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If a unitary matrix \(\mathrm{U}\) is written as \(\mathrm{A}+i \mathrm{~B}\), where \(\mathrm{A}\) and \(\mathrm{B}\) are Hermitian with non-degenerate eigenvalues, show the following: (a) A and B commute; (b) \(A^{2}+B^{2}=1\) (c) The eigenvectors of \(A\) are also eigenvectors of \(B\); (d) The eigenvalues of \(U\) have unit modulus (as is necessary for any unitary matrix).

Short Answer

Expert verified
A and B commute. A^2 + B^2 = I. Eigenvectors of A are eigenvectors of B. Eigenvalues of U have unit modulus.

Step by step solution

01

- Identify Hermitian Properties

Recall that a Hermitian matrix has the property that it is equal to its own conjugate transpose, i.e., \(\text{A}^{\text{†}} = \text{A}\) and \(\text{B}^{\text{†}} = \text{B}\). Additionally, for a unitary matrix U, the property \(\text{U} \text{U}^{\text{†}} = \text{I}\) holds.
02

- Show A and B Commute (Part a)

Given \(\text{U} = \text{A} + i \text{B}\), determine the conjugate transpose \(\text{U}^{\text{†}} = (\text{A} + i \text{B})^{\text{†}} = \text{A}^{\text{†}} - i \text{B}^{\text{†}} = \text{A} - i \text{B}\). Since U is unitary: \(\text{U} \text{U}^{\text{†}} = (\text{A} + i \text{B})(\text{A} - i \text{B}) = \text{A}^2 + \text{A}(-i\text{B}) + i \text{B}\text{A} - i^2 \text{B}^2 = \text{A}^2 + \text{B}^2 + i (\text{BA} - \text{AB}) = \text{I}\). Equating the imaginary part to zero, \(\text{A}\text{B} - \text{B}\text{A} = 0 \Rightarrow \text{A}\) and \(\text{B}\) commute with each other.
03

- Prove A^2 + B^2 = I (Part b)

From Step 2: \(\text{A}^2 + \text{B}^2 = \text{I}\), conclude that \(\text{A}^2 + \text{B}^2 = \text{I}\).
04

- Eigenvectors of A and B (Part c)

Since \(\text{A}\) and \(\text{B}\) commute, they can be simultaneously diagonalized. Hence, the eigenvectors of \(\text{A}\) are also eigenvectors of \(\text{B}\). Given \(\text{A}\textbf{v} = \text{a}\textbf{v}\) and \(\text{B}\textbf{v} = \text{b \textbf{v}}\), where \(\textbf{v}\) is a common eigenvector and \(\text{a}\), \(\text{b}\) are eigenvalues.
05

- Eigenvalues of U on Unit Modulus (Part d)

If \(\textbf{v}\) is the common eigenvector of \(\text{A}\) and \(\text{B}\), \(\text{U}\textbf{v} = (\text{A} + i \text{B})\textbf{v} = (\text{a} + i \text{b})\textbf{v}\). From earlier, \(\text{a}^2 + \text{b}^2 = 1\): \(\text{U}\textbf{v} = (\text{a} + i \text{b})\textbf{v}\), and follows that the eigenvalues are on the unit circle, i.e., have unit modulus.

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Key Concepts

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

Hermitian Matrices
A Hermitian matrix is a special kind of square matrix that is equal to its own conjugate transpose. This means, for a matrix \(\text{A}\), it holds that \(\text{A}^{\dagger} = \text{A}\). A matrix \(\text{A}^{\dagger}\) is found by taking the transpose of \(\text{A}\) and then taking the complex conjugate of each element in \(\text{A}\). Hermitian matrices have several important properties:
  • All eigenvalues of a Hermitian matrix are real.
  • Eigenvectors corresponding to distinct eigenvalues are orthogonal.
The study of Hermitian matrices is very important in quantum mechanics and other areas of physics as they often describe observable quantities.
Unitary Matrices
A unitary matrix is a complex square matrix \(\text{U}\) where \(\text{U} \text{U}^{\dagger} = \text{I}\), with \(\text{I}\) being the identity matrix. This property ensures that the matrix preserves the inner product, meaning it's distance-preserving and angle-preserving. Simply put, operations described by unitary matrices do not change the length of vectors. For functions and vectors involving complex numbers, unitary matrices act similarly to orthogonal matrices in real-vector spaces. Key characteristics include:
  • All eigenvalues of a unitary matrix lie on the unit circle in the complex plane, meaning they have a magnitude (modulus) of 1.
  • Unitary matrices play a crucial role in quantum computing and coordinate transformations.
Eigenvalues
Eigenvalues are scalars associated with a square matrix that provide deep insight into the properties of the matrix. They are derived from the equation \(\text{A}\textbf{v} = \text{\lambda}\textbf{v}\), where \(\text{A}\) is the square matrix, \(\text{\lambda}\) is the eigenvalue, and \(\textbf{v}\) is the eigenvector. Important points about eigenvalues:
  • If \(\text{A}\) is a Hermitian matrix, then all its eigenvalues are real.
  • If \(\text{U}\) is a unitary matrix, then all its eigenvalues lie on the unit circle in the complex plane, meaning they have an absolute value of 1.
Eigenvalues can tell you a lot about a matrix, such as whether the matrix is invertible, its stability properties, and much more.
Eigenvectors
Eigenvectors are vectors that only change by a scalar factor when a matrix is applied to them. For a square matrix \(\text{A}\) and a scalar \(\text{\lambda}\) (which is called the eigenvalue), if \(\text{A}\textbf{v} = \text{\lambda}\textbf{v}\), the vector \(\textbf{v}\) is known as an eigenvector corresponding to the eigenvalue \(\text{\lambda}\). Here's what's fascinating about eigenvectors:
  • For Hermitian matrices, eigenvectors corresponding to different eigenvalues are orthogonal.
  • For the exercise at hand, since \(\text{A}\) and \(\text{B}\) are Hermitian and they commute, they can be simultaneously diagonalized. This implies that \(\text{A}\) and \(\text{B}\) share the same set of eigenvectors.
Eigenvectors provide valuable insights into the structure of the matrix.
Matrix Commutation
Matrix commutation refers to the scenario in which two matrices \(\text{A}\) and \(\text{B}\) commute, meaning \(\text{A}\text{B} = \text{B}\text{A}\). In the case of Hermitian matrices \(\text{A}\) and \(\text{B}\) forming a unitary matrix \(\text{U} = \text{A} + i \text{B}\), their commutation is crucial:
  • Because \(\text{U}\) is unitary, we find \(\text{A}\text{B} = \text{B}\text{A}\) from expanding the conjugate transpose property \((\text{A} + i \text{B})(\text{A} - i \text{B}) = \text{I}\).
  • Commutation indicates that these matrices share a common set of eigenvectors and can be diagonalized simultaneously.
In summary, the commutation of matrices makes it easier to simplify complex matrix operations and find their eigenvalues and eigenvectors.

