
#1
Mar3110, 12:18 PM

P: 27

1. The problem statement, all variables and given/known data
The Question: The map is given: [tex]L\rightarrow \Re_{2} \rightarrow \Re_{3}, p \rightarrow[/tex] p' + q*p , with q(x) = x. Now i should find the representing matrix for L with respect to the bases {1+x, x+x^{2}, 1+x^{2}} for [tex] \Re_{2}[/tex] and {1,x,x+x^{2},1+x^{3}} for [tex] \Re_{3}[/tex]. 3. The attempt at a solution I dont know what i should do here ^^. But i tried to calculate it. p is the polynomial, so i thought its like this: p = a_{0}(1+x) + a_{2}(x+x^2) + a_{3}(1+x^2) So i can calculate [tex]p\rightarrow p' +q*p:[/tex] [tex]p\rightarrow a_{0}+a_{1} + 3a_{1}x + a_{0}x + 2a_{2}x^2 + 2a_{1}x^2 + 2a_{2}x^3[/tex] Then i just counted the coressponding values together, means a_{0} ~ 1+x > 1 + 1 + 0 + 0 a_{1} ~ x+x^{2} > 1 + 3 + 2 + 0 a_{2} ~ 1+x^{2} > 0 + 0 + 2 + 2 Puting this all together gives the repr. Matrix  1 1 0   1 3 0   0 2 2   0 0 2  Is this somehow correct or completely wrong? ^^ I'm missing the second base here.. Does anyone knows a good website (linear algebra 1) ? I have exam in little more than week and still a lot to learn. Thx Mumba 



#2
Mar3110, 12:28 PM

Mentor
P: 20,937





#3
Mar3110, 12:32 PM

P: 27

yes, we called it R.
sorry i meant the same. A transformations from <=2 to <=3.... 



#4
Mar3110, 01:00 PM

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P: 20,937

Linear Algebra  Representing Matrix
But R^{2} is not the same as P_{2}, nor is R^{3} the same as P_{3}. The dimension of R^{2} is 2, while the dimension of P_{2} is 3.
For this problem I would advise you to evaluate L at each of the three basis polynomials. Then write those three output polynomials in terms of the second basis. That should give you an idea of what the matrix for this transformation is. 



#5
Mar3110, 01:10 PM

P: 27

But i dont know how. I ve never seen this before... Thats why i asked is that Polynomial correct the way i have written it down? Maybe you can give me an example, lets say for 1+x... What should i do with this? Sorry but i really dont know...:( 



#6
Mar3110, 02:20 PM

Mentor
P: 20,937

Your formula is L(p) = p' + xp, so L(1 + x) = 1 + x(1 + x) = 1 + x + x^2. Do the same thing for the other two basis polynomials.




#7
Mar3110, 02:33 PM

P: 27

Thx!!!!!!!!!!!!!!!!!!
then i get for: L(x+x^2)=1+2x+x(x+^2)= 1 +2x+x^2+x^3 L(1+x^2)=2x+x(1+x^2)=3x+x^3 You said: write those three output polynomials in terms of the second basis. so if i did this correct i get for 1+x > 1 0 1 0 x+x^2 > 0 1 1 1 1+x^2 > 1 3 0 1 so my matrix is 1 0 1 0 1 3 1 1 0 0 1 1 is this ok? 



#8
Mar3110, 03:53 PM

P: 27

yeah its good
thx mate!!! ;) 


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