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Chapter 11: Problem 10

Let \(A, B\) be normal operators such that \(A B=B A\). Show that \(A B\) is normal.

Short Answer

Expert verified
Given that \(A\) and \(B\) are normal operators such that \(AB = BA\), we want to show that \(AB\) is also normal. We compute the adjoint of \(AB\) as \((AB)^* = B^* A^*\). Using the properties of normal and commuting operators, we evaluate \((AB)(AB)^*\) and \((AB)^*(AB)\) as follows: \[(AB)(AB)^* = A(B^*B)A^* = (AA^*)(BB^*)\] \[(AB)^*(AB) = (B^*B)(A^*A) = (BB^*)(AA^*)\] Since both expressions are equal, we have shown that \(AB\) is a normal operator: \((AB)(AB)^* = (AB)^*(AB)\).

Step by step solution

01

Recall the definition of a normal operator

An operator \(O\) is said to be normal if it commutes with its adjoint, that is, if the following condition holds: \[O O^* = O^* O\] Here, we are given that \(A\) and \(B\) are normal operators, which means: \[A A^* = A^* A\] \[B B^* = B^* B\] We aim to show that the operator \(AB\) is normal, i.e., that \((AB)(AB)^* = (AB)^*(AB)\).
02

Use the fact that \(A\) and \(B\) are normal operators and that they commute

We are given that \(AB = BA\). We will use this property to manipulate expressions involving \(A\), \(B\), and their adjoints.
03

Compute the adjoint of \(AB\)

We need to find the adjoint of \(AB\), which we denote by \((AB)^*\). Recall that for any operators \(C\) and \(D\), we have: \[(CD)^* = D^* C^*\] Applying this to our problem, we have: \[(AB)^* = B^* A^*\]
04

Show that \(AB\) commutes with its adjoint

Now, we need to show that \((AB)(AB)^* =(AB)^*(AB)\). Using the fact that \(AB = BA\), let's compute both sides of this equation and see if they are equal: 1. Compute \((AB)(AB)^*\): \[(AB)(AB)^* = (AB)(B^* A^*) = A(BB^*)A^*\] Since \(B\) is normal, we know that \(BB^* = B^*B\): \[(AB)(AB)^* = A(B^*B)A^*\] Now use the fact that \(A\) and \(B\) commute: \[(AB)(AB)^* = A(B^*A)B = (AA^*)(BB^*)\] 2. Compute \((AB)^*(AB)\): \[(AB)^*(AB) = (B^*A^*)(AB)\] Since \(A\) is normal, we know that \(A^*A = AA^*\): \[(AB)^*(AB) = (B^*)(A^*A)(B) = (B^*B)(A^*A)\] Comparing our results, we see that: \[(AB)(AB)^* = (AA^*)(BB^*) = (B^*B)(A^*A) = (AB)^*(AB)\] Hence, we have shown that \(AB\) is a normal operator since it commutes with its adjoint.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Commuting Operators
When we talk about commuting operators in linear algebra, we're referring to operators that behave nicely with one another. That is, if you multiply them together in different orders, you get the same result. More specifically, two operators \(A\) and \(B\) commute if \(AB = BA\). This property can often simplify complex equations and is quite valuable, particularly when dealing with matrices or transformations in higher mathematics.

For example, if you have two matrices that represent linear transformations and they commute, you can interchange their application without changing the outcome. Knowing that operators commute aids in solving equations and proving properties of other operators, just like in the exercise above where this property helped demonstrate that the product \(AB\) is a normal operator.
Adjoint Operators
The concept of an adjoint operator is essential in understanding the symmetry and behavior of operators in linear algebra. The adjoint of an operator \(O\), denoted as \(O^*\), is akin to taking a kind of 'transpose' and 'conjugate' of a matrix in mathematical terms. The operation is particularly relevant in contexts involving complex vector spaces.

The pedagogical importance of adjoint operators lies in how they help define and identify normal operators. A normal operator satisfies the condition \(O O^* = O^* O\), meaning it commutes with its own adjoint. This symmetry is critical in many calculations and proofs, as seen in our initial exercise. Understanding how to compute and manipulate adjoint operators is foundational for higher-level linear algebra studies.
Linear Algebra Operators
Linear algebra is the study of linear operators, which are essential for understanding the structures and transformations in mathematics. Operators in this context can be thought of as functions that take one vector in a vector space and map it to another vector in the same space. The various types of operators, including normal, unitary, and self-adjoint operators, each have unique properties and uses.

In our initial exercise, we explored the notion of normal operators, which are a special class of operators that have significant implications for stability and consistency in mathematical solutions. Normal operators can be diagonalized in specific ways, which greatly simplifies working with them. In general, understanding linear algebra operators equips students with the tools necessary to navigate and resolve complex mathematical problems. These concepts are not only crucial in pure mathematics but also in applied fields such as quantum mechanics and computer science.

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Most popular questions from this chapter

Find an orthogonal basis of \(\mathbf{R}^{2}\) consisting of eigenvectors of the given matrix. (a) \(\left(\begin{array}{ll}1 & 3 \\ 3 & 2\end{array}\right)\) (b) \(\left(\begin{array}{rr}-1 & 1 \\ 1 & 2\end{array}\right)\) (c) \(\left(\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right)\) (d) \(\left(\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right)\) (e) \(\left(\begin{array}{rr}1 & -1 \\ -1 & 1\end{array}\right)\) (f) \(\left(\begin{array}{rr}2 & -3 \\ -3 & 1\end{array}\right)\)

Let \(V\) be as in \(82 .\) Let \(A: V \rightarrow V\) be a symmetrie linear map. Referring back to Bylvester's theorem, show that the index of nullity of the form $$ (v, w) \mapsto\langle A v, w\rangle $$ is equal to the dimension of the kernel of \(A\). Show that the index of positivity is equal to the number of eigenvectors in a spectral basis having a positive eigenvalue.

Let \(V\) be as in \(82 .\) Let \(A: V \rightarrow V\) be a symmetric linear map. Let \(v_{1}, v_{2}\) be eigenvectors of \(A\) with eigenvalues \(\lambda_{1}, \lambda_{2}\) respeetively. If \(\lambda_{1} \neq \lambda_{2}\), show that \(v_{1}\) is perpendicular to \(v_{2}\).

Let \(V\) be as in 82 , and let \(A: V \rightarrow V\) be a symmetrie operator. Let \(\lambda_{1}, \ldots, \lambda_{r}\), be the distinet eigenvalues of \(A .\) If \(\lambda\) is an eigenvalue of \(A\), let \(V_{\lambda}(A)\) eonsist of the set of all \(v \in V\) such that \(A v=\lambda v\). (a) Show that \(V_{\lambda}(A)\) is a subspace of \(V\), and that \(.1\) maps \(V_{\lambda}(A)\) into itself. (b) Show that \(V\) is the direet sum of the spaces $$ V=V_{\lambda_{1}}(A) \oplus \cdots \oplus V_{\lambda,}(A) $$ and that any two of these subspaces are mutually orthogonal. We call \(V_{\lambda}(A)\) the eigenspace of \(A\) belonging to \(\lambda\).

Show that a positive hermitian operator \(A\) has a square root, i.e. a positive hermitian operator \(B\) sueh that \(B^{2}=A\), and \(B\) is uniquely determined.
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