Question

Find all the characteristic values and vectors of the matrix. $$ \left(\begin{array}{lll} 1 & 1 & -1 \\ 2 & 3 & -4 \\ 4 & 1 & -4 \end{array}\right) $$

Short answer

In conclusion, the eigenvalues of the given matrix are -1, 2, and 3. The corresponding eigenvectors are given by the general forms k\(\left(\begin{array}{l} -1 \\\ 1 \\\ 1 \end{array}\right),\) k\(\left(\begin{array}{l} 3 \\\ 2 \\\ 1 \end{array}\right),\) and k\(\left(\begin{array}{l} 2 \\\ 3 \\\ 1 \end{array}\right)\) respectively, where k is any scalar.

Step-by-step solution

Find the characteristic equation

We are given the matrix A: \( A = \left(\begin{array}{lll} 1 & 1 & -1 \\\ 2 & 3 & -4 \\\ 4 & 1 & -4 \end{array}\right) \) We need to find the determinant of (A - λI), where λ is an eigenvalue and I is the identity matrix: \( A - \lambda I = \left(\begin{array}{lll} 1-\lambda & 1 & -1 \\\ 2 & 3-\lambda & -4 \\\ 4 & 1 & -4-\lambda \end{array}\right) \) Now, we calculate the determinant: \( \det{(A - \lambda I)} = (1-\lambda)((3-\lambda)(-4-\lambda) - (-4)(1)) - (1)(2(-4-\lambda) - (-4)(4)) - (-1)(2(1) - (3-\lambda)(4)) \)

Solve the characteristic equation

We will now simplify the determinant and find the roots (eigenvalues) of the characteristic equation: \( \det{(A - \lambda I)} = -\lambda^{3} + 8\lambda^{2}+\lambda - 12 \) By solving this cubic equation, we get: \( \lambda_{1} = -1, \lambda_{2} = 2, \lambda_{3} = 3 \) These are the eigenvalues of the given matrix.

Find the eigenvectors

For each eigenvalue, we will compute the eigenvectors by solving (A - λI)X = 0. a) For λ₁ = -1, we have: \( (A - (-1)I)X = \left(\begin{array}{lll} 2 & 1 & -1 \\\ 2 & 4 & -4 \\\ 4 & 1 & -3 \end{array}\right) X = 0 \) Row reduce to echelon form: \( \left(\begin{array}{lll} 1 & 0 & 1 \\\ 0 & 1 & -1 \\\ 0 & 0 & 0 \end{array}\right) X = 0 \) Thus, the eigenvectors associated with λ₁ = -1 have the general form: X = k\(\left(\begin{array}{l} -1 \\\ 1 \\\ 1 \end{array}\right)\) b) For λ₂ = 2, we have: \( (A - 2I)X = \left(\begin{array}{lll} -1 & 1 & -1 \\\ 2 & 1 & -4 \\\ 4 & 1 & -6 \end{array}\right) X = 0 \) Row reduce to echelon form: \( \left(\begin{array}{lll} 1 & -1 & 1 \\\ 0 & 1 & -2 \\\ 0 & 0 & 0 \end{array}\right) X = 0 \) Thus, the eigenvectors associated with λ₂ = 2 have the general form: X = k\(\left(\begin{array}{l} 3 \\\ 2 \\\ 1 \end{array}\right)\) c) For λ₃ = 3, we have: \( (A - 3I)X = \left(\begin{array}{lll} -2 & 1 & -1 \\\ 2 & 0 & -4 \\\ 4 & 1 & -7 \end{array}\right) X = 0 \) Row reduce to echelon form: \( \left(\begin{array}{lll} 1 & 0 & -2 \\\ 0 & 1 & -3 \\\ 0 & 0 & 0 \end{array}\right) X = 0 \) Thus, the eigenvectors associated with λ₃ = 3 have the general form: X = k\(\left(\begin{array}{l} 2 \\\ 3 \\\ 1 \end{array}\right)\) In conclusion, the eigenvalues are -1, 2, and 3, with corresponding eigenvectors given by the general forms k\(\left(\begin{array}{l} -1 \\\ 1 \\\ 1 \end{array}\right),\) k\(\left(\begin{array}{l} 3 \\\ 2 \\\ 1 \end{array}\right),\) and k\(\left(\begin{array}{l} 2 \\\ 3 \\\ 1 \end{array}\right)\) respectively, where k is any scalar.

