Geometric Properties from Eigenvectors

[Daniel323]

Homework Statement

Hi! I just used MATLAB to find the eigenvalues and eigenvectors of A=[0 -1; 1 0]
I obtained the eigenvalues of 0 +/- i
and eigenvectors of v(1) = [ 0.7071; 0 - 0.7071i] and v(2) = [ 0.7071; 0 + 0.7071i]

Homework Equations

I'm having trouble interpreting these results in relation to the geometric properties of the linear transformation T(x; y) = [0 -1; 1 0] [x; y]

The Attempt at a Solution

As far as what the transformation does this is my explanation: It changes the sign of the y values i.e. positive to negative or negative to positive.
As far as interpreting the MATLAB results in relation to the geometric properties of the linear transformation T, I can only explain it as: the eigenvalues illustrate that the x values remain the same but the y values change in sign.

Is such an explanation correct. Is this what the transformation does.
Sorry I haven't done much mathematical explanations and not quite sure if this is how to go about it.

Thank you in advance.

 

[tiny-tim]

I just used MATLAB to find the eigenvalues and eigenvectors of A=[0 -1; 1 0]
I obtained the eigenvalues of 0 +/- i
and eigenvectors of v(1) = [ 0.7071; 0 - 0.7071i] and v(2) = [ 0.7071; 0 + 0.7071i]
…
I'm having trouble interpreting these results in relation to the geometric properties of the linear transformation T(x; y) = [0 -1; 1 0] [x; y]

Hi Daniel323!

(have plus-or-minus: ± and a square-root: √ )

i] You can multiply eiegenvectors (but not eigenvalues, of course) by any factor … in this case, it's much neater if you divide by .7071 (= 1/√2 ), to give eigenvectors of v(1) = [ 1; -i] and v(2) = [1; i].

ii] If a complex number is purely imaginary, you don't have to write it as "0 + …": in this case, you can write that the eigenvalues are ±i.

iii] Hint: draw the lines through the origin along [ 1; -i] and [1; i]. What does T do to these lines? Is there a rotation? Is there a reflection?

 

[Daniel323]

Hi tiny-tim, thanks for that.

I drew the lines on a graph and I think this is the effect (and this would be a much better explanation then my previous one);
The linear transformation T reflects the line or each point through the x axis.

Hopefully I'm right, thanks again! :)

 

[tiny-tim]

Hi Daniel323!

The linear transformation T reflects the line or each point through the x axis.

Nooo … y = x, x = -y is a … ?

Hint: a reflection will have a negative determinant.

A rotation (in 2D) has no fixed line, and therefore no real eigenvalues.

 

[Daniel323]

Ahhhh... here's goes nothing:
The linear transformation T rotates the line or each point through the x axis?

Thanks again.

 

[tiny-tim]

"nothing shall come of nothing …"

Ahhhh... here's goes nothing:
The linear transformation T rotates the line or each point through the x axis?

I'm not following you … what's a rotation through the x axis?

(this is only 2D …)

 

[Daniel323]

Sorry about that, my mistake. I'm just horrible at mathematical explanations.

The linear transformation T causes a rotation for each vector. Is this generally a good sort of explanation for such findings?

 

[tiny-tim]

The linear transformation T causes a rotation for each vector. Is this generally a good sort of explanation for such findings?

Yes, that's fine!

Rotation reflection and translation are the correct mathematical terms for what all 2D or 3D transformations are made of.

But can I just check … a rotation of how much, and about what axis?

 

[Daniel323]

Of course you can! I believe it is a rotation about the x-axis, but as far as how much I'm not completely sure how to say. If I was to guess I'd say a rotation of 180 degrees about the x-axis.

 

[tiny-tim]

Of course you can! I believe it is a rotation about the x-axis, but as far as how much I'm not completely sure how to say. If I was to guess I'd say a rotation of 180 degrees about the x-axis.

hmm … this is 2D, so any rotation must be about the z-axis (or a line parallel to the z-axis).

Just try a couple of test points, to see where they go.

Rotation reflection and translation are the correct mathematical terms for what all 2D or 3D transformations are made of.

(Actually, I should have added expansion, either generally or in one direction.)

 

[Daniel323]

Hmm I'm kind of confused now, because it is 2D I thought we were only dealing with a x and y axis.

So now the rotation is about the z-axis, but again not sure how to describe how much of a rotation.

I tried out some points and generally understand what the transformation is doing, just can't explain it mathematically, so to speak.

 

[tiny-tim]

Hmm I'm kind of confused now, because it is 2D I thought we were only dealing with a x and y axis.

ok, I agree, technically there's no z-axis, but you can rotate about the origin (or any other point), which is the same as a rotation about the z-axis.

In other words, stick a pin in the paper, to fix it to the desk, and rotate!

As for a proper mathematical explanation, any rotation about a particular axis (or point, in 2D) is defined by its angle … so how through many degrees is everything rotated?

 

[Daniel323]

Ok so I will say it is a rotation about the origin or point by 180 degrees.
I suspect it is 180 degrees, but maybe I'm looking at it wrong and it is only 90 degrees.

 

[tiny-tim]

Ok so I will say it is a rotation about the origin or point by 180 degrees.
I suspect it is 180 degrees, but maybe I'm looking at it wrong and it is only 90 degrees.

Yes, it's one or the other!

Hint: stop trying to be abstract …

do a concrete example …

what happens to (1,0)?
​

 

[Daniel323]

Umm I'm going to go with a rotation about the origin by 180 degrees. Please tell me I'm finally right!

 

[tiny-tim]

Well, 180º would send (1,0) to (-1,0), but A(1,0) = (0,1), so … ?

 

[HallsofIvy]

No need to guess. What does that transformation change <1, 0> to?

 

[Daniel323]

Ahh the transformation changes (1,0) to (0,1) hence the x and y value swaps at this point and therefore a rotation about the origin by 90 degrees.

I believe I am right here, thanks very much!