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Let V be the space of real polynomials of degree at most 1 with inner product defined by $\langle p,q \rangle = \frac12\int_{-1}^1p(t)q(t)dt.$ Define $\alpha\in End(V)$ by $$\alpha(p)=p(0)+p(1)t.$$ Find the adjoint endomorphism $\alpha^*.$

For this problem, I am wondering about several things. First, I know that $\langle\alpha(p),q\rangle= \langle p,\alpha^*(q)\rangle$. I wasn't sure if I could assume that $\alpha^*(q)=r+st$ for constants $r,s$. But I think I can since the adjoint is also an endomorphism. So, I attempted to compute $\alpha^*(q)$ by first defining $q(t)=m+nt,$ and I found $\alpha^*(q)=(m+\frac13n)+nt$. If this is even correct, I wouldn't know how to turn this into an expression for $\alpha^*(q)$ in the way that $\alpha(p)$ is defined.

Or should I determine a matrix representation for $\alpha$ and take its conjugate transpose? If so, how does that new matrix translate to an expression for $\alpha^*(q)$?

Thank you for your time in helping.

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1 Answer 1

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Note $V=(1,t)$. And $\langle a+bt, c+dt \rangle = \frac12\int_{-1}^1 (ac +(b+d)t+ bd)t^2 = ac+bd \frac{1}{3}$. So we have the expression $\langle a+bt, c+dt \rangle = (a,b)^T A (c,d)$ where $$A =\left(\begin{array}{cc} 1 & 0 \\ 0 & \frac{1}{3}\end{array}\right)$$

And $\alpha((a,b))=(a, a+b) $ so that $$ \alpha = \left(\begin{array}{cc} 1 & 0 \\ 1& 1 \end{array}\right)$$

From $$ v^T \left(\begin{array}{cc} 1 & 0 \\ 1& 1 \end{array}\right)^T \left(\begin{array}{cc} 1 & 0 \\ 0 & \frac{1}{3}\end{array}\right) w = v^T \left(\begin{array}{cc} 1 & 0 \\ 0 & \frac{1}{3}\end{array}\right) A w $$, then $$A = \left(\begin{array}{cc} 1 & 0 \\ 0 & \frac{1}{3}\end{array}\right)^{-1} \left(\begin{array}{cc} 1 & 0 \\ 1& 1 \end{array}\right)^T \left(\begin{array}{cc} 1 & 0 \\ 0 & \frac{1}{3}\end{array}\right) = \left(\begin{array}{cc} 1 & \frac{1}{3} \\ 0 & 1\end{array}\right)$$

So $A$ is the matrix we want Hence $\alpha^* ((a+bt) )= a+\frac{b}{3} + bt$

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Thank you. I obtained this same result, as noted above, only I used constants m,n rather than a,b. But I am not sure if the adjoint can be expressed in terms of the linear function's coefficients. Note how alpha is originally defined; the definition does not depend on its particular coefficients, but rather p(0) and p(1). –  CHG Nov 12 '12 at 14:41
    
$V$ is a finite dimensional vector space. So a linear transformation on $V$ has a matrix representation. This is an elementary fact in linear algebra. –  Hee Kwon Lee Nov 21 '12 at 1:28

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