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Linear algebra
Unit 3: Lesson 5
Eigen-everything- Introduction to eigenvalues and eigenvectors
- Proof of formula for determining eigenvalues
- Example solving for the eigenvalues of a 2x2 matrix
- Finding eigenvectors and eigenspaces example
- Eigenvalues of a 3x3 matrix
- Eigenvectors and eigenspaces for a 3x3 matrix
- Showing that an eigenbasis makes for good coordinate systems
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Finding eigenvectors and eigenspaces example
Finding the eigenvectors and eigenspaces of a 2x2 matrix. Created by Sal Khan.
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In class, we learned [A-(lambda)I]=0, yet here you teach [(lambda)I-A]=0. I see how these are both valid from Av=(lambda)v, however I am unsure if both formulas are equivalent. Does using one formula or the other effect your eigen-vectors or a general solution in a significant way?(47 votes)- They are the same! det[A] is not the same as det[-A], (it is equal to -det[A]), but in this case we are testing against 0 and 0 = -0, they are equivalent. More slowly, using m for lambda:
-(A-mI) = -A + mI = mI - A,
det[mI - A] = det[-(A-mI)] = -det[A-mI]
det[mI - A] = 0 equiv. -det[A-mI] = 0 equiv. det[A-mI] = 0.
So they are equivalent and either way is fine.(107 votes)
What is a null space?(24 votes)
A null space is commonly referred to as the "kernel" of some function and is the set of all points that send a function to zero i.e all x's such that f(x)= 0 is true.
In terms of linear algebra/linear transformation/matrix algebra you can think of a null space (or kernel) as the set of all vectors v such that
Av=0 where A is a mxn matrix and 0 is the zero vector.(35 votes)
- When Sal picks v_2 to be t, is this arbitrary? Could he also have picked v_1 to be t, making [1,2] the eigenvector for eigenvalue 5? Similarly, this would make [1,-1] the eigenvector for eigenvalue -1. N.B. I think I've just answered my own question using Wolfram Alpha: http://www.wolframalpha.com/input/?i=eigenvectors%28[[1%2C2]%2C[4%2C3]]%29. It appears this is a valid thing to do. I'll post this incase anyone has something interesting to add...(7 votes)
You are absolutely right. You could've pick v_1 to be t and solve the system with v_1 to get v_2. This would be an eigenvector as well(5 votes)
at7:45, why does the Eigenvector equal span (1/2, 1), not span (1, - 1/2)?(9 votes)- these are equivalent, since (1,-1/2) is in the span of (1/2,1) and vice versa. So it doesn't matter which one you choose, both statements are correct.(3 votes)
I understand most of the process, but I feel like the initial question was unanswered. Could someone help me?
So, if I understand correctly, our initial question was "What are the Eigenvalues (lambda) and Eigenvectors (v) that satisfy the equation T(v) = A*v = lambda*v?"
We found the values of lambda that are possible in the previous video (link at bottom).
We then used each distinct possible value of lambda, and plugged it back in to the equation [A-(lambda*I)]v = 0 to determine all possible vectors v that would make that work (the null space).
This is where I get confused. Sal ends up talking in the end of the video (starting about11:00) about the span of these vectors, or the Eigenspace. He acts as if the Eigenspace IS the answer.
Well, is it?
The initial question was "What are the Eigenvalues (lambda) and Eigenvectors (v) that satisfy the equation T(v) = A*v = lambda*v?"
I think that the Eigenspaces would accommodate all combinations of possible Eigenvalues and Eigenvectors, but am I wrong in assuming that? Would we have to specify what the Eigenvalues are? I feel comfortable listing a span as an answer to the set of all possible Eigenvectors, but I feel like I'm not accounting for the 2 distinct Eigenvalues.
Or am I just wrong in what the initial question was?
Previous video link: https://www.khanacademy.org/math/linear-algebra/alternate_bases/eigen_everything/v/linear-algebra--example-solving-for-the-eigenvalues-of-a-2x2-matrix(6 votes)
one point of finding eigenvectors is to find a matrix "similar" to the original that can be written diagonally (only the diagonal has nonzeroes), based on a different basis.
