Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 0 down vote accepted

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 $\alpha^* ((a+bt) )= a+\frac{b}{3} + bt$

share|cite|improve this answer
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. – HK Lee Nov 21 '12 at 1:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.