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Linear Algebra 2 - Representing Matrix

  • Thread starter Mumba
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  • #1
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Sry, this will be the last question^^

10muavl.jpg


Its a similiar to the problem i ve postet before. Maybe if i can solve the first problem i can solve this to. But what i dont understand is that notation. I ve circled it with a red line.
Does anyone know what this means?

Thx
Mumba
 

Answers and Replies

  • #2
dx
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It means the representation of L in the basis B = {1, x, x²}. They're just calling that matrix AB
 
  • #3
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So this A is the representing matrix?
But what does then Ap=q mean? Or how can i calculate L for 1, x and x^2?
 
  • #4
dx
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Ap = p(x+1) defines the action of L.

So for p = 1, you get L(1) = x + 1 and so on.
 
  • #5
dx
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Actually, it looks like q(x+1) is not intended as multiplication of q with (x+1). The action of L on the function p(x) is given by p(x+1).

So, if p = 1 + x, then L(p) = 1 + (x + 1) = 2 + x.
 
  • #6
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But so i get for
L(x) = x^2 + x,
L(x^2)=x^3 + x^2

So should i choose a new basis, for example {1+x,x^2,x^3} to get the repr matrix?
 
  • #7
dx
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See my post above.

L(x) will be x + 1 instead of x² + x.
 
  • #8
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L(p)=q(p)=p(x+1)?

wie kommst du da auf L(x)=1+x?
Was hast du denn dann für L(1)?
 
  • #9
dx
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Treat a polynomial as the function p(x). The action of L is to take that function and return the function q(x) = p(x+1)

So, if p(x) = 1 + x, then q(x) = p(x + 1) = 1 + x + 1 = x + 2 (i.e. L[1 + x] = 2 + x)

If p(x) = 1, then q(x) = p(x + 1) = 1. (i.e L[1] = 1 )
 
  • #10
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So i get then
L(1) = 1
L(x) = x+1
L(x^2)=x^2+1

?
 
  • #11
dx
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L[x²] = (x+1)² = x² + 2x + 1
 
  • #12
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and then for
L(1) --> 1 0 0
L(x) --> 1 1 0
L(x^2) --> 1 0 1

So my matrix would be
1 1 1
0 1 0
0 0 1

??
 
  • #13
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ahh ok so if u change this fpr L(x^2) --> 1 2 1
matrix:
1 1 1
0 1 2
0 0 1

correct? ^^
 
  • #14
dx
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Yes, thats correct.
 
  • #15
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:D:D
cool thx alooooot
 
  • #16
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Treat a polynomial as the function p(x). The action of L is to take that function and return the function q(x) = p(x+1)

So, if p(x) = 1 + x, then q(x) = p(x + 1) = 1 + x + 1 = x + 2 (i.e. L[1 + x] = 2 + x)

If p(x) = 1, then q(x) = p(x + 1) = 1. (i.e L[1] = 1 )
Should it be like that:

If p(x) =1 then, q(x) = p(x+1) = p(x)+p(1) = 1 + 1 = 2, so L[1]= 2!?
 
  • #17
dx
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No, p(x) is a constant function, i.e. it has the same value for any x. So p(x + 1) would still be 1.
 
  • #18
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hmm, ok
but why ist L(x) = x+1?

--> p(x) = x
if its a constant function and p(x)=x, shouldnt be P(x+1)=x too???
 
  • #19
dx
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p(x) = 1 is a constant function, p(x) = x is not!

For a given polynomial p, we get the L[p] by replacing every x with x + 1.

So if p = 1, we do nothing since there is no x, and L[1] = 1

If p = x, we replace x by x + 1, and we get L[x] = x + 1

If p = x2, we replace x by x + 1 to get (x + 1)2, i.e. L[x2] = (x + 1)2 = 1 + 2x + x2
 
  • #20
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oh man
yes sure thx again ^^
 

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