Question

Let P2 be the real vector space of polynomials in a of degree at most 2, and let T be the real vector space of upper triangular 2 x 2 matrica b,cERThe vector space P2 is equipped with the inner product 〈p(x), q(x)-1 p(z)q(z) dr, and the vector space T is equipped with the inner product 〈A.B)=tr(AB), where tr denotes trace. Let L: P2→T be 1.p(z)dr]. Find L 0 c given by L(p(z)):-17(1) .CE :J ) 1 2 0 p(-1) 0 3

Answer 1

Note

,2· 1 13 1 00 00 10 10 20 20

Therefore,

( ·0 0 0 0 23 23 12 00 3 0

1 1-2 1 ( 1 0,1 0 23 )の122 12 12 3 0

) 131 13 1 (a 2020 23 23 12 24 17-3 1 3 0

Let

3 3)=a+bx + cx2 0

We need to find $\begin{align*}a,b,c\end{align*}$ . By above, we have

5 3

(a+b) =1 (ar + br2 + cr3) dr ジジ 23 + 10b +1 1b (L (a J 2-3

1 22 17 (a + br+r) 1

Thus, we get the following system of linear equations:

17-3 1 3-5 252323 5 3

The second equation gives

$\begin{align*}b&=- {\frac 32}\end{align*}$

From first and third we get

$\begin{align*}{\frac 23}a+{\frac 25}c&={\frac{17}3}\\ {\frac23}a+{\frac 29}c&={\frac 53}\end{align*}$

Subtracting,

$\begin{align*}{\frac 8{45}}c&=4\\ \Rightarrow~~~~~~~~~~~c&={\frac {45}2}\end{align*}$

Hence, $\begin{align*}a=-5\end{align*}$ . This shows

$\begin{align*}L^\ast\begin{pmatrix}1&2\\ 0&3\end{pmatrix}=-5-1.5x+22.5x^2\end{align*}$