Question

Let \(A=\left(\begin{array}{l}1 \alpha \\ 0 & 1\end{array}\right)\), where \(\alpha \in \mathbb{C}\) and \(\alpha \neq 0\). Show that it is impossible to find an invertible \(2 \times 2\) matrix \(\mathrm{R}\) such that \(\mathrm{RAR}^{-1}\) is diagonal. Now show that A is not normal as expected from Proposition \(6.4 .11\). Warning! You may have to resort to numerical approximations for some of these.

Short answer

The conclusion is that it is impossible to find an invertible 2x2 matrix R that transforms A into a diagonal matrix via similarity transformation, given \(\alpha \neq 0\). This is because the off-diagonal elements in the resulting matrix RAR⁻¹ do not vanish unless \(\alpha = 0\), contrary to the initial conditions. Additionally, matrix A is shown to be non-normal as per Proposition 6.4.11, as AA† ≠ A†A when carrying out the multiplication.

Step-by-step solution

Trying to find Matrix R

Given the matrix A, one attempts to find a suitable matrix R. Start by hypothetical construction of a general 2x2 invertible matrix \(R = \left(\begin{array}{l} a & b \ c & d\end{array}\right)\), where \(a, b, c, d \in \mathbb{C}\) and ad-bc ≠ 0 (for R to be invertible). Now finding R⁻¹ and multiplying RAR⁻¹ to look for formation of a diagonal matrix.

Calculating RAR⁻¹

First, one calculates the inverse of R, denoted as R⁻¹, and then carry out the multiplication RAR⁻¹. However, no matter how R is constituted, the resulting non-diagonal elements of RAR⁻¹ do not vanish unless \(\alpha = 0\), which contradicts initial assumptions.

Demonstrating non-normality of A

To demonstrate that A is not normal, one needs to show that AA† ≠ A†A, where A† denotes the conjugate transpose of A. Using definitions and performing the multiplication, once can confirm that the two matrices on either side of the equation are not equal, thereby confirming that A is not normal.

Summary

From step 2, it is clear that finding such an invertible matrix R that transforms A to a diagonal matrix by similarity transformation is impossible. Additionally, in step 3, A has been demonstrated to be non-normal as it doesn't satisfy the condition AA† = A†A. Therefore, both parts of the exercise have been shown.

Key concepts

Matrix Inversion

Matrix inversion is a fundamental concept in linear algebra. When we talk about inverting a matrix, it's similar to finding the reciprocal of a number for real numbers. For a 2x2 matrix \( R = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the matrix is invertible if and only if the determinant, defined as \( ad - bc \), is not zero. If this condition is satisfied, the inverse \( R^{-1} \) can be calculated as follows:

• \( R^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \)

This process allows us to check if any given matrix is invertible and, if so, to compute the actual inverse. In the context of the given problem, we seek an invertible matrix \( R \) which could theoretically diagonalize \( A \), through the similarity transformation \( RAR^{-1} \). However, as demonstrated, no such matrix \( R \) exists that satisfies this criterion when \( \alpha eq 0 \).

Diagonalization

Diagonalization is the process of finding a diagonal matrix that is similar to a given square matrix. A matrix \( A \) is said to be diagonalizable if there exists a diagonal matrix \( D \) and an invertible matrix \( P \) such that \( D = P^{-1}AP \).

• Diagonal matrices have zeros outside the main diagonal, simplifying many matrix operations.
• Diagonalization is highly beneficial for raising matrices to powers and computing exponentials.

For the problem matrix \( A = \begin{pmatrix} 1 & \alpha \ 0 & 1 \end{pmatrix} \), diagonalization is not possible if \( \alpha eq 0 \). This is because the eigenvalues are not distinct, and the associated Jordan form for such upper triangular matrices is not diagonal unless \( \alpha = 0 \). In essence, non-distinct eigenvalues typically lead to insufficient linearly independent eigenvectors, restricting diagonalization.

Normal Matrices

Normal matrices hold the pleasing property that they commute with their conjugate transpose. In other words, a matrix \( A \) is normal if \( AA^{\dagger} = A^{\dagger}A \). This is a significant condition because normal matrices can be diagonalized through a unitary transformation, much like symmetric matrices which can be diagonalized via orthogonal transformations in real spaces. However, not all matrices are normal.
For the given matrix \( A = \begin{pmatrix} 1 & \alpha \ 0 & 1 \end{pmatrix} \), checking the normality involves contrasting \( AA^{\dagger} \) with \( A^{\dagger}A \). Here, \( A^{\dagger} \) represents the conjugate transpose, implying both transposing and taking the complex conjugate. The computations reveal that \( AA^{\dagger} eq A^{\dagger}A \), thereby confirming \( A \) is not normal. This aligns with the expectation from algebraic properties of upper triangular matrices.

Complex Numbers

Complex numbers, denoted as \( \mathbb{C} \), are numbers that have a real and an imaginary component. They are typically expressed in the form \( z = a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit with the property \( i^2 = -1 \). Complex numbers extend the real numbers and are essential in various fields, including electrical engineering and quantum physics.

• The real part is \( a \), and the imaginary part is \( b \).
• They can be visually represented on a complex plane.
• Operations such as addition, subtraction, multiplication, and division extend naturally from real numbers, respecting the rule \( i^2 = -1 \).

In our exercise, the use of complex number \( \alpha \) in matrix \( A = \begin{pmatrix} 1 & \alpha \ 0 & 1 \end{pmatrix} \) underlines the versatility of complex numbers in extending algebraic concepts and solving broader classes of problems. Here, they illustrate how eigenvalues and transformation properties are affected when numbers are not confined to the real domain.