Key concepts

Characteristic Equation

Finding the characteristic equation is the first step to determining the eigenvalues of a matrix. It involves the expression obtained after the determinant of the matrix \(A\) minus \(\lambda\) times the identity matrix \(I\) is set to zero. Here, \(\lambda\) represents the eigenvalues we are looking to find. The characteristic equation connects these eigenvalues directly to the matrix itself.

• The expression \(A - \lambda I\) is crucial as it modifies our matrix by subtracting a scalar transformation.
• The determinant of this new matrix format \([\det(A - \lambda I)]\) gives us the characteristic polynomial, which is solved to find the roots, and thereby, the eigenvalues.

In the given exercise, simplifying \(\det(A - \lambda I)\) results in a cubic equation i.e. \(-\lambda^{3} + 8\lambda^{2}+\lambda - 12 = 0\), from which eigenvalues can be extracted with correct calculation.

Matrix Determinant

The determinant is a specific value that can be computed from the elements of a square matrix. It offers information about the matrix properties, such as whether the matrix is invertible or not (a nonzero determinant indicates an invertible matrix).

• In its basic form, finding the determinant involves arithmetic operations across all elements of the matrix.
• For a 3x3 matrix, this involves breaking it down into smaller determinants, processing row by row.

In our context, to find the determinant of \(A - \lambda I\), we perform operations that reduce these into more manageable sub-expressions, allowing us to solve the characteristic equation.

Row Echelon Form

Row echelon form simplifies a matrix into a form that is easier to work with, especially for solving systems of linear equations. A matrix is in row echelon form when:

• All non-zero rows are above any rows of all zeros.
• The leading coefficient (also called a pivot) of a non-zero row is always strictly to the right of the leading coefficient of the row above it.
• All entries in a column below a pivot are zeros.

Achieving row echelon form involves row operations such as switching, scaling, and pivoting available to make matrix handling simpler. In the calculation of eigenvectors, converting a matrix to row echelon form allows for clear identification of vector dependencies, and isolated solutions, which are instrumental in outlining general forms for eigenvectors.

Cubic Equation Solutions

Solving a cubic equation, like the one derived from our characteristic polynomial, involves identifying and calculating the roots (real or complex). Each solution represents a potential eigenvalue for the matrix.

• Real solutions indicate eigenvalues that describe linearly independent directions in real space which matrix transformation holds.
• Special methods like factoring or the synthetic division can expedite finding solutions without exhaustive calculation.

In the exercise, solving the cubic polynomial \(-\lambda^{3} + 8\lambda^{2}+\lambda - 12 = 0\) gives us eigenvalues \(-1, 2,\) and \(3\). These eigenvalues are crucial inputs used for finding corresponding eigenvectors.

Linear Algebra Concepts

Understanding eigenvalues and eigenvectors sits at the heart of several linear algebra applications. They are crucial for simplifying matrix operations and solving linear systems, leading to applications in physics, engineering, data science, and more.

• Eigenvalues represent scales of transformation and show how much a space gets expanded or contracted along directions of eigenvectors.
• Eigenvectors, sharing a direct relationship with eigenvalues, define directions which are invariant under transformation by the matrix.

The process of determining eigenvalues and eigenvectors from a matrix involves various linear algebra techniques, each reinforcing concepts like matrix transformations, determinants, and solutions to polynomial roots, culminating in a comprehensive approach for efficient computation.