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Eigenvalues of a 3x3 matrix

Determining the eigenvalues of a 3x3 matrix. Created by Sal Khan.

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  • blobby green style avatar for user 4F534D414E
    How can I find the determinant for a matrix bigger than 3 x 3? The video player on this video doesn't give me the option of seeing the time with numbers but about 1/3 of the video the rule of Zaurus is mentioned. Is that the easier way for something bigger than 3 x 3. Thanks
    (8 votes)
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    • blobby green style avatar for user AlexHuang861
      The method used in this video ONLY works for 3x3 matrices and nothing else. Finding the determinant of a matrix larger than 3x3 can get really messy really fast. There are many ways of computing the determinant. One way is to expand using minors and cofactors. I don't know if Khan has explained that in one of his videos but it works well if there are a lot of zeros in a matrix. It is a little confusing to describe in words in such a small space so i leave it to you to look it up if you don't know that method already. Another method is by doing row operations. Switching two rows in a matrix does not change the determinant. Adding a multiple of one row to another row does not change the determinant. You can also factor a constant out of one row and the determinant becomes the determinant of the new matrix multiplied by the constant. Row reduce your matrix until you get a diagonal one and it should be easy to compute.
      (18 votes)
  • blobby green style avatar for user Simon Benjamin Moffatt
    Isn't the equivalency Ax = [lambda]x <==> (A - I[lambda])x = 0 and not (I[lambda] - A)x = 0?
    (9 votes)
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    • blobby green style avatar for user AlexHuang861
      Actually both work. the characteristic polynomial is often defined by mathematicians to be det(I[λ] - A) since it turns out nicer. The equation is Ax = λx. Now you can subtract the λx so you have (A - λI)x = 0. but you can also subtract Ax to get (λI - A)x = 0. You can easily check that both are equivalent.
      (12 votes)
  • leafers tree style avatar for user Draztak
    At , wouldn't it have been way easier to factor the quadratic polynomial by grouping? We could have factored out a [lambda]^2 from the first two terms and a -9 from the last two, giving us ([lambda]^2 - 9) * ([lambda] - 3) = 0 which easily leads us to find that [lambda] can be 3 or - 3.
    (5 votes)
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    • leaf green style avatar for user Lucas Van Meter
      Yes,

      As Sal mentions there is (long and tedious) formula to factoring but in practice it is more of an art. The method you describe is elegant and when you can you should do it. The way Sal describes might be a little more algorithmic so if you can't see a clever way to do it is method is nice to know. But by all means use your way and continue down the path to becoming an algebra ninja.
      (4 votes)
  • blobby green style avatar for user dannyboy
    Can't you simply apply row operations to make it a triangular matrix, before subtracting? Seems like saving tons of work, since the eigenvalues of a triangular matrix are on the main diagonal
    (5 votes)
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    • leaf green style avatar for user Lucas Van Meter
      Good question Danny,

      The short answer is no, while it is true that row operations preserve the determinant of a matrix the determinant does not split over sums. We want to compute det(M-lambda I_n) which does not equal det(M)-det(lambda n).

      The best way to see what problem comes up is to try it out both ways with a 2x2 matrix like ((1,2),(3,4)).
      (4 votes)
  • leaf green style avatar for user SteveSargentJr
    Is there a way to calculate eigenvalues & eigenvectors with a regular graphing calculator? For instance, how would you calculate eigenvalues for a matrix larger than 3x3 in practice (without resorting to extremely tedious algebra by hand)?
    (2 votes)
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    • old spice man green style avatar for user newbarker
      I have a Texas Instruments TI-85 (quite an old calculator and superseded now). It's the same as the calculator Sal uses a lot on the screen. On that there is a MATRX area where you can enter a matrix. Then you can choose the MATH submenu and choose the eigVl and eigVc menu items for eigenvalue and eigenvector respectively. Do you have a graphing calculator already? If so, what make?

      Wolfram Alpha is great for doing these computations too. If you give it a 3x3 matrix, it'll tell you some properties (including characteristic polynomial, eigenvalues/vectors):
      http://www.wolframalpha.com/input/?i=%7B%7B1%2C2%2C3%7D%2C%7B3%2C2%2C1%7D%2C%7B2%2C1%2C3%7D%7D&lk=4&num=1
      (5 votes)
  • blobby green style avatar for user Erik Martines Sanches
    At , shouldn't the possible divisors of 27 be +/-1, +/-3, +/-9, +/-27?
    (3 votes)
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  • starky ultimate style avatar for user 鄭聖文
    What name is the method that Sal use to factor out the characteristic polynomial in the video?
    (3 votes)
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  • aqualine ultimate style avatar for user William Chargin
    Does the fact that λ = 3 is a double root imply that the eigenspace E_3 will be of dimension 2? In other words, is the dimension of an eigenspace E_λ equal to the multiplicity of the λ root of the characteristic polynomial?
    (3 votes)
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  • blobby green style avatar for user wajacaranda
    Lucky for us, no need to calculate it by hand anymore, any mathematical software (matlab, mathematica, etc) or math lib of any programming language can do it for you, just type Eig(A)
    (3 votes)
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  • blobby green style avatar for user Michael Dahms
    in this video lesson the original matrix A row reduced to Isub3 identity matrix according to my trusty TI-84 buddy. So this makes me feel like sal sorta pulled a fast one on me. can someone explain and reassure me this isn't slight of math hand?
    (2 votes)
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Video transcript