T(v) = A*v = lambda*v is the right relation. the eigenvalues are all the lambdas you find, the eigenvectors are all the v's you find that satisfy T(v)=lambda*v, and the eigenspace FOR ONE eigenvalue is the span of the eigenvectors cooresponding to that eigenvalue.
suppose for an eigenvalue L1, you have T(v)=L1*v, then the eigenvectors FOR L1 would be all the v's for which this is true. the eigenspace of L1 would be the span of the eigenvectors OF L1, in this case it would just be the set of all the v's because of how linear transformations transform one dimension into another dimension. the (entire) eigenspace would be the span of all the eigenvectors from all the eigenvalues.(3 votes)
Is it always necessary to have row of zeros?(4 votes)
Yes, otherwise the vector from that matrix wouldn't be an eigenvector.(6 votes)
- I feel I may have missed a video as I don't know what a "reduced row echelon form of a matrix" is and how to make one?(1 vote)
- Is there a general rule for the relationships of eigenvalues graphically? Eg. Are the eigenvectors of the corresponding eigenvalue perpendicular...?(3 votes)
- The eigenvalues don't have any direction because they're scalars. For some 2x2 matrices the eigenspaces for different eigenvalues are orthogonal, for others not.
An nxn matrix always has n eigenvalues, but some come in complex pairs, and these don't have eigenspaces in R^n, and some eigenvalues are duplicated; so there aren't always n eigenspaces in R^n for an nxn matrix. Some eigenspaces have more than one dimension.(4 votes)
- For E5 shouldnt it be t{2;1] since v1=.05v2 which would mean 2v1=1v2(3 votes)
- Yes, v1 = (v2)/2. What's the solution space (all the vectors v = (v1, v2) | (lambda*I - A)v = 0)? First, what's the free variable in the rref?
(v2 (because it's not a pivot variable - it's not constrained by the solution - it can be any real number). What is v1 in terms of v2?
(v1 is constrained by the solution to be (v2)/2.) What's {v = (v1, v2)} (the entire solution space) in terms of the free variable(s)?
({(v1, v2)} = {((v2)/2, v2) | v2 is a real number}.) How do you express that as a span of basis vectors?
({(v2)/2, v2)} = {v2(1/2, 1)} = span(1/2, 1) = span(1, 2) = {t(1, 2) | t is real}.(2 votes)
- So what if we know what the eigenvectors and eigenvalues are? What practical purpose does this serve? I mean where would you use this?(1 vote)
Eigen values and vectors are used extensively in more advanced economics.(4 votes)
7:45
Video transcript
In the last video, we started
with the 2 by 2 matrix A is equal to 1, 2, 4, 3. And we used the fact that lambda
is an eigenvalue of A, if and only if, the determinate
of lambda times the identity matrix-- in this
case it's a 2 by 2 identity matrix-- minus A
is equal to 0. This gave us a characteristic
polynomial and we solved for that and we said, well, the
eigenvalues for A are lambda is equal to 5 and lambda
is equal to negative 1. That's what we saw in
the last video. We said that if you were trying
to solve A times some eigenvector is equal to lambda
times that eigenvector, the two lambdas, which this equation
can be solved for, are the lambdas 5 and minus 1. Assuming nonzero eigenvectors. So we have our eigenvalues,
but I don't even call that half the battle. What we really want is our
eigenvectors and our eigenvalues. So let's see if we
can do that. So if we manipulate this
equation a little bit and we've manipulate it in the past.
