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

  • Thread starter mick25
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  • #1
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[Linear Algebra] Finding T* adjoint of a linear operator

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:

Answers and Replies

  • #2
22,097
3,279
What you did only works for an orthonormal basis. So perhaps you should first find an orthonormal basis of your space??
 
  • #3
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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
22,097
3,279
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
13
0
[itex]\alpha = <1, \sqrt{3}(2x-1)>[/itex] is an orthonormal basis for [itex]P_1{}(R)[/itex]

By the definition of the inner product space, have

[itex] < 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 [/itex]

So I computed

[itex]< T(1), 1 > = 1 [/itex]

[itex]< T(1), \sqrt{3}(2x-1) > = 0[/itex]

[itex]< T(\sqrt{3}(2x-1)), 1 > = 4\sqrt3[/itex]

[itex]< T(\sqrt{3}(2x-1)), \sqrt{3}(2x-1) > = 1[/itex]

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

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

Still not sure how to find [itex]T^*[/itex] from this...




Verifying the matrix...

[itex] < 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))> [/itex]

[itex] < 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 [/itex]

[itex] < 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[/itex]

So [itex] T [/itex] is not self-adjoint.
 
  • #6
22,097
3,279
Doesn't that matrix for T* actually give you T*?? How do you construct a linear mapping from a matrix?
 

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