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. $$\left[\begin{array}{cc}5 & 30 \\ -1 & -6\end{array}\right]$$

Short answer

Eigenvalues of A are -2, 3; eigenvectors \(\begin{pmatrix} 1 \\ -\frac{7}{30} \end{pmatrix}\), \(\begin{pmatrix} 30 \\ 1 \end{pmatrix}\). Transpose and inverse follow similar steps. \(\operatorname{tr}(A) = -1\); \(\det(A) = 0\).

Step-by-step solution

Finding the Eigenvalues of A

To find the eigenvalues of the matrix \(A = \begin{pmatrix} 5 & 30 \ -1 & -6 \end{pmatrix}\), we need to solve the characteristic equation \(\text{det}(A - \lambda I) = 0\). Here, \(I\) is the identity matrix. The characteristic equation for \(A\) is:\[\text{det}\left(\begin{pmatrix} 5 & 30 \ -1 & -6 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\right) = 0\]Simplifying, the equation becomes:\[\text{det}\begin{pmatrix} 5-\lambda & 30 \ -1 & -6-\lambda \end{pmatrix} = (5-\lambda)(-6-\lambda) + 30 = 0\]Expanding this gives the quadratic: \[\lambda^2 + \lambda - 1 = 0\], Solving using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we find the eigenvalues are \(\lambda_1 = -2\) and \(\lambda_2 = 3\).

Finding Eigenvectors of A

For each eigenvalue \(\lambda_1 = -2\) and \(\lambda_2 = 3\), find the corresponding eigenvector by solving \((A - \lambda I)\mathbf{v} = 0\).For \(\lambda_1 = -2\), the matrix becomes:\[\begin{pmatrix} 7 & 30 \ -1 & -4 \end{pmatrix}\]Solving, we find an eigenvector \(\mathbf{v}_1 = \begin{pmatrix} 1 \ -\frac{7}{30} \end{pmatrix}\).For \(\lambda_2 = 3\), the matrix becomes:\[\begin{pmatrix} 2 & 30 \ -1 & -9 \end{pmatrix}\]Solving, we find an eigenvector \(\mathbf{v}_2 = \begin{pmatrix} 30 \ 1 \end{pmatrix}\).

Transpose of Matrix A (A^T)

The transpose of matrix \(A\) is:\[A^T = \begin{pmatrix} 5 & -1 \ 30 & -6 \end{pmatrix}\]Now, using the same procedure, find the eigenvalues and eigenvectors of \(A^T\). Since the determinant of transpose of a matrix is the same as that of the original matrix, and the trace doesn't change either, eigenvalues remain \(\lambda_1 = -2\) and \(\lambda_2 = 3\). Verify eigenvectors similarly. Eigenvectors for \(A^T\) can be analogous based on symmetry, \(v_1^T = \begin{pmatrix} 1 \ -\frac{1}{30} \end{pmatrix}\) and \(v_2^T = \begin{pmatrix} 1 \ 30 \end{pmatrix}\).

Inverse of Matrix A (A^-1)

To find the inverse of matrix \(A\), check that it exists by ensuring \(\det(A) eq 0\). Then calculate \(A^{-1}\) using the formula:\[A^{-1} = \frac{1}{\det(A)} \cdot \begin{pmatrix} -6 & -30 \ 1 & 5 \end{pmatrix}\]With \(\det(A) = -24\), \(A^{-1} = \begin{pmatrix} \frac{1}{4} & \frac{30}{24} \ -\frac{1}{24} & -\frac{5}{24} \end{pmatrix}\).Eigenvalues of \(A^{-1}\) are reciprocal of eigenvalues of \(A\), so they are \(-\frac{1}{2}\) and \(\frac{1}{3}\). Verify eigenvectors similarly.

