- #1
mick25
- 13
- 0
[Linear Algebra] Finding T* adjoint of a linear operator
Consider [itex]P_1{}(R)[/itex], the vector space of real linear polynomials, with inner product
[itex] < p(x), q(x) > = \int_0^1 \! p(x)q(x) \, \mathrm{d} x [/itex]
Let [itex]T: P_1{}(R) \rightarrow P_1{}(R)[/itex] be defined by [itex]T(p(x)) = p'(x) + p(x)[/itex]. Find [itex] T^*(p(x)) [/itex] for an arbitrary
[itex]p(x) = a + bx^2 \in P_1{}(R)[/itex].
[itex] < T(p(x)), q(x) > = < p(x), T^*(q(x)) > [/itex]
Using the standard basis of [itex]P_1{}(R), \alpha = <1, x>[/itex]
[itex]<T(1),1> = 1[/itex]
[itex]<T(1), x> = 1/2[/itex]
[itex]<T(x), 1> = 3/2[/itex]
[itex]<T(x), x> = 5/6[/itex]
[itex]
[T]_\alpha^{\alpha} =
\left[ {\begin{array}{cc}
1 & 3/2 \\
1/2 & 5/6 \\
\end{array} } \right]
[/itex]
[itex]
[T^*]_\alpha^{\alpha} =
\left[ {\begin{array}{cc}
1 & 1/2 \\
3/2 & 5/6 \\
\end{array} } \right]
[/itex]
Not sure how to find [itex]T^*[/itex] from here...
Homework Statement
Consider [itex]P_1{}(R)[/itex], the vector space of real linear polynomials, with inner product
[itex] < p(x), q(x) > = \int_0^1 \! p(x)q(x) \, \mathrm{d} x [/itex]
Let [itex]T: P_1{}(R) \rightarrow P_1{}(R)[/itex] be defined by [itex]T(p(x)) = p'(x) + p(x)[/itex]. Find [itex] T^*(p(x)) [/itex] for an arbitrary
[itex]p(x) = a + bx^2 \in P_1{}(R)[/itex].
Homework Equations
[itex] < T(p(x)), q(x) > = < p(x), T^*(q(x)) > [/itex]
The Attempt at a Solution
Using the standard basis of [itex]P_1{}(R), \alpha = <1, x>[/itex]
[itex]<T(1),1> = 1[/itex]
[itex]<T(1), x> = 1/2[/itex]
[itex]<T(x), 1> = 3/2[/itex]
[itex]<T(x), x> = 5/6[/itex]
[itex]
[T]_\alpha^{\alpha} =
\left[ {\begin{array}{cc}
1 & 3/2 \\
1/2 & 5/6 \\
\end{array} } \right]
[/itex]
[itex]
[T^*]_\alpha^{\alpha} =
\left[ {\begin{array}{cc}
1 & 1/2 \\
3/2 & 5/6 \\
\end{array} } \right]
[/itex]
Not sure how to find [itex]T^*[/itex] from here...
Last edited: