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# Eigenvalues of a 3x3 matrix

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

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• 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.
• 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.
• 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
• 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)).
• 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)?
• I have a Texas Instruments TI-85 (quite an old calculator and superseded now).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
• At , shouldn't the possible divisors of 27 be +/-1, +/-3, +/-9, +/-27?
• Yes, Sal should've checked the negative divisors if none of the positive divisors had worked.
• 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)
• What name is the method that Sal use to factor out the characteristic polynomial in the video?
• 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?
• the dimension is at least one and up to and including the multiplicity. in this case, either 1 or 2.
• 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?
• You are correct, it does row reduce to identity: http://www.wolframalpha.com/input/?i=rowreduce%5B%7B-1%2C2%2C2%7D%2C%7B2%2C2%2C-1%7D%2C%7B2%2C-1%2C2%7D%5D.

This means A is invertible. The matrix that has to be singular however is (λI - A).
• in my book it says the det( A- lambda*I)=0
im guessing both are correct, but why is this?

book is from Pearson so pretty sure that it is correct, also my proffesor taught it that way as well
• Yes, they mean the same thing. If det(λI - A) = 0, then det(A - λI) = 0. This is because, if you multiply a matrix, like λI - A, by a scalar, like -1, so that you get A - λI, the determinant of the new matrix is just the determinant of the old matrix times that scalar raised to the power of the number of dimensions of the matrix. Since in this case, it is a 3 by 3 matrix, the new determinant is the old determinant, 0, multiplied by the scalar, -1, raised to the 3rd power, which just leaves 0 again.
• Hi,
1) is there a way to verify the eigenvalues somehow? That way I could see if I didn't make a silly mistake.

2) When the multiplicity are all one for the eigenvalues (not this example), is that proof enough to say that the matrix is diagonalizable? for example (l - 2) (l - 3) (l + 3)

3) if instead of doing Saurus I just calculate three determinants and at a certain point I have
(lambda = L)
-1.(L-4)(-L-6) this would be (-L+4)(-L-6). Would it be ok to write (L-4)(L+6)?

Help with any of the 3 questions is highly appreciated!
Thanks
(1 vote)
• For the first question, I'm going to assume you're taking det(A-lambda*I) correctly. If you solve the characteristic equation set = 0, the roots you find you should be able to double check by plugging them back into the characteristic equation itself and getting 0. Besides that, just make sure you're taking the determinant correctly. Also, if you take that eigenvalue and find an associated eigenvector, you should be able to use the original matrix (lets say A) and multiple A by the eigenvector found and get out the SAME eigenvector (this is the definition of an eigenvector).

For the second question: Yes. If you have 3 distinct eigenvalues for a 3x3 matrix, it is diagonalizable because we will have 3 distinct associated eigenvectors. To generalize, an nxn matrix will be diagonalizable if you have n distinct eigenvectors *Note:* we could have less than n distinct eigenvalues and end up with n eigenvectors through multiplicities, meaning A would still be diagonalizable in that scenario.