We figured out the eigenvalues for a 2 by 2 matrix, so let's see if we can figure out the eigenvalues for a 3 by 3 matrix. And I think we'll appreciate that it's a good bit more difficult just because the math becomes a little hairier. So lambda is an eigenvalue of A. By definition, if and only if-- I'll write it like this. If and only if A times some non-zero vector v is equal to lambda times that non-zero vector v. Let we write that for some non-zero. I could call it eigenvector v, but I'll just call it for some non-zero vector v or some non-zero v. Now this is true if and only if, this leads to-- I'll write it like this. This is true if and only if-- and this is a bit of review, but I like to review it just because when you do this 10 years from now, I don't want you to remember the formula. I want you to just remember the logic of how we got to it. So this is true if and only if-- let's just subtract Av from both sides-- the 0 vector is equal to lambda- instead of writing lambda times v, I'm going to write lambda times the identity matrix times v. This is the same thing. The identity matrix times v is just v. Minus Av. I just subtracted Av from both sides, rewrote v as the identity matrix times v. Well this is only true if and only if the 0 vector is equal to lambda times the identity matrix minus A times v. I just factored the vector v out from the right-hand side of both of these guys, and I'm just left with some matrix times v. Well this is only true-- let me rewrite this over here, this equation just in a form you might recognize it. Lambda times the identity matrix times A. This is just some matrix. This matrix times v has got to be equal to 0 for some non-zero vector v. That means that the null space of this matrix has got to be nontrivial. Or another way to think about it is that its columns are not linearly independent. Or another way to think about it is it's not invertible, or it has a determinant of 0. So lambda is the eigenvalue of A, if and only if, each of these steps are true. And this is true if and only if-- for some at non-zero vector, if and only if, the determinant of lambda times the identity matrix minus A is equal to 0. And that was our takeaway. I think it was two videos ago or three videos ago. But let's apply it now to this 3 by 3 matrix A. We're going to use the 3 by 3 identity matrix. So we want to concern ourselves with-- lambda times the identity matrix is just going to be-- times the 3 by 3 identity matrix is just going to be-- this is, let me write this. This is lambda times the identity matrix in R3. So it's just going to be lambda, lambda, lambda. And everything else is going to be 0's. The identity matrix had 1's across here, so that's the only thing that becomes non-zero when you multiply it by lambda. Everything else was a 0. So that's the identity matrix times lambda. So lambda times the identity matrix minus A is going to be equal to-- it's actually pretty straightforward to find. Everything along the diagonal is going to be lambda minus-- let's just do it. Lambda minus minus 1-- I'll do the diagonals here. Lambda minus minus 1 is lambda plus 1. And then 0 minus 2-- I'll do that in a different color. 0 minus 2 is minus 2. 0 minus 2 is minus 2. 0 minus 2 is minus 2. Let's do this one. 0 minus 2 is minus 2. 0 plus or minus minus 1 is 0 plus 1, which is 1. And then let's just do this one. 0 minus minus 1. That's one. Let me finish up the diagonal. And then you have lambda minus 2. And then you have lambda minus 2. So lambda is an eigenvalue of A if and only if the determinant of this matrix right here is equal to 0. Let's figure out its determinate. And the easiest way, at least in my head to do this, is to use the rule of Sarrus. So let's use the rule of Sarrus to find this determinant. So I just rewrite these rows right there. I could just copy and paste them really. I just take those two rows. And then let me paste them, put them right there. It's a little bit too close to this guy, but I think you get the idea. And now the rule of Sarrus I just take this product plus this product plus this product and then I subtract out this product times this product times this product. We'll do that next. So this product is lambda plus 1 times lambda minus 2 times lambda minus 2. That's that one there. And then plus, let's see, minus 2 times minus 2. That's plus 4. And then we have minus 2 times minus 2 plus 4 times 1. So that is plus 4 again. And then we do minus this column times this column. Minus this column minus this column and then-- or I shouldn't say column, but diagonal really. So we say minus 2 times minus 2. Let me write this. Minus 2 times minus 2, which is 4. Times lambda minus 2. That was this diagonal. And then we have minus-- what is this going to be? Going to be minus 1 times lambda plus 1. So minus lambda plus 1. And then you go down this diagonal. Minus 2 times minus 2 is 4. So it's going to be 4 times lambda minus 2 and we're subtracting. So minus 4 times lambda minus 2. And let's see if we can simplify this. So this blue stuff over here-- let's see, these guys right here become an 8 and then this becomes-- this becomes lambda plus 1. Times-- if I multiply these two guys out, lambda squared minus 4 lambda. Minus 2 lambda and then minus 2 lambda. So minus 4 lambda. Plus 