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Chapter 13: Problem 50

Let \(V\) be a complex inner product space. Suppose \(\langle T(v), v\rangle\) is real for every \(v \in V\). Show that \(T\) is selfadjoint.

Short Answer

Expert verified
We are given that \(\langle T(v), v\rangle\) is real for every \(v \in V\). To prove that T is self-adjoint, we ought to show \(\langle T(u), v\rangle = \langle u, T^*(v)\rangle\). By exploiting the properties of inner products and adjoints, we derived the following equation: \(\langle T(v), v\rangle = \langle v, T(v)\rangle + \langle T(v), u \rangle - \langle u, T(v) \rangle\). After rearranging and isolating \(\langle T(u), v\rangle\), we have \(\langle T(u), v\rangle = \langle u, T(v)\rangle\), which shows that T is a self-adjoint operator.

Step by step solution

01

Write the given condition as an equation

We are given that \(\langle T(v), v\rangle\) is real for every \(v \in V\). In terms of inner product, this can be written as: \(\langle T(v), v\rangle = \overline{\langle T(v), v\rangle}\)
02

Add and subtract the term we want to prove

We want to prove that \(\langle T(u), v\rangle = \langle u, T^*(v)\rangle\). Let's add and subtract it from the right-hand side of the equation in step 1: \(\langle T(v), v\rangle = \overline{\langle T(v), v\rangle} + \langle T(u), v\rangle - \langle u, T^*(v)\rangle\)
03

Use the properties of inner product and adjoints

We know that \(\langle u, w\rangle = \overline{\langle w, u\rangle}\) for any vectors u, w. Apply this to what we got in step 2: \(\langle T(v), v\rangle = \overline{\langle v, T(v)\rangle} + \langle T(u), v\rangle - \overline{\langle T^*(v), u\rangle}\) Now, we can use the properties of adjoint operator that states that for any linear operator T, its adjoint T* satisfies: \(\langle u, T^*(v)\rangle = \langle T(u), v\rangle\) Apply this to the equation above: \(\langle T(v), v\rangle = \overline{\langle v, T(v)\rangle} + \langle T(u), v\rangle - \langle u, T^*(v)\rangle\).
04

Prove T is self-adjoint

Now, let \(w = T(v)\), then using the properties of the adjoint operator, we have: \(\langle T(v), v\rangle = \langle v, w\rangle + \langle w, u \rangle - \langle u, w \rangle\) Now we can substitute for \(w = T(v)\): \(\langle T(v), v\rangle = \langle v, T(v)\rangle + \langle T(v), u \rangle - \langle u, T(v) \rangle\) Rearranging to isolate the \(\langle T(u), v\rangle\) term, we get: \(\langle T(u), v\rangle = \langle u, T(v)\rangle\) As we have proven that \(\langle T(u), v\rangle = \langle u, T(v)\rangle\) for all u, v in the complex inner product space V, we can conclude that T is a self-adjoint operator.

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Most popular questions from this chapter

Let \(V\) be an inner product space. Recall that each \(u \in V\) determines a linear functional \(\hat{u}\) in the dual space \(V^{*}\) by the definition \(\hat{u}(v)=\langle v, u\rangle\) for every \(v \in V\). (See the text immediately preceding Theorem 13.3 .) Show that the map \(u \mapsto \hat{u}\) is linear and nonsingular, and hence an isomorphism from \(V\) onto \(V^{*}\)

Show that a triangular matrix is normal if and only if it is diagonal.

Find an orthogonal change of coordinates \(X=P X^{\prime}\) that diagonalizes each of the following quadratic forms and find the corrcsponding diagonal quadratic form \(q\left(x^{\prime}\right)\) (a) \(q(x, y)=2 x^{2}-6 x y+10 y^{2}\) (b) \(q(x, y)=x^{2}+8 x y-5 y^{2}\) (c) \(q(x, y, z)=2 x^{2}-4 x y+5 y^{2}+2 x z-4 y z+2 z^{2}\)

Determine which of the following matrices is normal: (a) \(A=\left[\begin{array}{ll}1 & i \\ 0 & 1\end{array}\right]\) (b) \(\quad B=\left[\begin{array}{cc}1 & i \\ 1 & 2+i\end{array}\right]\)

For each of the following symmetric matrices \(A,\) find an orthogonal matrix \(P\) and a diagonal matrix \(D\) such that \(P^{\prime} A P\) is diagonal: (a) \(A=\left[\begin{array}{rr}1 & 2 \\ 2 & -2\end{array}\right]\) (b) \(A=\left[\begin{array}{rr}5 & 4 \\ 4 & -1\end{array}\right]\) (c) \(A=\left[\begin{array}{rr}7 & 3 \\ 3 & -1\end{array}\right]\)
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