Actually, we've even come up with this statement
over here. We can rewrite this over here
as the 0 vector is equal to lambda times my eigenvector
minus A times my eigenvector. I just subtracted Av
from both sides. We know lambda times some
eigenvector is the same thing as lambda times the identity
matrix times that eigenvector. So all I'm doing is rewriting
this like that. You multiply the identity matrix
times an eigenvector or times any vector, you're just
going to get that vector. So these two things
are equivalent. Minus Av. That's still going to be
able to the 0 vector. So far all I've done is
manipulated this thing. This is really how we got
to that thing up there. You factor out the v so to speak
because we know that matrix vector products exhibit
the distributive property. And we get lambda times the
identity matrix minus A times my eigenvector have got
to be equal to 0. Or another way to say it is, for
any lambda eigenvalue, and let me write it for any
eigenvalue lambda, the eigenvectors that correspond
to that lambda, we can call that the eigenspace
for a lambda. So that's a new word,
eigenspace. Eigenspace just means all
of the eigenvectors that correspond to some eigenvalue. The eigenspace for some
particular eigenvalue is going to be equal to the set
of vectors that satisfy this equation. Well, the set of vectors that
satisfy this equation is just the null space of that
right there. So it's equal to the
null space of this matrix right there. The null space of lambda times
the identity matrix. And by an identity
matrix minus A. And so everything I've done
here, this is true-- this is the general case. But now we can apply
this notion to this matrix A right here. So we know that 5 is
an eigenvalue. Let's say for lambda is equal
to 5, the eigenspace that corresponds to 5 is equal
to the null space of? Well, what is 5 times
the identity matrix? It's going to be the 2
by 2 identity matrix. 5 times the identity matrix is
just 5, 0, 0, 5 minus A. That's just 1, 2, 4, 3. So that is equal to the null
space of the matrix. 5 minus 1 is 4. 0 minus 2 is minus 2. 0 minus 4 is minus 4. And then, 5 minus 3 is 2. So the null space of this matrix
right here-- and this matrix is just an actual
numerical representation of this matrix right here. The null space of this matrix
is the set of all of the vectors that satisfy this or all
of the eigenvectors that correspond to this eigenvalue. Or, the eigenspace that
corresponds to the eigenvalue 5. These are all equivalent
statements. So we just need to figure out
the null space of this guy is all of the vectors that satisfy
the equation 4 minus 2, minus 4, 2 times some
eigenvector is equal to the 0 vector. And the null space of a matrix
is equal to the null space of the reduced row echelon
form of a matrix. So what's the reduced row
echelon form of this guy? Well, I guess a good starting
point-- let me keep my first row the same, 4 minus 2. And let me replace my second
row with my second row plus my first row. So minus 4 plus 4 is 0. 2 plus minus 2 is 0. Now, let me divide my
first row by 4 and I get 1, minus 1/2. And then I get 0, 0. So what's the null
space of this? This corresponds to v. This times v1, v2-- that's just
another way of writing my eigenvector v-- has got to
be equal to the 0 vector. Or another way to say it is that
my first entry v1, which corresponds to this pivot
column, plus or minus 1/2 times my second entry has
got to be equal to that 0 right there. Or, v1 is equal to 1/2 v2. And so if I wanted to write all
of the eigenvectors that satisfy this, I could
write it this way. My eigenspace that corresponds
to lambda equals 5. That corresponds to the
eigenvalue 5 is equal to the set of all of the vectors, v1,
v2, that are equal to some scaling factor. Let's say it's equal
to t times what? If we say that v2 is equal to t,
so v2 is going to be equal to t times 1. And then, v1 is going to be
equal to 1/2 times v2 or 1/2 times t. Just like that. For any t is a member
of the real numbers. If we wanted to, we could
scale this up. We could say any real
number times 1, 2. That would also be the span. Let me do that actually. It'll make it a little
bit cleaner. Actually, I don't
have to do that. So we could write that the
eigenspace for the eigenvalue 5 is equal to the span of
the vector 1/2 and 1. So it's a line in R2. Those are all of the
eigenvectors that satisfy-- that work for the equation
where the eigenvalue is equal to 5. Now what about when
the eigenvalue is equal to minus 1? So let's do that case. When lambda is equal to minus
1, then we have-- it's going to be the null space. So the eigenspace for lambda is
equal to minus 1 is going to be the null space of lambda
times our identity matrix, which is going to be minus
1 and 0, 0, minus 1. It's going to be minus 1 times