4. And then I have this plus 8 here. And then I have-- let's see. I have minus 4 times lambda. Let me just multiply everything out. So I have minus 4 lambda plus 8 minus lambda minus 1 minus 4 lambda plus 8. And then let me simplify this up a little bit. So this guy over here-- let's see. The constant terms, I have an 8, I have a minus 1, I have an 8 and I have an 8. So that's 24 minus 1. So that is a 23. And then the lambda terms I have a minus 4 lambda. I have a minus lambda and I have a minus 4 lambda. So it's minus 8, minus 1. So I have minus 9 lambda. Plus 23. And now I have to simplify this out. So first I can take lambda and multiply it times this whole guy right there. So it's going to be lambda cubed minus 4 lambda squared plus 4 lambda. And then I can take this one and multiply it times that guy. So plus lambda squared. Minus 4 lambda plus 4. And now of course, we have these terms over here. So we're going to have to simplify it again. So what are all of our constant terms? We have a 23 and we have a plus 4. So we have a 27. Plus 27. And then, what are all of our lambda terms? We have a minus 9 lambda and then we have a-- let's see. We have a minus 9 lambda, we have a plus 4 lambda, and then we have a minus 4 lambda. So these two cancel out. So I just have a minus 9 lambda. And then, what are my lambda squared terms? I have a plus lambda squared and I have a minus 4 lambda squared. So if you add those two that's going to be minus 3 lambda squared. And then finally, I have only one lambda cubed term, that right there. So this is the characteristic polynomial for our matrix. So this is the characteristic polynomial and this represents the determinant for any lambda. The determinant of this matrix for any lambda. And we said that this has to be equal to 0 if any only if lambda is truly an eigenvalue. So we're going to set this equal to 0. And unlucky or lucky for us, there is no real trivial-- there is no quadratic. Well there is, actually, but it's very complicated. And so it's usually a waste of time. So we're going to have to do kind of the art of factoring a quadratic polynomial. I got this problem out of a book and I think it's fair to say that if you ever do run into this in an actual linear algebra class or really, in an algebra class generally-- it doesn't even have to be in the context of eigenvalues, you probably will be dealing with integer solutions. And if you are dealing with integer solutions, then your roots are going to be factors of this term right here. Especially if you have a 1 coefficient out here. So your potential roots-- in this case, what are the factors of 27? So 1, 3, 9 and 27. So all these are potential roots. So we can just try them out. 1 cubed is 1 minus 3. So let me try 1. So if we try a 1, it's 1 minus 3 minus 9 plus 27. That does not equal 0. It's minus 2 minus 9 is minus 11. Plus 16. That does not equal 0. So 1 is not a root. If we try 3 we get 3 cubed, which is 27. Minus 3 times 3 squared is minus 3 times 3, which is minus 27. Minus 9 times 3, which is minus 27. Plus 27. That does equal 0. So lucky for us, on our second try we were able to find one 0 for this. So if 3 is a 0, that means that x minus 3 is one of the factors of this. So that means that this is going to be x minus 3 times something else. Or I should say, lambda minus 3. So let's see what the other root is. So if I take lambda minus 3 and I divide it into this guy up here, into lambda cubed minus 3 lambda squared minus 9 lambda plus 27, what do I get? Lambda goes into lambda cubed lambda squared times. Lambda squared times that. Lambda squared times lambda is lambda cubed. Lambda squared times minus 3 is minus 3 lambda squared. You subtract these guys, you get a 0. You get 0. And then we can put here-- well, we could do it either way. We could put it down the minus 9. We could bring down everything really. So now you have minus 9 lambda plus 27. You can almost imagine we just subtracted this from this whole thing up here. And we're just left with these terms right here. And so lambda minus 3 goes into this. Well lambda minus 3 goes into 9 lambda. It goes into 9 lambda minus 9 times. So I'll just write minus 9 here. Minus 9 times lambda minus 3 is minus 9 lambda plus 27. So it went in very nicely. So you get to 0. Our characteristic polynomial has simplified to lambda minus 3 times lambda squared minus 9. And of course, we're going to have to set this equal to 0 if lambda is truly an eigenvalue of our matrix. And this is very easy to factor. So this becomes lambda minus 3 times-- lambda squared minus 9 is just lambda plus 3 times lambda minus 3. And all of that equals 0. And these roots, we already know one of them. We know that 3 is a root and actually, this tells us 3 is a root as well. So the possible eigenvalues of our matrix A, our 3 by 3 matrix A that we had way up there-- this matrix A right there-- the possible eigenvalues are: lambda is equal to 3 or lambda is equal to minus 3. Those are the two values that would make our characteristic polynomial or the determinant for this matrix equal to 0, which is a condition that we need to have in order for lambda to be an eigenvalue of a for some non-zero vector v. In the next video, we'll actually solve for the eigenvectors, now that we know what the eigenvalues are.