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Chapter 6: Problem 17

Prove that a matrix that is both unitary and upper triangular must be a diagonal matrix.

Short Answer

Expert verified
Given a matrix \(A\) that is both unitary and upper triangular, we know that \(A^*A = AA^* = I\) and \(A_{ij} = 0\) for \(i>j\). Calculating the product \(A^*A\), we have \(\sum_{k=1}^n A_{ki}^*A_{kj}\). If \(i>j\), then \(A_{kj} = 0\) for all \(k > j\), making the product zero. Since \(A\) already has its lower triangular part as zero, we have shown that all off-diagonal elements must also be zero. Therefore, matrix \(A\) must be a diagonal matrix.

Step by step solution

01

(Step 1: Represent the given matrix)

(Let \(A\) represent the given matrix, which is assumed to be a square matrix since both upper triangular and unitary matrices are square matrices. To represent its general form, we need to use the notation \(A_{ij}\) where \(i\) is the row number and \(j\) is the column number.)
02

(Step 2: Define properties of unitary and upper triangular matrix)

(For A to be unitary, we must have \(A^*A = AA^* = I\), where \(I\) is the identity matrix. For \(A\) to be upper triangular, all elements below the diagonal must be zero, i.e., \(A_{ij} = 0\) for \(i>j\).)
03

(Step 3: Calculate the product \(A^*A\))

(Next, we will compute the product of \(A^*\) and \(A\) and the result must be equal to the identity matrix. The elements of the resulting matrix are given by \(\sum_{k=1}^n A^*_{ik}A_{kj} = \delta_{ij}\), where \(\delta_{ij}\) is Kronecker's delta and \(\delta_{ij} = 1\) if \(i=j\) and \(\delta_{ij} = 0\) if \(i \neq j\).)
04

(Step 4: Apply the given constraints to the matrix product and simplify)

(Taking the definition of a conjugate transpose, the product of matrix elements \((A^*A)_{ij}\) turns into \(\sum_{k=1}^n A_{ki}^*A_{kj}\). Now notice that if \(i>j\), then \(A_{kj} = 0\) for all \(k > j\). This implies that if \(i>j\), the sum will not have any non-zero terms, so the product must be zero, i.e., \((A^*A)_{ij} = 0\).)
05

(Step 5: Conclude that A must be a diagonal matrix)

(We have shown that \((A^*A)_{ij} = 0\) when \(i > j\), i.e., all off-diagonal elements must be zero. Since \(A\) is upper triangular, it already has its lower triangular part as zero, and now the off-diagonal part of the upper triangular section is also zero. The only remaining non-zero elements are those on the diagonal. Thus, given the constraints of unitary and upper triangular, matrix \(A\) must be a diagonal matrix.)

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Key Concepts

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

Diagonal Matrix
A diagonal matrix is a special type of square matrix where all elements outside the main diagonal are zero. This means that if you were to write out a diagonal matrix, you'd see non-zero entries running from the top left to the bottom right corner, and all other positions would contain zeros.
Matrix Properties
To fully understand matrices, it's essential to know certain matrix properties. For instance, the transpose of a matrix is found by flipping it over its diagonal, so the row and column indices of each element are switched. A square matrix is called symmetric if it's equal to its transpose.When you hear 'unitary matrix', this refers to a square matrix whose conjugate transpose is also its inverse. This implies the lengths of column vectors, and hence, row vectors are preserved after transformation. Another interesting class is the 'upper triangular matrix'. Here, all elements below the main diagonal are zero.
Conjugate Transpose
The conjugate transpose, also known as the adjoint or Hermitian transpose, is perhaps one of the most interesting matrix operations. It involves taking the transpose of the matrix and then taking the complex conjugate of each element. That's swapping rows with columns and changing the sign of the imaginary part of complex numbers within the matrix. For real-valued matrices, this just reduces to taking the transpose, as there are no complex elements to conjugate.
Kronecker's Delta
Kronecker's delta is a function that may seem simple at first but plays a pivotal role in various areas of mathematics, including matrix algebra. It is defined as \( \delta_{ij} \) where \( \delta_{ij} = 1 \) if \( i = j \) and \( \delta_{ij} = 0 \) otherwise. In essence, Kronecker's delta 'picks out' the diagonal elements of a matrix by equating to one when the row and column indices match and nullifying the rest.

