Question

Demonstrate that the matrix $$ \mathrm{A}=\left(\begin{array}{ccc} 2 & 0 & 0 \\ -6 & 4 & 4 \\ 3 & -1 & 0 \end{array}\right) $$ is defective, i.e. does not have three linearly independent eigenvectors, by showing the following: (a) its eigenvalues are degenerate and, in fact, all equal; (b) any eigenvector has the form \(\left(\begin{array}{lll}\mu & (3 \mu-2 v) & v\end{array}\right)^{\mathrm{T}}\). (c) if two pairs of values, \(\mu_{1}, v_{1}\) and \(\mu_{2}, v_{2}\), define two independent eigenvectors \(\mathrm{v}_{1}\) and \(\mathrm{v}_{2}\) then any third similarly defined eigenvector \(\mathrm{v}_{3}\) can be written as a linear combination of \(\mathrm{v}_{1}\) and \(\mathrm{v}_{2}\), i.e. $$ \mathbf{v}_{3}=a \mathbf{v}_{1}+b \mathbf{v}_{2} $$ where $$ a=\frac{\mu_{3} v_{2}-\mu_{2} v_{3}}{\mu_{1} v_{2}-\mu_{2} v_{1}} \quad \text { and } \quad b=\frac{\mu_{1} v_{3}-\mu_{3} v_{1}}{\mu_{1} v_{2}-\mu_{2} v_{1}} $$ Illustrate (c) using the example \(\left(\mu_{1}, v_{1}\right)=(1,1),\left(\mu_{2}, v_{2}\right)=(1,2)\) and \(\left(\mu_{3}, v_{3}\right)=\) \((0,1)\) Show further that any matrix of the form $$ \left(\begin{array}{ccc} 2 & 0 & 0 \\ 6 n-6 & 4-2 n & 4-4 n \\ 3-3 n & n-1 & 2 n \end{array}\right) $$ is defective, with the same eigenvalues and eigenvectors as \(A\).

Short answer

Matrix A is defective with repeated eigenvalues and linearly dependent eigenvectors. The general form of the matrix shows the same properties.

Step-by-step solution

Find the Eigenvalues

To find the eigenvalues of matrix \(A\), solve the characteristic equation \(\text{det}(A - \, \boldsymbol{\text{I}}\boldsymbol{\text{I}}) = 0\). Calculate \(A - \lambda \boldsymbol{\text{I}}\) and then the determinant of the resulting matrix.

Calculate the Characteristic Polynomial

For matrix \(A\): \[ A - \lambda \boldsymbol{\text{I}} = \begin{pmatrix} 2 - \lambda & 0 & 0 \ -6 & 4 - \lambda & 4 \ 3 & -1 & -\lambda \end{pmatrix} \] Calculate the determinant: \[ \text{det}(A - \lambda \boldsymbol{\text{I}}) = (2 - \lambda) \begin{vmatrix} 4 - \lambda & 4 \ -1 & -\lambda \end{vmatrix} \] Solve the determinant inside the parentheses.

Solve for the Eigenvalues

Solve the determinant calculation and set it equal to zero to find the eigenvalues. Simplify the equation to: \[ (2 - \lambda)(\lambda^2 - 4 \lambda) = 0 \] Thus, the eigenvalues are: \[ \lambda = 2, 0, 0\]. Note the degeneracy of the eigenvalues.

Verify the Eigenvector Form

To verify the eigenvector form, solve the equation \((A - \lambda \boldsymbol{\text{I}}) \mathbf{v} = 0\). Substitute \(\lambda = 2\) and solve for \(\mathbf{v}\).

Show Dependent Eigenvectors

Assume \(\mathbf{v}_i\) are eigenvectors with corresponding components \((\mu_i, v_i)\). Demonstrate that any new eigenvector can be written as a linear combination of two others, proving dependency.

Example Eigenvectors

Use the given pairs \((\mu_1, v_1) = (1, 1), (\u03BC_2, v_2) = (1, 2), (\u03BC_3, v_3) = (0, 1)\). Show how \(\mathbf{v}_3 = a\mathbf{v}_1 + b\mathbf{v}_2 \). Calculate: \(a = \frac{\rho_3 u_2 - \rho_2 u_3}{\rho_1 u_2 - \rho_2 u_1}\) and \(b\frac{\rho_1 u_3 - \rho_3 u_1}{\rho_1 u_2 - \rho_2 u_1}\). Substituting gives \[a = 1, b = -1\].

Construct General Matrix Form

Show that the given matrix form also results in the same eigenvalues and eigenvectors by repeating steps for it.

Key concepts

Eigenvalues

Understanding eigenvalues is crucial in linear algebra. They are special numbers associated with a square matrix that provide insights into the matrix's properties.
The eigenvalues of a matrix arise from the characteristic equation, which is found by solving the determinant of the matrix subtracted by a scalar multiple of the identity matrix.
For our given matrix \(A\), the equation is defined as \(\text{det}(A - \lambda\text{I}) = 0\).
The eigenvalues \(\lambda\) for matrix \(A\) indicate where the squared term in the polynomial from the determinant calculation equals zero. These signify the scaling factors by which eigenvectors of the matrix are stretched or contracted.
In this problem, solving \((2 \lambda - \lambda)(\lambda^2 - 4 \lambda) = 0\) gives us the eigenvalues: \(2, 0, 0\). Notice the repetition of zero (degenerate eigenvalues), which hints at the defectiveness of the matrix.

Eigenvectors

Eigenvectors correspond to each eigenvalue and represent the directions along which the transformation defined by the matrix acts as a simple scalar multiplication.
For a matrix \(A\), an eigenvector \(\mathbf{v}\) with eigenvalue \(\lambda\) solves the equation \((A - \lambda I) \mathbf{v} = 0\).
In our example, consider an eigenvalue of \(2\). The corresponding eigenvectors can be found by solving \((A - 2I) \mathbf{v} = 0\).
The assumed form of eigenvector \(\mathbf{v}\) is \(\begin{pmatrix} \mu \ 3 \mu - 2 u \ u \end{pmatrix}\). This form demonstrates dependency among such vectors since they can be expressed in terms of each other using coefficients \(\mu\) and \(u\).
Thus, the presence of just two linearly independent eigenvectors for a 3 x 3 matrix with three eigenvalues proves the matrix’s defectiveness.

Linear Dependence

Linear dependence among vectors implies that some vectors can be written as linear combinations of others. This concept is crucial to verify if a matrix is defective.
A matrix is said to be defective if it does not possess a complete set of linearly independent eigenvectors.
In our exercise, the eigenvectors corresponding to \(\lambda = 0\) and \(\lambda = 2\) must be checked for linear independence.
For example, given sets \((\mu_1, u_1) = (1,1)\), \((\mu_2, u_2) = (1,2)\), and \((\mu_3, u_3) = (0, 1)\), we show that \(\mathbf{v}_3\) can be expressed as \(\mathbf{v}_3 = a \mathbf{v}_1 + b \mathbf{v}_2\).
By substituting the values, \(a\) and \(b\) can be computed to show the relationship among these vectors.
This verifies that the eigenvectors are linearly dependent, proving that the matrix is indeed defective.