Question

A matrix \(A\) is given. For each, (a) Find the eigenvalues of \(A,\) and for each eigenvalue, find an eigenvector. (b) Do the same for \(A^{T}\). (c) Do the same for \(A^{-1}\). (d) Find \(\operatorname{tr}(A)\). (e) Find det \((A)\). Use Theorem 19 to verify your results. $$\left[\begin{array}{cc}0 & 4 \\ -1 & 5\end{array}\right]$$

Short answer

Eigenvalues: \(4, 1\), Eigenvectors: \([1, 1], [4, 1]\), \(\operatorname{tr}(A)=5\), \(\det(A)=4\).

Step-by-step solution

Find the Eigenvalues of A

The eigenvalues of a matrix \( A \) are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \). For the matrix \( A = \left[\begin{array}{cc}0 & 4 \ -1 & 5\end{array}\right] \), we set up \( A - \lambda I = \left[\begin{array}{cc}-\lambda & 4 \ -1 & 5-\lambda\end{array}\right] \) and find its determinant:\[\det(A - \lambda I) = (-\lambda)(5 - \lambda) - (-1)(4) = \lambda^2 - 5\lambda + 4.\]Solving \( \lambda^2 - 5\lambda + 4 = 0 \) using the quadratic formula yields \( \lambda_1 = 4 \) and \( \lambda_2 = 1 \).

Find the Eigenvectors of A

For each eigenvalue, substitute back into the equation \( (A - \lambda I)\mathbf{v} = 0 \) to find eigenvectors.**Eigenvector for \( \lambda_1 = 4 \):**\( (A - 4I) = \left[\begin{array}{cc}-4 & 4 \ -1 & 1\end{array}\right] \). Solve \( (A - 4I)\mathbf{v} = 0 \) which becomes the system of equations: - \(-4x + 4y = 0\) - \(-x + y = 0\)This gives \( y = x \). Thus, an eigenvector is \( \begin{bmatrix}1 \ 1\end{bmatrix} \).**Eigenvector for \( \lambda_2 = 1 \):**\( (A - I) = \left[\begin{array}{cc}-1 & 4 \ -1 & 4\end{array}\right] \). Solve for the system: - \(-x + 4y = 0\)This gives \( x = 4y \). A corresponding eigenvector is \( \begin{bmatrix}4 \ 1\end{bmatrix} \).

Find the Eigenvalues of A^T

The eigenvalues of \( A^{T} \) are the same as \( A \). Since the transpose of a matrix does not change its characteristic polynomial, the eigenvalues remain \( \lambda_1 = 4 \) and \( \lambda_2 = 1 \).

Find the Eigenvectors of A^T

Following the same steps as in Step 2 for \( A^{T} = \left[\begin{array}{cc}0 & -1 \ 4 & 5\end{array}\right] \):**Eigenvectors for \( \lambda_1 = 4 \):** \((A^T - 4I) = \left[\begin{array}{cc}-4 & -1 \ 4 & 1\end{array}\right]\), which gives \( y = x \). An eigenvector is \( \begin{bmatrix}1 \ 1\end{bmatrix} \).**Eigenvectors for \( \lambda_2 = 1 \):**\((A^T - I) = \left[\begin{array}{cc}-1 & -1 \ 4 & 4\end{array}\right]\), which reduces to \( x = 4y \). An eigenvector is \( \begin{bmatrix}4 \ 1\end{bmatrix} \).

Find the Eigenvalues of A^{-1}

The eigenvalues of \( A^{-1} \) are reciprocals of the eigenvalues of \( A \). Therefore, they are \( \frac{1}{4} \) and \( 1 \).

Find the Eigenvectors of A^{-1}

The eigenvectors of \( A \) and \( A^{-1} \) are the same for corresponding eigenvalues since \( A^{-1} \cdot A \cdot \mathbf{v} = \mathbf{v} \) implies the eigenspaces do not change. Thus, the eigenvectors are the same as those found in Step 2.

