91影视

First, we need to determine the eigenvalues of the given matrix. We will do this by solving the characteristic equation, which is given by the determinant of (A - 位I) where A is the matrix, 位 is the eigenvalue, and I is the identity matrix. Let the given matrix A be defined as: \( A = \begin{pmatrix} a & 1 & 0 \\ 0 & a & 1 \\ 0 & 0 & b \end{pmatrix} \) The determinant |A - 位I| is: \(|A - 位I|= \begin{vmatrix} a - 位 & 1 & 0 \\ 0 & a - 位 & 1 \\ 0 & 0 & b - 位 \end{vmatrix} \) Now, let's evaluate the determinant using the first row for cofactor expansion: \(|A - 位I|= (a - 位) \begin{vmatrix} a - 位 & 1 \\ 0 & b - 位 \end{vmatrix} = (a - 位)((a - 位)(b - 位) - 0) = (a - 位)^2 (b - 位)\) To find the eigenvalues, we find the roots of |A - 位I|: \(|A - 位I| = (a - 位)^2 (b - 位) = 0\) This equation has two distinct eigenvalues 位鈧� = a, and 位鈧� = b, with algebraic multiplicities of 2 and 1 respectively.

Now let's find the eigenvectors for each eigenvalue by solving (A - 位I)v = 0, where v is the eigenvector. a) 位鈧� = a (algebraic multiplicity 2) (A - aI)v = 0 \(\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & b - a \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\) Row 1: \( y = 0 \) Row 2: \( z = 0 \) Row 3: \( (b - a)z = 0\) Since we are considering the case where 位鈧� = a, we have z = 0. Therefore, the only eigenvector corresponding to 位鈧� = a is \(v_1 = \begin{pmatrix} x \\ 0 \\ 0 \end{pmatrix}\), where x is any scalar. This indicates that 位鈧� has one linearly independent eigenvector. b) 位鈧� = b (algebraic multiplicity 1) (A - bI)v = 0 \(\begin{pmatrix} a - b & 1 & 0 \\ 0 & a - b & 1 \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\) Row 1: \( (a - b)x + y = 0 \) Row 2: \( (a - b)y + z = 0 \) Row 3: \( 0 = 0\) In this case, we get one eigenvector corresponding to 位鈧� = b. Let this eigenvector be \(v_2 = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a - b \\ b - a \\ 1 \end{pmatrix}\), which is the only linearly independent eigenvector.

We found 2 distinct eigenvalues for the given 3x3 matrix. The eigenvalues 位鈧� = a has one linearly independent eigenvector and the eigenvalue 位鈧� = b has its own linearly independent eigenvector. Since the total number of linearly independent eigenvectors for the given matrix is 2, which is less than the size of the matrix (3), the given matrix is defective.