# [Linear Algebra] Finding T*; complex conjugate linear transformation

1. Nov 28, 2011

### mick25

[Linear Algebra] Finding T* adjoint of a linear operator

1. The problem statement, all variables and given/known data

Consider $P_1{}(R)$, the vector space of real linear polynomials, with inner product

$< p(x), q(x) > = \int_0^1 \! p(x)q(x) \, \mathrm{d} x$

Let $T: P_1{}(R) \rightarrow P_1{}(R)$ be defined by $T(p(x)) = p'(x) + p(x)$. Find $T^*(p(x))$ for an arbitrary
$p(x) = a + bx^2 \in P_1{}(R)$.

2. Relevant equations

$< T(p(x)), q(x) > = < p(x), T^*(q(x)) >$

3. The attempt at a solution

Using the standard basis of $P_1{}(R), \alpha = <1, x>$

$<T(1),1> = 1$
$<T(1), x> = 1/2$
$<T(x), 1> = 3/2$
$<T(x), x> = 5/6$

$[T]_\alpha^{\alpha} = \left[ {\begin{array}{cc} 1 & 3/2 \\ 1/2 & 5/6 \\ \end{array} } \right]$
$[T^*]_\alpha^{\alpha} = \left[ {\begin{array}{cc} 1 & 1/2 \\ 3/2 & 5/6 \\ \end{array} } \right]$

Not sure how to find $T^*$ from here...

Last edited: Nov 29, 2011
2. Nov 28, 2011

### micromass

Staff Emeritus
What you did only works for an orthonormal basis. So perhaps you should first find an orthonormal basis of your space??

3. Nov 28, 2011

### mick25

I just got the new matrices with the orthonormal basis but I'm still at the same sticking point

The matrices aren't equal so T isn't self-adjoint; that's all I can conclude right now

4. Nov 28, 2011

### micromass

Staff Emeritus
If you work in an orthonormal basis, then the adjoint is given by the hermition conjugate. So find the hermitian conjugate and express the matrix as a linear transformation. And there you have your adjoint!!

5. Nov 29, 2011

### mick25

$\alpha = <1, \sqrt{3}(2x-1)>$ is an orthonormal basis for $P_1{}(R)$

By the definition of the inner product space, have

$< T(p(x)), q(x) > = \int_0^1 \! p'(x)q(x) \, \mathrm{d} x + \int_0^1 \! p(x)q(x) \, \mathrm{d} x$

So I computed

$< T(1), 1 > = 1$

$< T(1), \sqrt{3}(2x-1) > = 0$

$< T(\sqrt{3}(2x-1)), 1 > = 4\sqrt3$

$< T(\sqrt{3}(2x-1)), \sqrt{3}(2x-1) > = 1$

$[T]_\alpha^{\alpha} = \left[ {\begin{array}{cc} 1 & 4\sqrt{3} \\ 0 & 1 \\ \end{array} } \right]$

$[T^*]_\alpha^{\alpha} = \left[ {\begin{array}{cc} 1 & 0 \\ 4\sqrt{3} & 1 \\ \end{array} } \right]$

Still not sure how to find $T^*$ from this...

Verifying the matrix...

$< T(p(x)), q(x) > = \int_0^1 \! p'(x)q(x) \, \mathrm{d} x + \int_0^1 \! p(x)q(x) \, \mathrm{d} x = < p(x), T^*(q(x))>$

$< T(p(x)), q(x) > = \int_0^1 \! p'(x)q(x) \, \mathrm{d} x + \int_0^1 \! p(x)q(x) \, \mathrm{d} x$

$< p(x), T(q(x)) > = \int_0^1 \! p(x)q'(x) \, \mathrm{d} x + \int_0^1 \! p(x)q(x) \, \mathrm{d} x = p(1)q(1) - p(0)q(0) - \int_0^1 \! p'(x)q(x) \, \mathrm{d} x + \int_0^1 \! p(x)q(x) \, \mathrm{d} x$

So $T$ is not self-adjoint.

6. Nov 29, 2011

### micromass

Staff Emeritus
Doesn't that matrix for T* actually give you T*?? How do you construct a linear mapping from a matrix?