 2014 at 11:44 am

Real-life superhero!

• July 31, 2014 at 11:38 am

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

• October 29, 2014 at 3:02 am

Dear Khan,
You da real MVP.

• November 17, 2014 at 7:06 pm

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.

• June 28, 2015 at 10:17 pm

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

• March 10, 2016 at 10:06 pm

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.

• April 11, 2016 at 9:55 pm

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

• October 12, 2016 at 4:44 pm

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!!

• January 27, 2017 at 8:02 am

10:32 "Free real estate"

Awesome video btw!!

• February 19, 2017 at 1:41 am

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!

• June 2, 2017 at 2:36 am

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.

• December 10, 2017 at 9:22 am

Thank You

• December 20, 2017 at 1:46 pm

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

• December 20, 2017 at 1:48 pm

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

• April 27, 2018 at 10:15 pm

Thanks Sal!

• June 5, 2018 at 9:15 am

10:32 It's free real estate.

• December 13, 2018 at 5:37 am

so eigen vectors and eigen space is the same thing?

• January 1, 2019 at 2:46 pm

didnt understand

• January 30, 2019 at 12:52 am
• 