FP1 Invariant Points Watch

mattallica
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Hi, I was wondering if anyone could help me with finding invariant points. The question is to find the invariant points for transformations of the following matrix:

\begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix}

Here's what I did:

\begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}

\begin{pmatrix} -y \\ x+2y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}

-x = x+2y
-2x=2y
-x=y

This is the only method I can figure out of solving it, and as far as I can tell I've made no mistakes, but then the answer given in the book is (\lambda, -\lambda), and I can't for the life of me figure out what Lambda is supposed to represent! Can somebody please explain this, and if I've not solved this properly, I'd really appreciate a step-by-step solution if anyone can spare the time. Thanks in advance! <3 xoxo
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Hasufel
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take the corresponding entries in the top "row" - -y=x OR: y=-x which implies that the solution is points of the form

(x,-x) or, as they put it: (\lambda, -\lambda)

(the lower rows equated just confirms the top one, once you sub y=-x into it)

they just use lambda as it is arbitrary, to let you see that the solution is any one number, the y of which is the negation.
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brianeverit
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(Original post by mattallica)
Hi, I was wondering if anyone could help me with finding invariant points. The question is to find the invariant points for transformations of the following matrix:

\begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix}

Here's what I did:

\begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}

\begin{pmatrix} -y \\ x+2y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}

-x = x+2y
-2x=2y
-x=y

This is the only method I can figure out of solving it, and as far as I can tell I've made no mistakes, but then the answer given in the book is (\lambda, -\lambda), and I can't for the life of me figure out what Lambda is supposed to represent! Can somebody please explain this, and if I've not solved this properly, I'd really appreciate a step-by-step solution if anyone can spare the time. Thanks in advance! <3 xoxo
 \lambda just indicates that it is a line of invariant points. You could just have correctly given it as (k,-k) or (t,-t) the letter used is arbitrary.
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mattallica
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(Original post by brianeverit)
 \lambda just indicates that it is a line of invariant points. You could just have correctly given it as (k,-k) or (t,-t) the letter used is arbitrary.
So if I was to say (\lambda, \lambda) would that just be the same as x=y?
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davros
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(Original post by mattallica)
So if I was to say (\lambda, \lambda) would that just be the same as x=y?
Yes
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