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Linear Algebra

Anonymous Tue 14 Jan 14:29:55 2020 No.11301272 View Reply Original Report

Quoted By: >>11301281 >>11301294 >>11301318 >>11301337 >>11301447 >>11304695

Are all matrices 3x3 with 3 different eigenvalue diagonalisable?

Anonymous

Anonymous Tue 14 Jan 2020 14:36:22 No.11301279 Report

Quoted By: >>11301284 >>11301287 >>11301318 >>11303499 >>11303809

iirc yes. if it has fewer distinct eigenvalues then they need to have combined multiplicity equal to the dimension

Anonymous

Anonymous Tue 14 Jan 2020 14:39:36 No.11301281 Report

Quoted By: >>11301284

>>11301272
Yes, and having 3 linearly independent eigenvectors yields the same result.

Anonymous

Anonymous Tue 14 Jan 2020 14:41:50 No.11301284 Report

Quoted By:

>>11301279
>>11301281
thanks

Anonymous

Anonymous Tue 14 Jan 2020 14:43:24 No.11301287 Report

Quoted By:

>>11301279
cont.

and i remember finding it weird that some books emphasize this point. afaik there's nothing special about matrices with n *distinct* eigenvalues other than that you can conveniently say that they must have n linearly independent eigenvectors. it might still have n linearly independent eigenvectors with some sharing the same eigenvalues.

Anonymous

Anonymous Tue 14 Jan 2020 14:47:09 No.11301294 Report

Quoted By:

>>11301272
>>11300734 /sqt/

Anonymous

Anonymous Tue 14 Jan 2020 15:07:06 No.11301318 Report

Quoted By: >>11301328

>>11301272
>>11301279
Not if one of those distinct eigenvalues is zero.

Anonymous

Anonymous Tue 14 Jan 2020 15:11:40 No.11301328 Report

Quoted By: >>11301340

>>11301318

can't you still diagonalize it? you can pick out vectors from the null space to fill out your basis, right?

Anonymous

Anonymous Tue 14 Jan 2020 15:16:07 No.11301337 Report

Quoted By: >>11301367 >>11301372

>>11301272
Why do americans learn Linear Algebra with matrices instead of linear transformations or quadratic forms?

Anonymous

Anonymous Tue 14 Jan 2020 15:17:32 No.11301340 Report

Quoted By: >>11301493

>>11301328
Oof, my bad. You're right. As long as there's only one.

Anonymous

Anonymous Tue 14 Jan 2020 15:27:14 No.11301367 Report

Quoted By:

>>11301337
I live in Poland.

Anonymous

Anonymous Tue 14 Jan 2020 15:30:44 No.11301372 Report

Quoted By:

>>11301337
We do learn linear algebra with linear transformations (and we represent them with matrices upon choosing a basis like everyone else)
Engineering/computational linear algebra classes probably use matrices mostly because that's the form they interact with.
In my linear algebra course the first things we talked about were fields, vector spaces, and linear transformations, then we did bases and matrix representations.

Anonymous

Anonymous Tue 14 Jan 2020 15:59:57 No.11301447 Report

Quoted By: >>11301486 >>11301493 >>11301717 >>11303168

>>11301272
All matrices with three unique nonzero eigenvalues are diagonalizable. If any of your eigenvalues are zero, it is not guaranteed even if all three are unique.

Anonymous

Anonymous Tue 14 Jan 2020 16:34:55 No.11301486 Report

Quoted By:

>>11301447
All 3x3 matrices with "..."* Sorry

Anonymous

Anonymous Tue 14 Jan 2020 16:39:51 No.11301493 Report

Quoted By: >>11301506 >>11303809

>>11301340
>>11301447

wouldn't it work in any instance where your null space is big enough to fill out the rest of the basis? if you have a k-dimensional nullspace and j independent eigenvectors, then can you diagonalize it as long as j+k=n?

Anonymous

Anonymous Tue 14 Jan 2020 16:49:06 No.11301506 Report

Quoted By: >>11301513

>>11301493
cont.

for example, if you have a rank-1, NxN matrix with one eigenvector. you pick that eigenvector as the first element of your basis P, and the fill the rest with a basis of the null space. every element along the diagonal is zero except the one corresponding to the direction that your transformation preserves.

Anonymous

Anonymous Tue 14 Jan 2020 16:57:06 No.11301513 Report

Quoted By: >>11301520

>>11301506
Of course, but that doesnt always happen. Eg
0 1
0 0

Anonymous

Anonymous Tue 14 Jan 2020 17:01:26 No.11301520 Report

Quoted By:

>>11301513

but that's a rank-1 matrix with no eigenvalues. you can pick a vector from its null space, but you don't have an eigenvector to complete the change of basis.

Anonymous

Anonymous Tue 14 Jan 2020 17:05:10 No.11301526 Report

Quoted By: >>11303168

All n×n matrices with n distinct eigenvalues are eigenvalue diagonalisable. The eigenspace of every eigenvalue is always >=1, and <= to the algebraic multiplicity, which for n distinct eigenvalues is always 1. Therefore you have n linearly independent eigenvectors giving you a basis for decomposition.