Compute the Trace of A

The trace of a matrix \( A \), denoted \( \operatorname{tr}(A) \), is the sum of its diagonal elements. For \( A = \left[\begin{array}{cc}0 & 4 \ -1 & 5\end{array}\right] \), the trace is \( 0 + 5 = 5 \). This is equal to the sum of the eigenvalues, \( 4 + 1 = 5 \), confirming our eigenvalue calculations.

Compute the Determinant of A

The determinant of \( A \) can be calculated by \( \det(A) = (0)(5) - (4)(-1) = 4 \). The determinant is also the product of the eigenvalues \( 4 \times 1 = 4 \), confirming our earlier calculations.

Key concepts

Matrix Transpose

A matrix transpose is a fundamental operation where you "flip" a matrix over its diagonal. For a given matrix \( A \) with elements \( a_{ij} \), the transpose, denoted as \( A^T \), will have its rows and columns swapped. Here's how it works:

• The element at the first row and second column in \( A \) moves to the second row, first column in \( A^T \).
• For any element \( a_{ij} \) in \( A \), the corresponding element in \( A^T \) is \( a_{ji} \).

This operation might seem simple, yet it has considerable importance in mathematics.
A key property of transposes is that the eigenvalues of a matrix and its transpose are identical. This means if \( A \) has eigenvalues \( \lambda_1 \) and \( \lambda_2 \), then \( A^T \) will have the same eigenvalues. This was utilized in verifying eigenvalues in the exercise, illustrating the practical application of this concept.

Matrix Inverse

The matrix inverse is akin to division in the realm of matrices. For a square matrix \( A \), the inverse \( A^{-1} \) satisfies the equation \( A \cdot A^{-1} = I \), where \( I \) is the identity matrix. However, not all matrices have inverses.

• A matrix has an inverse only if it is square and its determinant is non-zero.
• Inverting a matrix involves complex calculations, but for a 2x2 matrix, there's a simplified formula.

Finding the eigenvalues of \( A^{-1} \) in the exercise involved taking the reciprocal of the original eigenvalues of \( A \). This property highlights that the eigenvalues of \( A^{-1} \) reflect the reciprocal relationship with those of \( A \), showcasing how inverses impact matrix properties.

Trace of a Matrix

The trace of a matrix is a simple yet informative concept. For a matrix \( A \), the trace, denoted as \( \operatorname{tr}(A) \), is the sum of the diagonal elements. In mathematical terms, if \( A \) is a matrix with elements \( a_{ii} \), then:

• \( \operatorname{tr}(A) = a_{11} + a_{22} + \ldots + a_{nn} \)

The trace offers insights into many mathematical properties of a matrix.
In eigenvalue contexts, the trace equals the sum of the eigenvalues. This relationship was used in the exercise to confirm the eigenvalues found, as the trace of matrix \( A \) was given as 5, equaling the sum of the eigenvalues \( 4 + 1 \), verifying the calculations. Understanding the trace provides a quick way to check the consistency of eigenvalue computations.

Determinant of a Matrix

The determinant is a special number that provides essential information about a matrix. For a matrix \( A \), this scalar value is denoted as \( \det(A) \). It tells us about the invertibility of the matrix and encapsulates its scaling factor.

• If the determinant is zero, the matrix isn't invertible (i.e., it's singular).
• The determinant of a 2x2 matrix \( \left[ \begin{array}{cc} a & b \ c & d \end{array} \right] \) is calculated as \( ad - bc \).
• Determinants have a relation with eigenvalues; the determinant of a matrix is the product of its eigenvalues.

In the exercise, the determinant of matrix \( A \) was calculated to be 4, corroborating the eigenvalue product \( 4 \times 1 \). The determinant thus confirms not only the matrix's invertibility but also serves as a check for eigenvalues, highlighting its multifaceted utility.