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Chapter 6: Problem 11

Prove the Frobenius theorem in the case where \(A\) is a \(2 \times 2\) matrix.

Short Answer

Expert verified
For a given 2x2 matrix \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), we find the conjugate transpose \(A^* = \begin{bmatrix} \bar{a} & \bar{c} \\ \bar{b} & \bar{d} \end{bmatrix}\), calculate \(A^*A\) and find the eigenvalues (位鈧�, 位鈧�). The singular values of \(A\) are the square roots of the eigenvalues, \(\sigma_1^2 = 位_1\) and \(\sigma_2^2 = 位_2\). We then show that the sum of squares of the singular values equals the sum of squares of the entries of \(A\), i.e., \(\sigma_1^2 + \sigma_2^2 = |a|^2 + |b|^2 + |c|^2 + |d|^2\). This proves the Frobenius theorem for a 2x2 matrix A.

Step by step solution

01

Define the 2x2 matrix A

Let A be a 2x2 matrix, where \[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] with a, b, c, and d being complex numbers.
02

Calculate the conjugate transpose A*

Find the conjugate transpose of A, denoted as A^*. To find A^*, take the transpose of A and then find the complex conjugate of each element: \[ A^* = \begin{bmatrix} \bar{a} & \bar{c} \\ \bar{b} & \bar{d} \end{bmatrix} \]
03

Calculate A^*A

Now, we will multiply A^* by A. Since both matrices are 2x2, this will be a simple matrix multiplication. \[ A^*A = \begin{bmatrix} \bar{a} & \bar{c} \\ \bar{b} & \bar{d} \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} \bar{a}a + \bar{c}c & \bar{a}b + \bar{c}d \\ \bar{b}a + \bar{d}c & \bar{b}b + \bar{d}d \end{bmatrix} \]
04

Find the eigenvalues of A^*A

To find the eigenvalues of A^*A, we need to solve the characteristic equation det(A^*A - 位I) = 0, where 位 is an eigenvalue and I is the identity matrix: \[ \begin{vmatrix} \bar{a}a + \bar{c}c - 位 & \bar{a}b + \bar{c}d \\ \bar{b}a + \bar{d}c & \bar{b}b + \bar{d}d - 位 \end{vmatrix} = 0 \] Expanding this determinant, we get \[ (\bar{a}a + \bar{c}c - 位)(\bar{b}b + \bar{d}d - 位) - (\bar{a}b + \bar{c}d)(\bar{b}a + \bar{d}c) = 0 \] This is a quadratic equation in 位. Since the matrix A^*A is Hermitian, its eigenvalues will be real.
05

Find the singular values of A

The singular values of A are the square roots of the eigenvalues of A^*A. Let 蟽_1 and 蟽_2 be the singular values of A. Then, \[ 蟽_1^2 = 位_1 \quad , \quad 蟽_2^2 = 位_2 \] where 位_1 and 位_2 are eigenvalues of A^*A.
06

Show that the sum of squares of the singular values equals the sum of squares of the entries of A

We need to show that 蟽_1^2 + 蟽_2^2 = |a|^2 + |b|^2 + |c|^2 + |d|^2. Using the results from step 4, we have \[ 蟽_1^2 + 蟽_2^2 = 位_1 + 位_2 = tr(A^*A) = \bar{a}a + \bar{c}c + \bar{b}b + \bar{d}d \] Since \(\bar{x}x = |x|^2\) for any complex number x, we have \[ 蟽_1^2 + 蟽_2^2 = |a|^2 + |b|^2 + |c|^2 + |d|^2 \] Therefore, we have proved the Frobenius theorem for a 2x2 matrix A.

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Most popular questions from this chapter

Given \(A \in \mathbb{R}^{m \times m}, B \in \mathbb{R}^{n \times n}, C \in \mathbb{R}^{m \times n},\) the equa- tion \\[ A X-X B=C \\] is known as Sylvester's equation. An \(m \times n\) matrix \(X\) is said to be a solution if it satisfies (3) (a) Show that if \(B\) has Schur decomposition \(B=U T U^{H},\) then Sylvester's equation can be transformed into an equation of the form \(A Y-Y T=G,\) where \(Y=X U\) and \(G=C U\) (b) Show that \\[ \begin{array}{l} \left(A-t_{11} I\right) \mathbf{y}_{1}=\mathbf{g}_{1} \\ \left(A-t_{j j} I\right) \mathbf{y}_{j}=\mathbf{g}_{j}+\sum_{i=1}^{j-1} t_{i j} \mathbf{y}_{j}, j=2, \ldots, n \end{array} \\] (c) Show that if \(A\) and \(B\) have no common eigenvalues, then Sylvester's equation has a solution.

Show that if a matrix \(U\) is both unitary and Hermitian, then any eigenvalue of \(U\) must equal either 1 or -1

Show that any \(3 \times 3\) matrix of the form \\[ \left(\begin{array}{lll} a & 1 & 0 \\ 0 & a & 1 \\ 0 & 0 & b \end{array}\right) \\] is defective.

Let \(A\) be a symmetric positive definite matrix. Show that the diagonal elements of \(A\) must all be positive.

Let \(A\) be an \(n \times n\) matrix with an eigenvalue \(\lambda\) of multiplicity \(n .\) Show that \(A\) is diagonalizable if and only if \(A=\lambda I\)
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