Question

Given that \(A\) is a real symmetric matrix with normalised eigenvectors \(\mathrm{e}^{i}\) obtain the coefficients \(\alpha_{i}\) involved when column matrix \(x\), which is the solution of $$ \mathrm{A} \mathrm{x}-\mu \mathrm{x}=\mathrm{v} $$ is expanded as \(x=\sum_{i} \alpha_{i} e^{i} .\) Here \(\mu\) is a given constant and \(v\) is a given column matrix. (a) Solve (*) when $$ \mathrm{A}=\left(\begin{array}{lll} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right) $$ \(\mu=2\) and \(\mathrm{v}=\left(\begin{array}{lll}1 & 2 & 3\end{array}\right)^{\mathrm{T}}\) (b) Would \((*)\) have a solution if \(\mu=1\) and (i) \(v=\left(\begin{array}{lll}1 & 2 & 3\end{array}\right)^{\mathrm{T}}\), (ii) \(\mathrm{v}=\) \(\left(\begin{array}{lll}2 & 2 & 3\end{array}\right)^{\mathrm{T}} ?\)

Short answer

For (a), the coefficients are \(\alpha_1 = -\sqrt{2}\), \(\alpha_2 = \sqrt{2}\), \(\alpha_3 = 1.\) For (b), no solutions exist for both cases.

Step-by-step solution

Eigenvalue Decomposition of A

Find the eigenvalues and eigenvectors of the matrix \(\mathrm{A}.\) The matrix \(\mathrm{A}\) is: \[ \mathrm{A}=\left(\begin{array}{lll} 2 & 1 & 0 \ 1 & 2 & 0 \ 0 & 0 & 3 \end{array}\right)\]

Eigenvalues of A

To find the eigenvalues, solve the characteristic equation \[ \text{det}(\mathrm{A} - \lambda \mathrm{I}) = 0. \] The eigenvalues are: \[ \lambda_1 = 1, \lambda_2 = 3, \lambda_3 = 3. \]

Eigenvectors of A

Find the corresponding eigenvectors for each eigenvalue. The normalized eigenvectors are: \[ \mathrm{e}^1 = \begin{pmatrix} -1/\sqrt{2} \ 1/\sqrt{2} \ 0 \end{pmatrix}, \quad \mathrm{e}^2 = \begin{pmatrix} 1/\sqrt{2} \ 1/\sqrt{2} \ 0 \end{pmatrix}, \quad \mathrm{e}^3 = \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix}. \]

Substitute Given Values

Substitute \( \mu = 2 \) and \( \mathrm{v} = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} \) into the equation \( \mathrm{A} \mathrm{x} - \mu \mathrm{x} = \mathrm{v}. \)

System of Equations

Rewrite the given equation as: \[ \mathrm{Ax} = \mu \mathrm{x} + \mathrm{v}. \]Substituting \( \mu = 2 \): \[ \begin{pmatrix} 2 & 1 & 0 \ 1 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = 2 \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} + \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix}. \]

Solve the System

Solve the system for \( \mathrm{x}. \) The solution is: \[ \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} -1 \ 1 \ 1 \end{pmatrix}. \]

Coefficients \( \alpha_i \)

Express \( \mathrm{x} \) as a linear combination of normalized eigenvectors: \[ \mathrm{x} = \sum_{i} \alpha_i \mathrm{e}^i. \]Solving for \( \alpha_i \), we get: \[ \begin{pmatrix} -1 \ 1 \ 1 \end{pmatrix} = \alpha_1 \begin{pmatrix} -1/\sqrt{2} \ 1/\sqrt{2} \ 0 \end{pmatrix} + \alpha_2 \begin{pmatrix} 1/\sqrt{2} \ 1/\sqrt{2} \ 0 \end{pmatrix} + \alpha_3 \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix}. \] By solving, we get: \[ \alpha_1 = -1 \sqrt{2}, \alpha_2 = \sqrt{2}, \alpha_3 = 1. \]

For (b) Analysis

Analyze if the system has a solution for different \( \mu \) and \( \mathrm{v} \): \( \mu = 1 \) and (i) \( \mathrm{v} = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} \) (ii) \( \mathrm{v} = \begin{pmatrix} 2 \ 2 \ 3 \end{pmatrix}.\)

Matrix Rank and Solution - Case (i)

Check if \( \mathrm{A}- \mu \mathrm{I} = \begin{pmatrix} 1 & 1 & 0 \ 1 & 1 & 0 \ 0 & 0 & 2 \end{pmatrix} \) is invertible. Since the second row is a linear combination of the first row, the matrix is not invertible. No solution for case (i).

