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Topic: Linear algebra with slope.
Replies: 2   Last Post: Feb 12, 2013 5:17 AM

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mina_world

Posts: 2,142
Registered: 12/13/04
Re: Linear algebra with slope.
Posted: Feb 12, 2013 5:17 AM
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"José Carlos Santos" ?? ??? ??? ??????.
ansb5mFmjobU1@mid.individual.net...

On 11-02-2013 10:38, mina_world wrote:

> Linear transformation f : R^2 -> R^2.
> Let M be the standard matrix of f.
> Let Rank(M) = 1 or 2.
>
> Given a straight line y = ax+b.
>
> Then f transforms this line(y=ax+b) into
> a line(y=cx+d) OR a fixed point.
>
> If y=cx+d line exists, show that (1,c) = f(1,a).


This can't be true. If f(x,y) = (2x,2y), then _f_ has rank 2 and
transforms the line y = x into
itself. But f(1,1) = (2,2).

----------------------------------------------------------------------
Oh, yes. you're right. In fact, my interest is slope.
If I revise original post,
"If y=cx+d line exists, show that c/1 = v/u when f(1,a)=(u,v)."

If, in this case,
pf)
y = ax+b ==> (x,y) = (1, a)*t + (0,b) (verctor)
so, f(x,y) = f{(1,a)*t} + f(0,b)
so, f(x,y) = t*f(1, a) + f(0,b) ==> this line y=cx+d
It means that c/1 = v/u when f(1,a)=(u, v)




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