will coincide.

I was just playing with it on Desmos: https://www.desmos.com/calculator/ujxsbj1gbi

You can always move the red point onto the purple point. Change the location of the purple point by changing the value of t, then try and make them coincide.

Not a proper proof, but just how I visualised it. Idea obviously expands to 3-space. And yes probably not the approach that was intended.

Page 10 of this doc is what OP wants I think. http://www.maths.gla.ac.uk/~ajb/dvi-ps/2q-notes.pdf

Reply 4

DFranklin

18

Original post by RDKGames

A

is a rotation. Choose some

\theta

so that

\cos \theta \neq 1

, then the image of

\mathbf{c}

rotated is definitely not equal to

\mathbf{c}

.

Now wherever

A \mathbf{c}

ends up, you can add fixed

\mathbf{t}

to it.

Now vary

\mathbf{c}

and eventually

A \mathbf{c} + \mathbf{t}

and

\mathbf{c}

will coincide.

I was just playing with it on Desmos: https://www.desmos.com/calculator/ujxsbj1gbi

You can always move the red point onto the purple point. Change the location of the purple point by changing the value of t, then try and make them coincide.

Not a proper proof, but just how I visualised it. Idea obviously expands to 3-space. And yes probably not the approach that was intended.

The key point here is that if 1 is not an eigenvalue then A-I is invertible (and so you can solve for c).

Reply 5

Feynboy

OP

10

Original post by DFranklin

The key point here is that if 1 is not an eigenvalue then A-I is invertible (and so you can solve for c).

How would you solve for c?

Original post by Feynboy

How would you solve for c?

\mathbf{c} = I \mathbf{c}

So

A\mathbf{c} + \mathbf{t} = \mathbf{c} \iff A \mathbf{c} - I \mathbf{c} = - \mathbf{t} \iff (A-I)\mathbf{c} = -\mathbf{t}

Then employ the fact that A-I is invertible.

Reply 7

Feynboy

OP

10

Original post by RDKGames

\mathbf{c} = I \mathbf{c}

So

A\mathbf{c} + \mathbf{t} = \mathbf{c} \iff A \mathbf{c} - I \mathbf{c} = - \mathbf{t} \iff (A-I)\mathbf{c} = -\mathbf{t}

Then employ the fact that A-I is invertible.

I see. What is 5.3 saying? Does it mean A*(the vector), or (A|the vector)?

Original post by Feynboy

I see. What is 5.3 saying? Does it mean A*(the vector), or (A|the vector)?

Looks to me like they just want you to evaluate

\begin{pmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} r \cos \theta \\ r \sin \theta \end{pmatrix}

Though I cheat a little bit... because I know what matrix A does. But even from the notation I think it's obvious that it's saying "apply A to this vector"

Reply 9

Feynboy

OP

10

Original post by RDKGames

Looks to me like they just want you to evaluate

\begin{pmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} r \cos \theta \\ r \sin \theta \end{pmatrix}

Though I cheat a little bit... because I know what matrix A does. But even from the notation I think it's obvious that it's saying "apply A to this vector"

How to show 6.1? I can't solve the characteristic eqn.

Original post by Feynboy

How to show 6.1? I can't solve the characteristic eqn.

Post your working.

Reply 11

DFranklin

18

Original post by Feynboy

How to show 6.1? I can't solve the characteristic eqn.

You don't need to worry about the characteristic equation - you solve it to find the eigenvalues, but in this case they've explicitly told you what the eigenvalues are.

Reply 12

DFranklin

18

Incidentally, given a matrix

A = \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}

and a vector

t = \begin{pmatrix}t_1 \\ t_2 \end{pmatrix}

, if you form the matrix

T = \begin{pmatrix}a_{11} & a_{12} & t_1 \\ a_{21} & a_{22} & t_2 \\ 0 & 0 & 1 \end{pmatrix}

, then you find that if

Au+t = v, where

u= \begin{pmatrix}u_1 \\ u_2 \end{pmatrix}

and

v= \begin{pmatrix}v_1 \\ v_2 \end{pmatrix}

then we have:

T \begin{pmatrix}u_1 \\ u_2 \\ 1\end{pmatrix} = \begin{pmatrix}v_1 \\ v_2 \\ 1 \end{pmatrix}

In other words, by embedding our 2-vectors in R^3 by setting "z=1", we can obtain a formulation where we can represent the Au+t transform by a single 3x3 matrix.

I'm not sure this is actually helpful to what you're doing in any way to be honest; but it's a common trick in computer graphics when we want to represent rotations, scales, and translations in a common format.

(edited 5 years ago)

Isometry of the plane

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