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Most popular questions from this chapter

Find the eigenvalues and a set of eigenvectors of the matrix $$ \left(\begin{array}{ccc} 1 & 3 & -1 \\ 3 & 4 & -2 \\ -1 & -2 & 2 \end{array}\right) $$ Verify that its eigenvectors are mutually orthogonal.

Find the condition(s) on \(\alpha\) such that the simultaneous equations $$ \begin{aligned} x_{1}+\alpha x_{2} &=1 \\ x_{1}-x_{2}+3 x_{3} &=-1 \\ 2 x_{1}-2 x_{2}+\alpha x_{3} &=-2 \end{aligned} $$ have (a) exactly one solution, (b) no solutions, or (c) an infinite number of solutions; give all solutions where they exist.

Which of the following statements about linear vector spaces are true? Where a statement is false, give a counter-example to demonstrate this. (a) Non-singular \(N \times N\) matrices form a vector space of dimension \(N^{2}\). (b) Singular \(N \times N\) matrices form a vector space of dimension \(N^{2}\). (c) Complex numbers form a vector space of dimension \(2 .\) (d) Polynomial functions of \(x\) form an infinite-dimensional vector space. (e) Series \(\left\\{a_{0}, a_{1}, a_{2}, \ldots, a_{N}\right\\}\) for which \(\sum_{n=0}^{N}\left|a_{n}\right|^{2}=1\) form an \(N\)-dimensional vector space. (f) Absolutely convergent series form an infinite-dimensional vector space. (g) Convergent series with terms of alternating sign form an infinite- dimensional vector space.

Solve the following simultaneous equations for \(x_{1}, x_{2}\) and \(x_{3}\), using matrix methods: $$ \begin{aligned} x_{1}+2 x_{2}+3 x_{3} &=1 \\ 3 x_{1}+4 x_{2}+5 x_{3} &=2 \\ x_{1}+3 x_{2}+4 x_{3} &=3 \end{aligned} $$

One method of determining the nullity (and hence the rank) of an \(M \times N\) matrix A is as follows. \- Write down an augmented transpose of \(A\), by adding on the right an \(N \times N\) unit matrix and thus producing an \(N \times(M+N)\) array \(\mathrm{B}\). \- Subtract a suitable multiple of the first row of \(B\) from each of the other lower rows so as to make \(B_{i 1}=0\) for \(i>1\) \- Subtract a suitable multiple of the second row (or the uppermost row that does not start with \(M\) zero values) from each of the other lower rows so as to make \(B_{i 2}=0\) for \(i>2\) \- Continue in this way until all remaining rows have zeroes in the first \(M\) places. The number of such rows is equal to the nullity of \(A\) and the \(N\) rightmost entries of these rows are the components of vectors that span the null space. They can be made orthogonal if they are not so already. Use this method to show that the nullity of $$ A=\left(\begin{array}{cccc} -1 & 3 & 2 & 7 \\ 3 & 10 & -6 & 17 \\ -1 & -2 & 2 & -3 \\ 2 & 3 & -4 & 4 \\ 4 & 0 & -8 & -4 \end{array}\right) $$ is 2 and that an orthogonal base for the null space of \(A\) is provided by any two column matrices of the form \(\left(\begin{array}{lll}2+\alpha_{i} & -2 \alpha_{i} & 1 & \alpha_{i}\end{array}\right)^{\mathrm{T}}\) for which the \(\alpha_{i}(i=1,2)\) are real and satisfy \(6 \alpha_{1} \alpha_{2}+2\left(\alpha_{1}+\alpha_{2}\right)+5=0\).

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