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

[mick25]

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

Homework Statement

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).

Homework Equations

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

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

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

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

 

[micromass]

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

 

[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

 

[micromass]

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!

 

[mick25]

\alpha = &lt;1, \sqrt{3}(2x-1)&gt; is an orthonormal basis for P_1{}(R)

By the definition of the inner product space, have

&lt; T(p(x)), q(x) &gt; = \int_0^1 \! p&#039;(x)q(x) \, \mathrm{d} x + \int_0^1 \! p(x)q(x) \, \mathrm{d} x

So I computed

&lt; T(1), 1 &gt; = 1

&lt; T(1), \sqrt{3}(2x-1) &gt; = 0

&lt; T(\sqrt{3}(2x-1)), 1 &gt; = 4\sqrt3

&lt; T(\sqrt{3}(2x-1)), \sqrt{3}(2x-1) &gt; = 1

<br /> [T]_\alpha^{\alpha} =<br /> \left[ {\begin{array}{cc}<br /> 1 &amp; 4\sqrt{3} \\<br /> 0 &amp; 1 \\<br /> \end{array} } \right]<br />

<br /> [T^*]_\alpha^{\alpha} =<br /> \left[ {\begin{array}{cc}<br /> 1 &amp; 0 \\<br /> 4\sqrt{3} &amp; 1 \\<br /> \end{array} } \right]<br />

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

Verifying the matrix...

&lt; T(p(x)), q(x) &gt; = \int_0^1 \! p&#039;(x)q(x) \, \mathrm{d} x + \int_0^1 \! p(x)q(x) \, \mathrm{d} x = &lt; p(x), T^*(q(x))&gt;

&lt; T(p(x)), q(x) &gt; = \int_0^1 \! p&#039;(x)q(x) \, \mathrm{d} x + \int_0^1 \! p(x)q(x) \, \mathrm{d} x

&lt; p(x), T(q(x)) &gt; = \int_0^1 \! p(x)q&#039;(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&#039;(x)q(x) \, \mathrm{d} x + \int_0^1 \! p(x)q(x) \, \mathrm{d} x

So T is not self-adjoint.

 

[micromass]

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