Anonymous

Anonymous Tue 14 Jan 2020 19:03:19 No.11301717 Report

Quoted By: >>11301738

>>11301447
Give counter example

Anonymous

Anonymous Tue 14 Jan 2020 19:20:15 No.11301738 Report

Quoted By: >>11301802

>>11301717
I mean, really any rotation matrix will work. There's an entire class of "defective" square matrices which are not diagonalizable over the reals.

Anonymous

Anonymous Tue 14 Jan 2020 19:54:56 No.11301802 Report

Quoted By: >>11303168

>>11301738
I mean give example of 3x3 matrix with three distinct eigenvalues, one of them zero, which is not diagonalizable

Anonymous

Anonymous Wed 15 Jan 2020 07:13:03 No.11303168 Report

Quoted By:

>>11301447
Wrong. Liar. Enough of your nonsense.

>>11301802
You can't because >>11301526

Anonymous

Anonymous Wed 15 Jan 2020 07:32:50 No.11303204 Report

Quoted By: >>11303255 >>11303809 >>11303966 >>11306579

/sci/ is consistently, without exception totally and utterly incompetent at understanding the inuition and mechanics of linear algebra, the single most elegant contribution of mathematics to the sciences since the advent of mathematical analysis with Newton, Lagrange, Laplace. These threads are an abomination, everyone who posts these idiotic answers should be sterilized, banned from the internet, barred from ever holding a job performing intellectual labor. Disgusting ignorance.

Anonymous

Anonymous Wed 15 Jan 2020 07:54:56 No.11303255 Report

Quoted By:

>>11303204
Yes, but this is also true for the vast majority of students at universities.

Anonymous

Anonymous Wed 15 Jan 2020 10:43:36 No.11303499 Report

Quoted By:

>>11301279

The dimension of the eigenspace MUST be equal to the multiplicity of the associated eigenvalue for a matrix to be diagonalisable.

Anonymous

Anonymous Wed 15 Jan 2020 14:22:42 No.11303809 Report

Quoted By:

>>11303204
am i wrong here?
>>11301279
>>11301493

it has been a while since i've thought of it, and as i am neither a math major or mathematician, i don't get to refresh my memory often.

iirc an eigen-decomposition is just changing your basis to the eigenvectors.

since the ith column of a matrix of a linear transformation is the transformed ith basis vector of the domain expressed in terms of the basis of the range, it just ends up having the eigenvalues along the diagonal.

Anonymous

Anonymous Wed 15 Jan 2020 16:02:38 No.11303955 Report

Quoted By: >>11303987 >>11304765 >>11304776

Can anyone give me some real life examples of the use of eiganvalues.

Anonymous

Anonymous Wed 15 Jan 2020 16:08:39 No.11303966 Report

Quoted By:

>>11303204
this but unironically

Anonymous

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Anonymous Wed 15 Jan 2020 16:15:58 No.11303987 Report

Quoted By:

>>11303955
the eigenvalues of the stress tensor are the principle stresses in a material (the maximum stresses that occur at a location)

Anonymous

Anonymous Wed 15 Jan 2020 22:04:58 No.11304695 Report

Quoted By: >>11304701

>>11301272
>Are all matrices 3x3 with 3 different eigenvalue diagonalisable?
ALL square matrices with the same number of different eigenvalues as the length of a square are diagonalisable.

Anonymous

Anonymous Wed 15 Jan 2020 22:05:59 No.11304701 Report

Quoted By:

>>11304695
sorry, meant side instead of square

Anonymous

Anonymous Wed 15 Jan 2020 22:09:24 No.11304706 Report

Quoted By: >>11304774

Why there seems to be a misconception that a 0 eigenvalue doesn't imply diagonalizability?

Anonymous

Anonymous Wed 15 Jan 2020 22:30:01 No.11304765 Report

Quoted By:

>>11303955
literally any physical system involves solving a system of differential equations, which reduces to diagonalizing a matrix.

Anonymous

Anonymous Wed 15 Jan 2020 22:32:39 No.11304774 Report

Quoted By: >>11304781 >>11304920

>>11304706
the matrix
$begin{pmatrix} 0 & 1 & 0 \\<br />0 & 0 & 0 \\<br />0 & 0 & 1 \end{pmatrix}$
has distinct eigenvalues of 0 and 1, and is not diagonalizable. it just doesn't have 3 of them.

Anonymous

Anonymous Wed 15 Jan 2020 22:32:51 No.11304776 Report

Quoted By:

>>11303955
Find a formula for fibonacci numbers
if A=(0 1
1 1 )
A(F_n
F_n+1)= (F_n+1
F_n+2)
Finding the eigenvalues of A will give you the general formula for F_n with the golden ratio

Anonymous

Anonymous Wed 15 Jan 2020 22:33:45 No.11304781 Report

Quoted By:

>>11304774
$\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$
sigh, fucking backslash

Anonymous

Anonymous Wed 15 Jan 2020 23:28:49 No.11304920 Report

Quoted By:

>>11304774
That's because the eigenvalues are not distinct. It's completely irrelevant if one of them is zero or not.

Anonymous

Anonymous Thu 16 Jan 2020 18:39:05 No.11306579 Report

Quoted By:

>>11303204
Waiting for your insight

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