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Chapter 13: Problem 60

Suppose \(T_{1}\) and \(T_{2}\) are normal and commute. Show that \(T_{1}+T_{2}\) and \(T_{1} T_{2}\) are also normal.

Short Answer

Expert verified
In summary, we showed that both \(T_1 + T_2\) and \(T_1 T_2\) are normal. We demonstrated this by first computing their conjugate transposes and then showing that they commute with the original operators, leveraging the given conditions that \(T_1\) and \(T_2\) are normal and commute with each other.

Step by step solution

01

Compute the Conjugate Transpose of \(T_1 + T_2\)

Since \(T_1\) and \(T_2\) are normal, we have that \(T_1^\dagger T_1 = T_1 T_1^\dagger\) and \(T_2^\dagger T_2 = T_2 T_2^\dagger\). Now let's compute the conjugate transpose of \(T_1 + T_2\): \((T_1 + T_2)^\dagger = T_1^\dagger + T_2^\dagger\)
02

Demonstrate that \(T_1 + T_2\) is normal

Now, let's compute the product of \((T_1 + T_2)\) and its conjugate transpose in both orders: \((T_1 + T_2)(T_1^\dagger + T_2^\dagger) = T_1 T_1^\dagger + T_1 T_2^\dagger + T_2 T_1^\dagger + T_2 T_2^\dagger\) \((T_1^\dagger + T_2^\dagger)(T_1 + T_2) = T_1^\dagger T_1 + T_1^\dagger T_2 + T_2^\dagger T_1 + T_2^\dagger T_2\) Notice that the terms in both expressions are the same, except for the middle two terms \(T_1 T_2^\dagger\) and \(T_2^\dagger T_1\). Since \(T_1\) and \(T_2\) commute, we have \(T_1 T_2^\dagger = T_2^\dagger T_1\), and the expressions are equal. Therefore, \(T_1 + T_2\) is normal.
03

Compute the Conjugate Transpose of \(T_1 T_2\)

Next, let's compute the conjugate transpose of \(T_1 T_2\): \((T_1 T_2)^\dagger = T_2^\dagger T_1^\dagger\)
04

Demonstrate that \(T_1 T_2\) is normal

Now, let's compute the product of \((T_1 T_2)\) and its conjugate transpose in both orders: \((T_1 T_2)(T_2^\dagger T_1^\dagger) = T_1 (T_2 T_2^\dagger) T_1^\dagger\) \((T_2^\dagger T_1^\dagger)(T_1 T_2) = T_2^\dagger (T_1^\dagger T_1) T_2\) Since \(T_2\) is normal, we have \(T_2 T_2^\dagger = T_2^\dagger T_2\). And since \(T_1\) is normal, we have \(T_1 T_1^\dagger = T_1^\dagger T_1\). Using these properties and the fact that \(T_1\) and \(T_2\) commute, we can rewrite our expressions: \((T_1 T_2)(T_2^\dagger T_1^\dagger) = T_1 (T_2^\dagger T_2) T_1^\dagger = T_1 T_1^\dagger T_2 T_2^\dagger\) \((T_2^\dagger T_1^\dagger)(T_1 T_2) = T_2^\dagger T_1^\dagger T_1 T_2 = T_2^\dagger T_2 T_1 T_1^\dagger\) The expressions are equal, and thus \(T_1 T_2\) is normal. We have shown that both \(T_1 + T_2\) and \(T_1 T_2\) are normal, as required.

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Most popular questions from this chapter

Let \(U\) be a unitary (orthogonal) operator on \(V\), and let \(W\) be a subspace invariant under \(U\). Show that \(W^{\perp}\) is also invariant under \(U\)

Prove Theorem 13.11: Let \(T\) be a symmetric operator on a real \(n\) -dimensional inner product space \(V\). Then there exists an orthonormal basis of \(V\) consisting of eigenvectors of \(T\). (Hence, \(T\) can be represented by a diagonal matrix relative to an orthonormal basis.)

Show that \((\mathrm{a}) \quad I^{*}=I,\) and \((\mathrm{b}) \quad 0^{*}=0\)

Show that there exists an orthonormal basis \(\left\\{u_{1}, \ldots, u_{n}\right\\}\) of \(V\) consisting of eigenvectors of \(T\) if and only if there exist orthogonal projections \(E_{1}, \ldots, E_{r}\) and scalars \(\lambda_{1}, \ldots, \lambda_{r}\) such that (i) \(T=\lambda_{1} E_{1}+\cdots+\lambda_{r} E_{r}\) (ii) \(\quad E_{1}+\cdots+E_{r}=I\) (iii) \(\quad E_{i} E_{j}=0 \quad\) for \(\quad i \neq j\)

Find the adjoint of \(G: \mathbf{C}^{3} \rightarrow \mathbf{C}^{3}\) defined by G(x, y, z)=[2 x+(1-i) y, \quad(3+2 i) x-4 i z, \quad 2 i x+(4-3 i) y-3 z]
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