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Chapter 13: Problem 34

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

Short Answer

Expert verified
In summary, we have proven that the products and inverses of unitary matrices are also unitary. We did this by demonstrating that the product of two unitary matrices is unitary, and that the inverse of a unitary matrix is also unitary. This shows that the unitary matrices form a group under multiplication, known as the unitary group.

Step by step solution

01

Part 1: Product of Unitary Matrices is Unitary

Suppose A and B are unitary matrices. We must show that their product (AB) is also a unitary matrix. First, we find the conjugate transpose of (AB), which we will denote as (AB)*. (AB)* = A*B* Now, we need to prove that (AB)* is the inverse of (AB). To do this, let's multiply (AB) and (AB)* and see if it gives us the identity matrix I: (AB)(AB)* = A(BB*)A* Since A and B are unitary, we know that AA* = I and BB* = I. So, A(BB*)A* = AIA* = AA* = I. Therefore, the product of two unitary matrices, A and B, is also a unitary matrix as (AB)* is the inverse of (AB).
02

Part 2: Inverse of a Unitary Matrix is Unitary

Suppose U is a unitary matrix, and U^(-1) is its inverse. We need to show that the inverse of U is also unitary. To prove that U^(-1) is unitary, we need to show that its conjugate transpose (U^(-1)*) is the inverse of U^(-1). To find the conjugate transpose of U^(-1), we need to use the property (A^-1)* = (A*)^-1. U^(-1)* = (U*)^(-1) Now, since U is unitary, its conjugate transpose U* is equal to its inverse U^(-1): U^(-1)* = (U*)^(-1) = U Hence, U^(-1)* is the inverse of U^(-1). So, we have proven that both the inverse and the product of unitary matrices are also unitary. Consequently, the unitary matrices form a group under multiplication, which is called the unitary group.

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Most popular questions from this chapter

Prove Theorem 13.3: Let \(\phi\) be a linear functional on an \(n\) -dimensional inner product space \(V\) Then there exists a unique vector \(u \in V\) such that \(\phi(v)=\langle v, u\rangle\) for every \(v \in V\)

Show that inner product spaces \(V\) and \(W\) over \(K\) are isomorphic if and only if \(V\) and \(W\) have the same dimension.

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.

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 \(V^{*}\)

Unitary and Orthogonal Operators and Matrices Find a unitary (orthogonal) matrix whose first row is (a) \(\quad(2 / \sqrt{13}, 3 / \sqrt{13})\) (b) a multiple of \((1,1-i)\) (c) a multiple of \((1,-i, 1-i)\)
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