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Treadstone 71

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Treadstone 71

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- #2

0rthodontist

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

- #3

Treadstone 71

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

- #4

TD

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

- #5

HallsofIvy

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Treadstone 71 said: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.

- #6

matt grime

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... with respect to some choice of basis.

Linear maps, basis free; matrices, with basis.

Linear maps, basis free; matrices, with basis.

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Treadstone 71

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But I'm not given any basis.

- #8

0rthodontist

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- #9

HallsofIvy

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Any matrix

<1, 0, 0,...>, <0, 1, 0,...>, <0, 0, 1,...>,...

- #10

Hurkyl

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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.

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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- #11

matt grime

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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?

- #12

HallsofIvy

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- #13

Treadstone 71

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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

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

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?

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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- #14

matt grime

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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?

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- #16

matt grime

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- #17

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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

I've reduced this to solving

[tex]20x^3+246x^2+769x+764=0[/tex]

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

Code:

```
15 -4 -2
27 -8 -3
58 -14 -9
```

I've reduced this to solving

[tex]20x^3+246x^2+769x+764=0[/tex]

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

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- #18

matt grime

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matt grime

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matt grime

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- #24

matt grime

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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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- #26

shmoe

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Treadstone 71 said:

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.

- #27

matt grime

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Treadstone 71 said:I'm supposed to show that the matrix is diagonalizable, but since I only have 2 distinct eigenvalues, this doesn't work out.

Consider the obviously diagonalizable identity matrix with 1s on the diagonal and zeroes elsewhere. This has only one eigenvalue, do you think that means that the identity matrix is not a diagonalizable matrix?

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