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Most popular questions from this chapter

Assume the notation of Theorem \(6.32\). (a) Prove that for any ordered basis \(\beta, \psi_{\beta}\) is linear. (b) Let \(\beta\) be an ordered basis for an \(n\)-dimensional space \(\mathrm{V}\) over \(F\), and let \(\phi_{\beta}: \mathrm{V} \rightarrow \mathrm{F}^{n}\) be the standard representation of \(\mathrm{V}\) with respect to \(\beta\). For \(A \in \mathbf{M}_{n \times n}(F)\), define \(H: \mathbf{V} \times \mathbf{V} \rightarrow F\) by \(H(x, y)=\) \(\left[\phi_{\beta}(x)\right]^{t} A\left[\phi_{\beta}(y)\right]\). Prove that \(H \in \mathcal{B}(\mathrm{V})\). Can you establish this as a corollary to Exercise 7? (c) Prove the converse of (b): Let \(H\) be a bilinear form on V. If \(A=\psi_{\beta}(H)\), then \(H(x, y)=\left[\phi_{\beta}(x)\right]^{t} A\left[\phi_{\beta}(y)\right]\).

Let \(\mathrm{T}\) and \(\mathrm{U}\) be self-adjoint linear operators on an \(n\)-dimensional inner product space \(\mathrm{V}\), and let \(A=[\mathrm{T}]_{\beta}\), where \(\beta\) is an orthonormal basis for V. Prove the following results. (a) \(\mathrm{T}\) is positive definite [semidefinite] if and only if all of its eigenvalues are positive [nonnegative]. (b) \(\mathrm{T}\) is positive definite if and only if $$ \sum_{i, j} A_{i j} a_{j} \bar{a}_{i}>0 \text { for all nonzero } n \text {-tuples }\left(a_{1}, a_{2}, \ldots, a_{n}\right) \text {. } $$ (c) \(\mathrm{T}\) is positive semidefinite if and only if \(A=B^{*} B\) for some square matrix \(B\). (d) If \(T\) and \(U\) are positive semidefinite operators such that \(T^{2}=U^{2}\), then \(\mathrm{T}=\mathrm{U}\). (e) If \(T\) and \(U\) are positive definite operators such that \(T U=U T\), then TU is positive definite. (f) \(\mathrm{T}\) is positive definite [semidefinite] if and only if \(A\) is positive definite [semidefinite]. Because of (f), results analogous to items (a) through (d) hold for matrices as well as operators.

Let \(V\) be a finite-dimensional inner product space over \(F\). (a) Parseval's Identity. Let \(\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) be an orthonormal basis for \(\mathrm{V}\). For any \(x, y \in \mathrm{V}\) prove that $$ \langle x, y\rangle=\sum_{i=1}^{n}\left\langle x, v_{i}\right\rangle \overline{\left\langle y, v_{i}\right\rangle} . $$ (b) Use (a) to prove that if \(\beta\) is an orthonormal basis for \(V\) with inner product \(\langle\cdot, \cdot\rangle\), then for any \(x, y \in \mathrm{V}\) $$ \left\langle\phi_{\beta}(x), \phi_{\beta}(y)\right\rangle^{\prime}=\left\langle[x]_{\beta},[y]_{\beta}\right\rangle^{\prime}=\langle x, y\rangle, $$ where \(\langle\cdot, \cdot\rangle^{\prime}\) is the standard inner product on \(\mathrm{F}^{n}\).

Let \(V\) be a finite-dimensional inner product space, and let \(T\) and \(U\) be self-adjoint operators on \(V\) such that \(T\) is positive definite. Prove that both TU and UT are diagonalizable linear operators that have only real eigenvalues. Hint: Show that UT is self-adjoint with respect to the inner product \(\langle x, y\rangle^{\prime}=\langle\mathrm{T}(x), y\rangle\). To show that TU is self-adjoint, repeat the argument with \(\mathrm{T}^{-1}\) in place of \(\mathrm{T}\).

Assume that \(\mathrm{T}\) is a linear operator on a complex (not necessarily finitedimensional) inner product space \(\mathrm{V}\) with an adjoint \(\mathrm{T}^{*}\). Prove the following results. (a) If \(\mathrm{T}\) is self-adjoint, then \(\langle\mathrm{T}(x), x\rangle\) is real for all \(x \in \mathrm{V}\). (b) If \(\mathrm{T}\) satisfies \(\langle\mathrm{T}(x), x\rangle=0\) for all \(x \in \mathrm{V}\), then \(\mathrm{T}=\mathrm{T}_{0}\). Hint: Replace \(x\) by \(x+y\) and then by \(x+i y\), and expand the resulting inner products. (c) If \(\langle\mathrm{T}(x), x\rangle\) is real for all \(x \in \mathrm{V}\), then \(\mathrm{T}\) is self-adjoint.
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