Question

Given a matrix $$ \mathrm{A}=\left(\begin{array}{lll} 1 & \alpha & 0 \\ \beta & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) $$ where \(\alpha\) and \(\beta\) are non-zero complex numbers, find its eigenvalues and eigenvectors. Find the respective conditions for (a) the eigenvalues to be real and (b) the eigenvectors to be orthogonal. Show that the conditions are jointly satisfied if and only if \(A\) is Hermitian.

Short answer

Eigenvalues are: \ \bm{\text{a}} = \bm{a(\text{Eigenvalues})}\bm{\text{value}}) their vectors, if Hermitian if and if only real values. Conditions: Hermitian > a real > < only. \

Step-by-step solution

Determine the characteristic polynomial

The characteristic polynomial of a matrix \(\bm{A}\) is given by \[ \text{det}(\bm{A} - \bm{\text{I}}\bm{\text{u}}) = 0 \] where \(\bm{\text{I}}\) is the identity matrix and \(\bm{\text{u}}\) is the eigenvalue. Substitute \(\bm{A}\) and solve for \(\bm{\text{u}}\).

Compute the determinant

Calculate the determinant of \(\bm{A} - \bm{\text{I}}\bm{\text{u}}\) by setting \[ \bm{A} - \bm{\text{I}}\bm{\text{u}} = \begin{pmatrix} 1-\bm{\text{u}} & \bm{\text{a}} & 0 \ \bm{\text{b}} & 1-\bm{\text{u}} & 0 \ 0 & 0 & 1-\bm{\text{u}} \ \right) \] and finding the determinant, which simplifies as \[ = (1 - \bm{\text{u}})^3 - \bm{\text{a}}\bm{\text{b}}(1 - \bm{\text{u}}), \]

Solve for eigenvalues

Simplify the determinant expression: \[ (1 - \bm{\text{u}})^3 - \bm{\text{a}} \bm{\text{b}} (1 - \bm{\text{u}}) = 0. \] Expand and distribute to arrive at the final calculations: \[ \bm{\text{u}} = 1 \bm{**(GE)**} \bm{\text{\text{steps}}}. \]

Solve for eigenvectors

For each eigenvalue \(\bm{\text{u}}\), solve \( \bm{A} \bm{\text{\text{x}}} = \bm{\text{u}} \bm{\text{x}} \) or equivalently, \( (\bm{A} - \bm{I}\bm{u}) \bm{x} = 0 \bm{\text{x}}. \bm{\text{Solve}}. \)

Check whether eigenvalues are real

Eigenvalues are real if the characteristic equation has only real roots.

Check orthogonality of eigenvectors

Two eigenvectors are orthogonal if their dot product is zero.

Analyze conditions for Hermitian matrix

Matrix \( \bm{\text{A}} \) is Hermitian if \( \bm{\text{A}} = \bm{\text{A}}^{\text{\text{H}}} \). By comparing, only possible if its conditions meet jointly.

Key concepts

Characteristic Polynomial

To find the eigenvalues of a matrix, we first need to determine its characteristic polynomial. The characteristic polynomial is derived from the matrix equation \( \text{det}(\bm{A} - \bm{I}u) = 0 \). Here, \( \bm{A} \) is the given matrix, \( \bm{I} \) is the identity matrix, and \( u \) represents the eigenvalues we seek.
For our matrix \( \bm{A} = \begin{pmatrix} 1 & \alpha & 0 \ \beta & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} \), the substitution leads to the matrix \( \bm{A} - \bm{I}u = \begin{pmatrix} 1 - u & \alpha & 0 \ \beta & 1 - u & 0 \ 0 & 0 & 1 - u \end{pmatrix} \).
This matrix equation will help us derive the characteristic polynomial necessary for finding the eigenvalues.

Hermitian Matrix

A Hermitian matrix is one that is equal to its own conjugate transpose: \( A = A^H \). This property implies that its entries are symmetric along the main diagonal, and its off-diagonal elements are complex conjugates of each other.
For matrix \( A \), to be Hermitian, \( \alpha \) must equal the complex conjugate of \( \beta \), i.e., \( \alpha = \bar{\beta} \).
If these conditions hold, we can ensure the joint satisfaction of real eigenvalues and orthogonal eigenvectors, making matrix \( A \) Hermitian.

Orthogonality of Eigenvectors

Eigenvectors are orthogonal if their dot product is zero. Given eigenvectors \( x \) and \( y \), they are orthogonal if \( x^Ty = 0 \).
This property is especially significant when dealing with Hermitian matrices, because their eigenvectors are guaranteed to be orthogonal.
In other words, for our eigenvector problem, if the matrix \( A \) is Hermitian, any resulting eigenvectors corresponding to different eigenvalues will naturally be orthogonal.

Determinant Calculation

The determinant of a matrix is a scalar value that is a function of its entries. It provides important properties of the matrix, particularly in solving the characteristic polynomial.
For the matrix \( A \), computing \( \text{det}(A - Iu) \) is crucial in finding eigenvalues. We set up the matrix \( A - Iu = \begin{pmatrix} 1 - u & \alpha & 0 \ \beta & 1 - u & 0 \ 0 & 0 & 1 - u \end{pmatrix} \) and calculate its determinant. This determinant simplifies to \( (1 - u)^3 - \alpha \beta (1 - u) \).
By solving \( (1 - u)^3 - \alpha \beta (1 - u) = 0 \), we arrive at the eigenvalues needed for further analysis.