Question

For each of the matrices below, compute the characteristic polynomial, find the real eigenvalues and bases for the corresponding eigenspaces. $$ \left. \begin{array}{r} \left(\begin{array}{rr} 7 & 5 -10 & -8 \end{array}\right) W \left(\begin{array}{rrr) -1 & -21 4 & 5 \end{array}\right) W \left(\begin{array}{rrr} -1 & 0& O -4 & 2 & -1 4 & 0 & 3 \end{array}\right) W \left(\begin{array}{rrr} -1 & 0 & 1 -7 & 2 & 5 3 & 0 & 1 \end{array}\right) W \left. \begin{array}{rrr} -7 & -5 1 16 & 17 \end{array}\right) \left(\begin{array}{rrr} 1 & -21 1 & 2 \end{array}\right) W 6 & 4 & -3 2 & 0 & 3 \end{array}\right) } \end{array} $$ 2. If $A$ is an $n \times n$ matrix prove that $\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ is an eigenvalue of $A^{T}$. 3. Suppose $A$ is an invertible $n \times n$ matrix and that $\lambda$ is an eigenvalue of $A$. Prove that $\lambda \\eq 0$ and that $1 / \lambda$ is an eigenvalue of $A^{-1}$.CS.SD. 119

Answer 1

To find the characteristic polynomial and eigenvalues of each matrix, determine the determinant of the matrix minus lambda times the identity matrix. The eigenspace can be found by solving the system of equations. Lambda is an eigenvalue of A if and only if it is an eigenvalue of A^T, which is proven by the characteristic polynomial. Lambda cannot be zero, and 1/lambda is an eigenvalue of A^-1.

Step-by-step explanation:

To find the characteristic polynomial and eigenvalues of each matrix, we need to find the determinant of the matrix minus lambda times the identity matrix. For the first matrix, the characteristic polynomial is (lambda-4)(lambda+1)(lambda-3). The eigenvalues are 4, -1, and 3. The eigenspace for each eigenvalue can be found by solving the system of equations (A-lambdaI)x=0, where A is the matrix and lambda is the eigenvalue.

To prove that lambda is an eigenvalue of A if and only if lambda is an eigenvalue of A^T, we can use the property that the characteristic polynomial of a matrix is the same as the characteristic polynomial of its transpose. This means that if lambda is a root of the characteristic polynomial of A, it will also be a root of the characteristic polynomial of A^T, and vice versa.

To prove that lambda is not equal to 0 and 1/lambda is an eigenvalue of A^-1, we can use the fact that the determinant of A is the product of its eigenvalues. Since A is invertible, its determinant is non-zero, which means none of its eigenvalues can be zero. And since A^-1 has the reciprocal of the eigenvalues as its eigenvalues, 1/lambda will be an eigenvalue of A^-1.