Finding Matrix adjoint

Question

Need someone to check my reasoning as I don't feel confident in this topic:

consider $P_2$(C) with inner product $$<p(x), q(x)> = \int {q(x)p(x) dx} $$

T is defied by T(p(x)) = p'(x) + p(x)

Find T*(p(x)) for an arbitrary polynomial of degree 2

1. first thing to do is to find a matrix of T, applying standard basis

   T($x^2$) = 4x

   T(x) = x + 1

   T(1) = 1

   So, $$[T]_\alpha = \begin{matrix} 0 & 0 & 0 \\ 4 & 1 & 0 \\ 0 & 1 & 1 \\ \end{matrix} $$

2. Find A* of A

   $$A^t = \begin{matrix} 0 & 4 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{matrix} $$

   $$A^* = \begin{matrix} 0 & 4 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{matrix} $$

But this is obviously not right, as I am given def of inner product, which I did not make any use of. So, please, point at the mistake

Answer 1

Given a square matrix representing a linear operator, the adjoint matrix is defined to be the conjugate transpose matrix only if your original matrix is with respect to an orthonormal basis. Try the question again with this information and you will realize that your given inner product will come into play.

Cheers

Answer 2

Alternatively, you can use the definition of the adjoint, as we will do below. First, let the basis $\alpha = (1,x,x^2)$. Then $$ T(a+bx+cx^2) = (a+bx+cx^2) + (a+bx+cx^2)' = (a+b) + (b+2c)x + cx^2, $$ and we can represent the linear operator $T$ in coordinates given by basis $\alpha$ as \begin{align} T_\alpha = \begin{bmatrix} 1&1&0\\ 0&1&2\\ 0&0&1 \end{bmatrix}. \end{align} The adjoint $T^*$ behaves as \begin{align} \langle T(p(x)), q(x) \rangle = \langle p(x), T^*(q(x)) \rangle \end{align} so we can find $T^*_\alpha$ from \begin{align} \langle T_\alpha(a,b,c), (a',b',c') \rangle &= \langle(a+b,b+2c,c),(a',b',c')\rangle\\ &=\langle (a,b,c), (a',a'+b',2b'+c')\rangle\\ &=\langle (a,b,c), T^*_\alpha (a',b',c') \rangle\rangle\\ \end{align} and conclude that $T^*_\alpha (a',b',c') = (a',a'+b',2b'+c')$ which can be expressed as \begin{align} T^*_\alpha = \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&2&1 \end{bmatrix}. \end{align} Thus $T^*_\alpha$ is not the conjugate transpose of $T_\alpha$; this only works when $\alpha$ is an orthonormal basis.