# Finding eigenvalues and eigenvectors of a matrix

I'm asked to find the eigenvalues and eigenvectors of an nxn matrix. Up until now I thought eigenvectors and eigenvalues are something that's related to linear transformations. The said matrix is not one of any linear transformation. What do I do?

Multiplication by any matrix is a linear transformation.

I'm sure your book gives detailed instructions for how to find eigenvalues and eigenvectors.

So an nxn matrix A is a linear transformation from V -> V where dim V = n?

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With a lineair transformation f:V->W you can always associate an m x n matrix.

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So an nxn matrix A is a linear transformation from V -> V where dim V = n?

Basically, yes. More generally, it could be V->W where V and W both have dimension n, but then they are isomorphic anyway.

Any matrix represents a linear transformation.

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... with respect to some choice of basis.

Linear maps, basis free; matrices, with basis.

But I'm not given any basis.

Treadstone, quit hanging around here and look in your book. It will have unambiguous instructions for how to find the eigenvalues of a matrix.

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What we said was that any linear transformation can be written as a matrix- in a given basis.

Any matrix is a linear transformation with "canonical" basis
<1, 0, 0,...>, <0, 1, 0,...>, <0, 0, 1,...>,...

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An mxn matrix is naturally associated with a linear transformation from the space of nx1 matrices to the space of mx1 matrices.

Bases have nothing to do with it! They only come into play when you're trying to translate the general case into the language of matrices.

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My follow up: so you're not given a basis explicitly but that doesn't matter. You don't need to have one to work on a matrix any more than you need to know whether I'm working in feet, metres, or lightyears to know that twice one unit of distance is two units of distance. The fact of the matter is that there is an implicitly used representation of vectors as Halls states, but you do not know whether the (1,0,0) of one question corresponds to the (0,1,0) of another and nor do you need to, everything is relative.

What are the eigenvalues of a refleciton in the y-axis (in 2-d)? What is the matrix representing this linear transformation? What basis did you use? Pick another basis, have the eigenvalues changed?

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It's hard for me to imagine a linear algebra course in which you have learned about eigenvalues for linear transformations without learning about eigenvalues for matrices first. Often it is difficult to convince students that linear algebra isn't simply about matrices!

We use a book called "Linear Algebra Done Right" by S. Axler. I don't know how his approach is different from anyone else's, but my professor seems to like this approach a lot.

Let me just point out the last remnant of my confusion:

Given some square matrix, say

Code:
<1 2 3>
<4 5 6>
<7 8 9>

If I want to find the eigenvalues of this matrix, I will have to solve

Code:
<1 2 3><x1>   <ex1>
<4 5 6><x2> = <ex2>
<7 8 9><x3>   <ex3>

Where e are the possible eigenvalues, correct?

A friend of mine did it wrt to bases, i.e., given some basis (b1, b2, b3),

b1+4b2+7b3=eb1
2b1+5b2+8b3=eb2
3b1+6b2+9b3=eb3

Is this approach equivalent?

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In no sense has your friend done it 'with respect to a basis' and you not. You have done exactly the same thing, you have put it together in matrix form, he has not, that is all. Neither of them has anything to do with picking a basis, or not picking one.

doing something in vector form is not the same as 'picking a basis'

Ok, suppose I say consider the vector space of polynomials of degree 2 or less (this is 3d over a base field, say R, it doesn't matter). I describe a linear map that sends 1 to 2, x to 3x and x^2 to 4x^2, what are the eigenvalues? They are 2,3,4 with the obvious eigenvectors. Note how I've not given you a basis at all? I could ask you what the matrix corresponding to that transformation is with respect to the natural basis {1,x,x^2}, or to another basis such as {1-x, x-x^2, x^2-1}. The eigenvalues would remain unchanged.

I can equally ask you what the eigenvalues are for some square matrix without telling you anything about the underlying vector space or the basis of it I picked, right?

True. Then I guess the difficulty lies in solving the system I wrote above, which I am unable to. I've reduced it to 3 equations with 2 unknowns each (plus the eigenvalue e). I'm guessing that's a technical problem and not a theoretical one?

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you know what the characteristic equation is? if you can do this for linear maps you can do it for matrices, which are linear maps.

I don't "know" what a characteristic equation is since we haven't learned it. Here's the problem. I'm asked to find the eigenvalues of

Code:
15 -4 -2
27 -8 -3
58 -14 -9

I've reduced this to solving

$$20x^3+246x^2+769x+764=0$$

But I don't know if I got this right, since Mathematica gave me this REALLY complicated solution.

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what method have you been given for fiding eigenvalues of your linear maps? Matrices *are* linear maps, so the method works here, surely.

Most theorems we learned so far at abstract existence proofs. The only concrete technique that I know of is finding the roots of the minimal polynomial of T, and we haven't learned anything about FINDING the said polynomial.

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Mv=kv iff (M-k)v=0, so det(M-k) is zero for eigenvalues, so find the roots of the poly P(x)=det(M-x) (this is the characteristic equation).

Thanks for the help, but once again we haven't "learned" what a determinant is. I'll just do it anyway, though, seems hopeless to get stuck on a single question.

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Well, the other thing to do is simply solve those simultaneous equations you wrote down. Solving simultaneous equations cannot be 'something you've not yet been officially taught', surely.

I did, but like I said, I got $$20x^3+246x^2+769x+764=0$$, which is impossible since the operator is on a vector space over the reals, and that equation has 2 complex roots.

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So? (There is nothing that requires eigenvalues to lie in the same field as the entries of the matrix; they lie in the algebraic closure.)

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Ok I redid my computions 4 times and I found that I left out a "-" sign somewhere. Now, however, I only have 2 eigenvalues, since I got $$(2+L)^2(31+17L)=0$$, where L is the eigenvalue. I'm supposed to show that the matrix is diagonalizable, but since I only have 2 distinct eigenvalues, this doesn't work out.

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Ok I redid my computions 4 times and I found that I left out a "-" sign somewhere. Now, however, I only have 2 eigenvalues, since I got $$(2+L)^2(31+17L)=0$$, where L is the eigenvalue. I'm supposed to show that the matrix is diagonalizable, but since I only have 2 distinct eigenvalues, this doesn't work out.

The matrix in post #17 has 3 distinct eigenvalues, you might want to check your work again.

In any case, it's not necessary to have 3 distinct eigenvalues for a 3x3 matrix to be diagonalizable. You've made no mention of the corresponding eigenspaces, or any eigenvectors at all for that matter. Can you find an eigenvector corresponding to L=-31/17? This should tell you something was off in your calculation.