Question

In each exercise, \(\lambda\) is an eigenvalue of the given matrix \(A\). Determine an eigenvector corresponding to \(\lambda\). $$ A=\left[\begin{array}{rrr} 1 & -7 & 3 \\ -1 & -1 & 1 \\ 4 & -4 & 0 \end{array}\right], \quad \lambda=-4 $$

Short answer

Question: Find the eigenvector associated with the eigenvalue λ = -4 for the matrix A = \(\begin{bmatrix} 1 & -7 & 3 \\ -1 & -1 & 1 \\ 4 & -4 & 0 \end{bmatrix}\). Answer: The eigenvector corresponding to the eigenvalue λ = -4 is \(v = \begin{bmatrix} -7 \\ -5 \\ 15 \end{bmatrix}\).

Step-by-step solution

Find the matrix (A - λI)

To find the matrix \((A - \lambda I)\), we subtract the eigenvalue \(\lambda\) multiplied by the identity matrix \(I\) from the given matrix \(A\): $$ (A - \lambda I) = \left[\begin{array}{rrr} 1 & -7 & 3 \\\ -1 & -1 & 1 \\\ 4 & -4 & 0 \end{array}\right] - (-4) \left[\begin{array}{rrr} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \end{array}\right] = \left[\begin{array}{rrr} 1+4 & -7 & 3 \\\ -1 & -1+4 & 1 \\\ 4 & -4 & 0+4 \end{array}\right] $$ $$ (A - \lambda I) = \left[\begin{array}{rrr} 5 & -7 & 3 \\\ -1 & 3 & 1 \\\ 4 & -4 & 4 \end{array}\right] $$

Solve the linear system

Now, we need to solve the linear system \((A - \lambda I)v = 0\) for the vector \(v\). We can represent the system in the augmented matrix form and reduce it to row echelon form: $$ \left[\begin{array}{rrr|r} 5 & -7 & 3 & 0 \\\ -1 & 3 & 1 & 0 \\\ 4 & -4 & 4 & 0 \end{array}\right] \rightarrow \left[\begin{array}{rrr|r} 1 & -7/5 & 3/5 & 0 \\\ 0 & 1 & 1/3 & 0 \\\ 0 & 0 & 0 & 0 \end{array}\right] $$ The system of equations represented by the row echelon form is: $$ x_1 - \frac{7}{5}x_2 + \frac{3}{5}x_3 = 0 \\ x_2 + \frac{1}{3}x_3 = 0 $$ Now, let \(x_3 = t\), where t is a free parameter. The second equation gives us: $$ x_2 = -\frac{1}{3}t $$ Substituting this into the first equation, we get: $$ x_1 - \frac{7}{5}(-\frac{1}{3}t) + \frac{3}{5}t = 0 \\ x_1 + \frac{7}{15}t = 0 $$ So, \(x_1 = -\frac{7}{15}t\). This gives us the eigenvector \(v\) in the form: $$ v = \left[\begin{array}{r} -\frac{7}{15}t \\\ -\frac{1}{3}t\\\ t \end{array}\right] = t \left[\begin{array}{r} -\frac{7}{15}\\\ -\frac{1}{3}\\\ 1 \end{array}\right] $$ Since eigenvectors can be multiplied by a constant without changing their significance, we can choose any non-zero value of \(t\). Here, we choose \(t = 15\) to avoid fractions: $$ v = 15 \left[\begin{array}{r} -\frac{7}{15}\\\ -\frac{1}{3}\\\ 1 \end{array}\right] = \left[\begin{array}{r} -7\\\ -5\\\ 15 \end{array}\right] $$ So, the eigenvector corresponding to the eigenvalue \(\lambda = -4\) is: $$ v = \left[\begin{array}{r} -7 \\\ -5 \\\ 15 \end{array}\right] $$

Key concepts

Eigenvalues

Eigenvalues are special numbers associated with a square matrix. They tell us about the scaling factor of eigenvectors when a matrix acts on them. In mathematical terms, if a matrix \( A \) has an eigenvalue \( \lambda \), it means there exists a non-zero vector \( v \) such that

• \( Av = \lambda v \)

This equation shows that the action of \( A \) on \( v \) results in scaling the vector by \( \lambda \).
Finding eigenvalues typically involves solving the characteristic equation, which is derived from the determinant:

• \( \det(A - \lambda I) = 0 \)

Here, \( I \) is the identity matrix of the same size as \( A \). Eigenvalues play a crucial role in many applications, such as stability analysis and quantum mechanics.

Matrix Algebra

Matrix algebra involves operations with matrices, including addition, multiplication, and finding inverses. When dealing with eigenvalues and eigenvectors, we often manipulate matrices to simplify calculations.
Some key concepts include:

• Matrix Addition: Add corresponding elements of matrices.
• Matrix Multiplication: Multiply rows by columns (the number of columns in the first matrix must match the number of rows in the second).
• Identity Matrix: A matrix with ones on the diagonal and zeroes elsewhere, often denoted by \( I \).

In the exercise, we used matrix subtraction to find \( A - \lambda I \), an essential part of determining eigenvectors.
Understanding these basics is crucial for solving linear systems and exploring deeper mathematical theories.

Linear Systems

Solving linear systems involves finding solutions for variables within a set of linear equations. In the context of eigenvalues and eigenvectors, we express the problem \( (A - \lambda I)v = 0 \). This linear system represents a homogeneous equation since it equals zero.
To find eigenvectors, we solve this system, often involving:

• Formulating the system as an augmented matrix
• Applying row operations to simplify the system

A solution exists when variables align in a way that satisfies all the system's equations. For eigenvectors, we often find a free parameter, resulting in infinitely many solutions, or scaling factors.

Row Echelon Form

Row Echelon Form (REF) is a simplified version of a matrix that makes solving linear systems straightforward. It involves arranging the matrix so that each leading (non-zero) entry is to the right of the leading entry in the previous row, and zero rows are at the bottom.
Here's how we achieve REF:

• Pivoting: Use leading entries (or pivots) to eliminate elements in lower rows.
• Row Swap: Adjust row order for optimal zero placement.
• Row Multiplication: Multiply or divide rows by non-zero scalars.

In the exercise, we transformed the matrix into REF to make the relationship between variables clearer. This helped isolate free variables and solve for the eigenvector. Understanding REF is foundational for deep insights into matrix solutions and linear algebra techniques.