Matrix Rank and Solution - Case (ii)

Similarly, for case (ii), check: \( \mathrm{A}- \mu \mathrm{I} \begin{pmatrix} 1 & 1 & 0 \ 1 & 1 & 0 \ 0 & 0 & 2 \end{pmatrix}.\) Again, the matrix is not invertible since the second row is a linear combination of the first row. No solution for case (ii).

Key concepts

Eigenvalues

Eigenvalues are critical numbers associated with a square matrix that provide important information about the matrix's properties.
They are found by solving the characteristic equation that results from the matrix. For a given matrix \(A\), the eigenvalues \(\lambda\) are the solutions to the equation \[ \text{det}(\mathrm{A} - \lambda \mathrm{I}) = 0. \]
Each eigenvalue represents a scaling factor that doesn't change the direction of its corresponding eigenvector.
For example, in the given matrix: \[ \mathrm{A} = \begin{pmatrix} 2 & 1 & 0 \ 1 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix} \] the eigenvalues can be calculated as \[ \lambda_1 = 1, \lambda_2 = 3, \lambda_3 = 3. \] This tells us about the transformation properties of the matrix A.

Eigenvectors

Eigenvectors are vectors that do not change direction when a linear transformation is applied to them via a matrix.
Instead of changing direction, they only get scaled by their corresponding eigenvalue.
For the matrix \(A\) given in our exercise, we found normalized eigenvectors for each eigenvalue:
\[ \mathrm{e}^1 = \begin{pmatrix} -1/\sqrt{2} \ 1/\sqrt{2} \ 0 \end{pmatrix}, \mathrm{e}^2 = \begin{pmatrix} 1/\sqrt{2} \ 1/\sqrt{2} \ 0 \end{pmatrix}, \mathrm{e}^3 = \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} \]
These eigenvectors are fundamental as they form a basis for new coordinate systems, helping us to analyze and solve system equations more efficiently.

Real Symmetric Matrix

A real symmetric matrix is a matrix that is equal to its transpose.
This property makes it particularly nice to work with, especially in the context of eigenvalue decomposition.
For instance, matrix \(A\) in our problem: \[ \mathrm{A} = \begin{pmatrix} 2 & 1 & 0 \ 1 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix} \]
is symmetric because \( \mathrm{A}^T = \mathrm{A} \). Symmetric matrices have real eigenvalues and their eigenvectors can always be chosen to be orthonormal.
This special property simplifies many linear algebra problems, including the ones related to solving linear systems and finding characteristic values.

Linear System Solution

Solving a linear system involves finding the values of variables that satisfy given matrix equations.
In the problem, we are given: \[ \mathrm{A} \mathrm{x} - \mu \mathrm{x} = \mathrm{v} \]
and asked to find the solution when given specific \( \mu \) and \( \mathrm{v} \).
By rewriting the equation as \[ \mathrm{Ax} = \mu \mathrm{x} + \mathrm{v}, \] and substituting the given values: \[ \mathrm{A} = \begin{pmatrix} 2 & 1 & 0 \ 1 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix}, \mu = 2, \mathrm{v} = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix}, \]
we solve for \ x\ where: \[ \mathrm{A} \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = 2 \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} + \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix}. \]
The solution to this system is found to be: \[ \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} -1 \ 1 \ 1 \end{pmatrix}. \] We then express \ x \ as a linear combination of the normalized eigenvectors to find the coefficients \ \alpha_i \. This process breaks down a complex problem into simpler parts, making it easier to understand.