1, 0, 0, 1, which is just minus 1 there. Minus A. So minus 1, 2, 4, 3. And this is equal to the null
space of-- minus 1, minus 1 is minus 2. 0 minus 2 is minus 2. 0 minus 4 is minus 4 and minus
1 minus 3 is minus 4. And that's going to be equal
to the null space of the reduced row echelon
form of that guy. So we can perform some row
operations right here. Let me just put it in reduced
row echelon form. So if I replace my second row
plus 2 times my first row. So I'll keep the first
row the same. Minus 2, minus 2. And then my second row, I'll
replace it with two times-- I'll replace it with it plus
2 times the first. Or even better, I'm going to replace it
with it plus minus 2 times the first. So minus
4 plus 4 is 0. And then if I divide the top
row by minus 2, the reduced row echelon form of this matrix
right here or this matrix right here is going
to be 1, 1, 0. So the eigenspace that
corresponds to the eigenvalue minus 1 is equal to the null
space of this guy right here It's the set of vectors
that satisfy this equation: 1, 1, 0, 0. And then you have v1,
v2 is equal to 0. Or you get v1 plus-- these
aren't vectors, these are just values. v1 plus v2 is equal to 0. Because 0 is just equal to
that thing right there. So 1 times v1 plus 1 times v2 is
going to be equal to that 0 right there. Or I could write v1 is
equal to minus v2. Or if we say that v2 is equal to
t, we could say v1 is equal to minus t. Or we could say that the
eigenspace for the eigenvalue minus 1 is equal to all of the
vectors, v1, v2 that are equal to some scalar t times v1 is
minus t and v2 is plus t. Or you could say this is equal
to the span of the vector minus 1 and 1. So let's just graph this a
little bit just to understand what we just did. We were able to find
two eigenvalues for this, 5 and minus 1. And we were able to find all
of the vectors that are essentially-- or, we were able
to find the set of vectors that are the eigenvectors that
correspond to each of these eigenvalues. So let's graph them. So if we go to R2, let
me draw my axes, this is my vertical axis. That's my horizontal axis. So all of the vectors that
correspond to lambda equal 5 are along the line 1/2, 1. Or the span of 1/2, 1. So that is 1. That is 1. So you go 1/2 and 1
just like that. So that's that vector,
spanning vector. But anything along the span of
this, all the multiples of this, are going to be
valid eigenvectors. So anything along that line,
all of the vectors when you draw them in standard
position, point to a point on that line. All of these vectors, any vector
on there is going to be a valid eigenvector and the
corresponding eigenvalue is going to be equal to 5. So you give me this
guy right here. When you apply the
transformation, it's going to be five times this guy. If this guy is x, t of
x is going to be five times this guy. Whatever vector you give along
this line, the transformation of that guy, the transformation
is literally, multiplying it by
the matrix A. Where did I have the matrix A? The matrix A right up there. You're essentially just scaling
this guy by 5 in either direction. This is for lambda equal 5. And for lambda equals 1, it's
the span of this vector, which is minus 1, 1. Which looks like this. So this vector looks
like that. We care about the span of it. Any vector that when you draw in
standard position lies, or points to, points on this line,
will be an eigenvector for the eigenvalue minus 1. So lambda equals minus 1. Let's say you take the
spanning vector here. You apply the transformation,
you're going to get minus 1 times it. So if this is x, the
transformation of x is going to be that right there. Same length, just in the
opposite direction. If you have this guy right
here, you apply the transformation, it's going to
be in the same spanning line just like that. So the two eigenspaces for the
matrix-- where did I write it? I think it was the matrix
1, 2, 3-- 1, 2, 4, 3. The two eigenvalues were
5 and minus 1. And then it has an infinite
number of eigenvectors, so they actually create
two eigenspaces. Each of them correspond to
one of the eigenvalues. And these lines represent
those two eigenspaces. You give me any vector in
either of these sets and they're going to be
an eigenvector. I'm using the word
vector too much. You give me any vector in either
of these sets, and they will be an eigenvector
for our matrix A. And then, depending on which
line it is, we know what their transformation is going to be. If it's going to be on
this guy, we take the transformation, the resulting
vector's going to be five times the vector. If you take one of these
eigenvectors and you transform it, the resulting transformation
of the vector's going to be minus 1
times that vector. Anyway, we now know what
eigenvalues, eigenvectors, eigenspaces are. And even better, we know how
to actually find them.