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Chapter 13: Problem 51

Suppose \(T_{1}\) and \(T_{2}\) are self-adjoint. Show that \(T_{1} T_{2}\) is self- adjoint if and only if \(T_{1}\) and \(T_{1}\) commute; that is, \(T_{1} T_{2}=T_{2} T_{1}\)

Short Answer

Expert verified
The product of self-adjoint operators \(T_1\) and \(T_2\) is self-adjoint if and only if they commute. This is proven by showing (i) if \(T_1\) and \(T_2\) commute, then \(T_1 T_2\) is self-adjoint, and (ii) if \(T_1 T_2\) is self-adjoint, then \(T_1\) and \(T_2\) must commute.

Step by step solution

01

1) If part: Proving that if \(T_1\) and \(T_2\) commute, then \(T_1 T_2\) is self-adjoint

Assume that \(T_1\) and \(T_2\) commute, so \(T_1 T_2 = T_2 T_1\). We want to show that \((T_1 T_2)^* = T_1 T_2\). Since both \(T_1\) and \(T_2\) are self-adjoint, we have \(T_1^* = T_1\) and \(T_2^* = T_2\). Now, let us consider the adjoint of \(T_1 T_2\): \((T_1 T_2)^* = T_2^* T_1^* = T_2 T_1\) By our assumption that \(T_1\) and \(T_2\) commute, we have \(T_1 T_2 = T_2 T_1\). Therefore: \((T_1 T_2)^* = T_1 T_2\) Thus, if \(T_1\) and \(T_2\) commute, the product \(T_1 T_2\) is self-adjoint.
02

2) Only if part: Proving that if \(T_1 T_2\) is self-adjoint, then \(T_1\) and \(T_2\) commute

Assume that the product of the two operators is self-adjoint, i.e., \((T_1 T_2)^* = T_1 T_2\). Now, by taking the adjoint of \(T_1 T_2\), we get: \((T_1 T_2)^* = T_2^* T_1^* = T_2 T_1\) Since we assumed \((T_1 T_2)^* = T_1 T_2\), we have: \(T_1 T_2 = T_2 T_1\) Thus, we have shown that if \(T_1 T_2\) is self-adjoint, then \(T_1\) and \(T_2\) commute. This completes the proof in both directions, and we have demonstrated that \(T_1 T_2\) is self-adjoint if and only if \(T_1\) and \(T_2\) commute.

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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 \(V^{*}\)

Prove that the products and inverses of orthogonal matrices are orthogonal. (Thus, the orthogonal matrices form a group under multiplication, called the orthogonal group.)

Let \(T\) be a symmetric operator. Show that (a) The characteristic polynomial \(\Delta(t)\) of \(T\) is a product of linear polynomials (over \(\mathbf{R})\); (b) \(T\) has a nonzero eigenvector. (a) Let \(A\) be a matrix representing \(T\) relative to an orthonormal basis of \(V\); then \(A=A^{T}\). Let \(\Delta(t)\) be the characteristic polynomial of \(A\). Viewing \(A\) as a complex self-adjoint operator, \(A\) has only real eigenvalues by Theorem 13.4. Thus, $$ \Delta(t)=\left(t-\lambda_{1}\right)\left(t-\lambda_{2}\right) \cdots\left(t-\lambda_{n}\right) $$ where the \(\lambda_{i}\) are all real. In other words, \(\Delta(t)\) is a product of linear polynomials over \(\mathbf{R}\). (b) By (a), \(T\) has at least one (real) eigenvalue. Hence, \(T\) has a nonzero eigenvector.

Detcrmine which of the following matrices are positive (positive definitc): (i) \(\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right],(\text { ii })\left[\begin{array}{rr}0 & i \\ -i & 0\end{array}\right],\) (iii) \(\left[\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right],\) (iv) \(\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right],\left(\begin{array}{ll}v & 1 \\ 1 & 2\end{array}\right],\) (vi) \(\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]\)

Suppose \(T\) is self-adjoint. Show that \(T^{2}(v)=0\) implies \(T(v)=0 .\) Using this to prove that \(T^{n}(v)=0\) also implics that \(T(v)=0\) for \(n>0\)
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