Question

2 0 -21 3. Let A= 1 3 2 LO 0 3 (a) Find the characteristic equation of A. in Find the other (b) One of the eigenvalues for A is ) = 2 with corresponding eigenvector 1 10 eigenvalue and a basis for the eigenspace associated to it. (e) Find matrices S and B that diagonalize A, if possible.

Answer 1

(a)

$\\\left(2 - \lambda\right) \left(3 - \lambda\right)^{2}=0\\\lambda^3-8\lambda^2+21\lambda-18=0$

(b)

Eigenvalue: 3, eigenvectors:

$\left[ \begin{array}{c} 0 \\\\ 1 \\\\ 0 \end{array} \right],~~\left[ \begin{array}{c} -2 \\\\ 0 \\\\ 1 \end{array} \right]$

(c)

$A=SBS^{-1}$

$S=\left[\begin{matrix}-1 & 0 & -2\\1 & 1 & 0\\0 & 0 & 1\end{matrix}\right]$

$B=\left[\begin{matrix}2 & 0 & 0\\0 & 3 & 0\\0 & 0 & 3\end{matrix}\right]$

Solution:

[2 0 -2] A = 132 [0 0 3 | Start from forming a new matrix by subtracting from the diagonal entries of the given matrix: [2 - 0 -2 13 - 22 [ 0 0 3 - | Find the determinant of the obtained matrix: | 2 - 0 1 3 2 |= (2 - 9 (3 - 2 | 0 0 3 - | This is a characteristic polynomial. Solve the equation (2 - 2) (3 - 2) = 0. The roots are: = 2 23 = 3 These are the eigenvalues. Next, find the eigenvectors. a. ) = 2 2 - 0 | 13 - [ 0 0 -2 ] [0 0 -27 2 = 1 1 2 3 - A] [0 0 1 | Perform row operations to obtain the rref of the matrix: 0 -2] [1 1 0] | 1 1 2 | | 0 0 1 [0 0 1 [ [1 1 0] [1 Now, solve the matrix equation | 0 0 1 Lo o o] [v3 If we take 2 = t, then 01 =-t, 02 = t, 03 = 0. | Therefore, y = |t |= ( 1 | t b. [2 - 1 0 3 - -2 ] -1 2 | =| 1 0 2 [ 0 0 [ 0 0 0 | Perform row operations to obtain the rref of the matrix: -1 0 -2] [1 0 2] 1 0 2 | | 0 0 0 [0 0 0 0 0 0 [10] [1] Now, solve the matrix equation | 0 0 0 || 2 | = || [0 0 0] [0] [0] If we take 12 = 5, 03 = t, then 01 =-2t, 02 = 5, 03 =t. T-2t] -2 Therefore, v= = | 1 | s+| 0 | t Eigenvalue: 2, eigenvector: | 1 Eigenvalue: 3, eigenvectors: 1 , 0

Form the matrix S, whose i-th column is the i-th eigenvector:

$S=\left[\begin{matrix}-1 & 0 & -2\\1 & 1 & 0\\0 & 0 & 1\end{matrix}\right]$

Form the diagonal matrix B, whose element at row i, column i is i-th eigenvalue:

$B=\left[\begin{matrix}2 & 0 & 0\\0 & 3 & 0\\0 & 0 & 3\end{matrix}\right]$