Problem 25 - Advanced Engineering Mathematics

Question

In Problems, find the eigenvalues and eigenvectors of the given nonsingular matrix \(\mathbf{A} .\) Then without finding \(\mathbf{A}^{-1}\), find its eigenvalues and corresponding eigenvectors. $$ \mathbf{A}=\left(\begin{array}{rrr} 4 & 2 & -1 \\ 0 & 3 & -2 \\ 0 & 0 & 5 \end{array}\right) $$

Step-by-Step Solution

Answer

Eigenvalues of \( \mathbf{A} \) are 4, 3, 5. Eigenvalues of \( \mathbf{A}^{-1} \) are \( \frac{1}{4} \), \( \frac{1}{3} \), \( \frac{1}{5} \). Corresponding eigenvectors are the same for both matrices.

Step 1: Find the Characteristic Polynomial

To find the eigenvalues, we first need to determine the characteristic polynomial of \( \mathbf{A} \). The characteristic polynomial is given by \( \det(\mathbf{A} - \lambda \mathbf{I}) \). The matrix \( \mathbf{A} - \lambda \mathbf{I} \) is:\[ \begin{pmatrix} 4 - \lambda & 2 & -1 \ 0 & 3 - \lambda & -2 \ 0 & 0 & 5 - \lambda \end{pmatrix} \]Calculate the determinant:\[ \det(\mathbf{A} - \lambda \mathbf{I}) = (4 - \lambda)((3-\lambda)(5-\lambda) - (0 \cdot 0)) \]\[ = (4-\lambda)((3-\lambda)(5-\lambda)) \]\[ = (4-\lambda)(15 - 8\lambda + \lambda^2) \]Simplifying gives the characteristic polynomial.

Step 2: Solve the Characteristic Equation

Set \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \) to find the eigenvalues:\[ (4-\lambda)((3-\lambda)(5-\lambda)) = 0 \]This equation clearly has the roots \( \lambda = 4, 3, 5 \). Thus, the eigenvalues are \( \lambda_1 = 4 \), \( \lambda_2 = 3 \), and \( \lambda_3 = 5 \).

Step 3: Find Eigenvectors for Each Eigenvalue

To find the eigenvectors, substitute each eigenvalue back into \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = \mathbf{0} \):1. For \( \lambda_1 = 4 \):\[ \begin{pmatrix} 0 & 2 & -1 \ 0 & -1 & -2 \ 0 & 0 & 1 \end{pmatrix} \mathbf{v} = \mathbf{0} \]This implies \( v_2 = \frac{1}{2}v_3 \). Choose \( v_3=1 \), then \( v_2 = \frac{1}{2} \). Thus, an eigenvector is \( \mathbf{v} = \begin{pmatrix} 0 \ 0.5 \ 1 \end{pmatrix} \).2. For \( \lambda_2 = 3 \):\[ \begin{pmatrix} 1 & 2 & -1 \ 0 & 0 & -2 \ 0 & 0 & 2 \end{pmatrix} \mathbf{v} = \mathbf{0} \]This leads to \( v_3 = 0 \) and \( v_1 = -2v_2 \). Choose \( v_2 = 1 \), hence \( v_1 = -2 \). The eigenvector is \( \mathbf{v} = \begin{pmatrix} -2 \ 1 \ 0 \end{pmatrix} \).3. For \( \lambda_3 = 5 \):\[ \begin{pmatrix} -1 & 2 & -1 \ 0 & -2 & -2 \ 0 & 0 & 0 \end{pmatrix} \mathbf{v} = \mathbf{0} \]This implies \( v_2 = -v_3 \) and \( v_1 = 2v_3 - v_2 \). Choose \( v_3 = 1 \), hence \( v_2 = -1 \) and \( v_1=3 \). The eigenvector is \( \mathbf{v} = \begin{pmatrix} 3 \ -1 \ 1 \end{pmatrix} \).

Step 4: Finding Eigenvalues of \( \mathbf{A}^{-1} \)

The eigenvalues of \( \mathbf{A}^{-1} \) are the reciprocals of those of \( \mathbf{A} \) since it is nonsingular. Hence, the eigenvalues of \( \mathbf{A}^{-1} \) are \( \lambda_1 = \frac{1}{4} \), \( \lambda_2 = \frac{1}{3} \), and \( \lambda_3 = \frac{1}{5} \).