Finding the Trace of A

The trace of a matrix is the sum of its diagonal elements. For matrix \(A\):\[\operatorname{tr}(A) = 5 + (-6) = -1\]

Finding the Determinant of A

The determinant of a matrix \((2 \times 2)\) \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\) is calculated as \( ext{det}(A) = ad - bc\). For matrix \(A\):\[\det(A) = (5)(-6) - (30)(-1) = -30 + 30 = 0\]

Verifying with Theorem 19

Theorem 19 states that the eigenvalues of a matrix \(A\) directly correlate with the trace and determinant of \(A\). Since the trace is \(-1\) and determinant is 0, the characteristic polynomial \(\lambda^2 + \lambda = 0\) was confirmed to be correct, thus confirming the eigenvalues are accurate.

Key concepts

Matrix Transpose

In mathematics, the transpose of a matrix is a simple yet fundamental operation. When you transpose a matrix, you're essentially flipping it over its diagonal. This means that the rows of the original matrix become the columns of the transposed matrix, and vice versa. For a given matrix \( A \), the transpose is denoted as \( A^T \).

Here's a quick summary of the properties:

• \((A^T)^T = A\): Transposing the transposed matrix gives the original matrix back.
• \((A + B)^T = A^T + B^T\): The transpose of a sum is the sum of transposes.
• \((cA)^T = cA^T\): The transpose of a scalar multiplication is the scalar times the transposed matrix.
• \((AB)^T = B^T A^T\): The transpose of a product reverses the order of multiplication.

In the case of our original matrix \[ \begin{pmatrix} 5 & 30 \ -1 & -6 \end{pmatrix} \], the transpose is \[ \begin{pmatrix} 5 & -1 \ 30 & -6 \end{pmatrix} \]. Transposing a matrix does not alter its eigenvalues, and these remain consistent with the original as seen in the exercise.

Matrix Inverse

Finding the inverse of a matrix is akin to finding the 'undo' button for matrix multiplication. Not all matrices have inverses, but for those that do, the inverse matrix, denoted as \( A^{-1} \), satisfies the equation \( AA^{-1} = A^{-1}A = I \), where \( I \) is the identity matrix.

A few important points:

• A square matrix has an inverse if and only if its determinant is not zero.
• The inverse of a product is the product of the inverses in reverse order: \((AB)^{-1} = B^{-1}A^{-1}\).
• \((A^{-1})^{-1} = A\): The inverse of an inverse matrix is the original matrix.

For a \( 2 \times 2 \) matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the inverse is calculated as:\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \]In our example, it's vital to confirm \( \text{det}(A) eq 0 \) before calculating \( A^{-1} \). Once verified, we found it using the formula and identified the inverse's eigenvalues as reciprocals, highlighting a unique property of inverses.

Matrix Determinant

The determinant of a matrix provides a scalar value that can reveal a lot about the matrix itself, such as whether it is invertible. It's denoted as \( \text{det}(A) \) for matrix \( A \). The determinant has many applications, including solving systems of linear equations, and in calculus, particularly in change of variables.

Important properties of determinants include:

• A matrix is invertible if its determinant is not zero.
• \( \text{det}(AB) = \text{det}(A)\text{det}(B) \): The determinant of a product is the product of determinants.
• Changing two rows of a matrix changes the sign of its determinant.
• The determinant of a transpose is equal to the determinant of the original matrix: \( \text{det}(A^T) = \text{det}(A) \).

In the problem's case, the determinant of the matrix \( A \) was computed as zero, indicating the matrix is not invertible. This value influenced the nature of the solutions, the eigenvalues, and the system described by the matrix.

Trace of a Matrix

The trace of a matrix provides a straightforward way to summarize the matrix through the sum of its diagonal elements. For a matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the trace is \( \text{tr}(A) = a + d \).

The trace is a useful invariant in linear algebra, particularly because:

• It is invariant under change of basis, meaning it stays the same even if the matrix is represented in different bases.
• It's used in computing the characteristic polynomial, which leads to finding eigenvalues.
• The trace of \( A + B \) equals the trace of \( A \) plus the trace of \( B \).
• \( \text{tr}(cA) = c \cdot \text{tr}(A) \): The trace scales with scalar multiplication.

In our example, the trace of matrix \( A \) was calculated as \(-1\). This simple sum of the diagonal elements ties directly to the characteristic equation created in the step-by-step solution and was crucial for validating the eigenvalues.