
#1
May608, 04:28 AM

P: 5

Im trying to figure out how to do this question. This is an example in the book i have. Im not sure how they got the answer.
Here is the example from the book: Find the Transition Matrix P from the basis B={t+1, 2t, t1} to B'={4t[tex]^{2}[/tex]6t, 2t[tex]^{2}[/tex]2, 4t} for the space R[t]. A little computataion shows that 4t[tex]^{2}[/tex]6t:(3, 2, 3), 2t[tex]^{2}[/tex]2: (1, 1, 1) and 4t:(2, 0, 2). Therefore [tex]P=\left(\begin{array}{ccc}3 & 1 & 2 \\ 2 & 1 & 0\\ 3 & 1 & 2\end{array}\right)[/tex] I'd like to know how they found 4t[tex]^{2}[/tex]6t to be(3, 2, 3), 2t[tex]^{2}[/tex]2 to be (1, 1, 1) and 4t to be(2, 0, 2). Any help is apprectated. 



#2
May608, 06:30 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886

I'm afraid you are going to have to clarify some things first. What is "R[t]"? My first thought would be "functions in thr variable t on the real numbers" but all you give for your basis are powers of t. Polynomials over the real numbers with variable t? But "B" only has linear polynomials so that can't be right. And, for that matter, B' has quadratic polynomials so B and B' can't possibly span the same space! Are you sure you didn't miss at least one square in B?
The only way I could interpret "4t^{2} 6t: (3, 2, 3)" is that 4t^{2} 6t, one of the basis vectors in B', can be written 3(t+1)+ 2(2t) 3(t1), in terms of the basis B. But that is equal to 3t 3+ 4t 3t+ 3. Trying to find f1, f2, f3 so that 3f1+ 2f2 3f3= 4^{2} 6t, f1+ f2+ f3= 2t^{2} 2, and 2f1+ 2f3= 4t, I arrive at f1= t+ 1, just as you have, f3= t 1, just as you have, but f2= 2t^{2}, not 2t you dropped the "square". Now, what they are saying is: 4t^{2} 6t can be represented as (3, 2, 3) in basis B because 3(t+1)+ 2(2t^{2}) 3(t1)= 4t^{2} 6t, etc. 


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