Step 5: Eigenvectors of \( \mathbf{A}^{-1} \)

Since \( \mathbf{A} \) is nonsingular and its eigenvectors are linearly independent, the eigenvectors of \( \mathbf{A}^{-1} \) are the same as those of \( \mathbf{A} \). Therefore, the corresponding eigenvectors of \( \mathbf{A}^{-1} \) are \( \begin{pmatrix} 0 \ 0.5 \ 1 \end{pmatrix} \), \( \begin{pmatrix} -2 \ 1 \ 0 \end{pmatrix} \), and \( \begin{pmatrix} 3 \ -1 \ 1 \end{pmatrix} \).

Key Concepts

Characteristic Polynomial

To find the eigenvalues of a matrix, we begin with the concept of the characteristic polynomial. The characteristic polynomial is essentially a polynomial whose roots are the eigenvalues of the matrix. To find this polynomial for a matrix \( \mathbf{A} \), we use the formula \( \det(\mathbf{A} - \lambda \mathbf{I}) \). Here, \( \lambda \) represents an unknown scalar, and \( \mathbf{I} \) is the identity matrix.

• The term \( \mathbf{A} - \lambda \mathbf{I} \) modifies \( \mathbf{A} \) by subtracting \( \lambda \) from each diagonal entry of \( \mathbf{A} \).
• By computing the determinant of this new matrix, \( \det(\mathbf{A} - \lambda \mathbf{I}) \), we arrive at a polynomial in \( \lambda \).
• The roots of this polynomial, obtained by setting \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), are the eigenvalues we are looking for.

The process of finding these roots involves standard polynomial solving techniques and gives us crucial insights into the matrix's behavior.

Nonsingular Matrix

A matrix is called nonsingular if it has an inverse, meaning there is a matrix \( \mathbf{A}^{-1} \) such that \( \mathbf{A} \mathbf{A}^{-1} = \mathbf{I} \), the identity matrix. One important feature of a nonsingular matrix is that its determinant is not zero.

• The nonsingularity of a matrix implies that it does not "squash" its eigenvectors upon transformation and retains full rank, equal to the number of rows or columns.
• Nonsingular matrices also imply that their eigenvalues are all non-zero. This attribute arises because, if any eigenvalue were zero, it would mean \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), resulting in a zero determinant and thus making \( \mathbf{A} \) singular.

Understanding nonsingular matrices helps us in various applications, including solving systems of linear equations and determining properties about eigenvectors and eigenvalues.

Inverse Matrix

The inverse of a matrix is a matrix that, when multiplied with the original matrix, yields the identity matrix. For a given nonsingular matrix \( \mathbf{A} \), its inverse \( \mathbf{A}^{-1} \) can be thought of as "undoing" the transformation \( \mathbf{A} \) imposes. In the context of eigenvalues, there are interesting relationships:

• The eigenvalues of \( \mathbf{A}^{-1} \) are simply the reciprocals of the eigenvalues of \( \mathbf{A} \).
• This reciprocal relationship arises because if \( \mathbf{A} \mathbf{v} = \lambda \mathbf{v} \), then it follows that \( \mathbf{A}^{-1} \mathbf{v} = \frac{1}{\lambda} \mathbf{v} \).
• Since eigenvectors are not changed by inversion, the eigenvectors of \( \mathbf{A}^{-1} \) are the same as those of \( \mathbf{A} \).

This property is quite useful, particularly in computational applications where inversions can be costly but understanding the inverse's eigenstructure is crucial.

Linear Independence

Linear independence is a fundamental concept in linear algebra that plays a crucial role in understanding eigenvectors. A set of vectors is said to be linearly independent if no vector in the set can be written as a linear combination of the others. In our context:

• The eigenvectors corresponding to distinct eigenvalues of a matrix are always linearly independent.
• This independence means that the eigenvectors form a basis that can span the entire vector space if the matrix is of full rank.
• For a nonsingular matrix, the linear independence of its eigenvectors is especially useful because it assures us that we have a complete basis for transformation and inversion operations.

Understanding linear independence helps in dimensional analysis, simplifying matrix operations, and solving linear systems by constructing solution spaces effectively.