Answer :
To solve this problem, we need to find the characteristic equation, the eigenvalues, and a basis for each of the corresponding eigenspaces of the given matrix:
[tex]\[ \mathbf{A} = \begin{bmatrix} 0 & -3 & 5 \\ -4 & 4 & -10 \\ 0 & 0 & 4 \end{bmatrix} \][/tex]
Let's go through the steps in detail:
(a) Finding the Characteristic Equation:
The characteristic equation of a matrix is found by solving the determinant of [tex]\(\mathbf{A} - \lambda \mathbf{I} = 0\)[/tex], where [tex]\(\mathbf{I}\)[/tex] is the identity matrix and [tex]\(\lambda\)[/tex] represents the eigenvalues. For our given matrix [tex]\(\mathbf{A}\)[/tex],
[tex]\[ \mathbf{A} - \lambda \mathbf{I} = \begin{bmatrix} 0 - \lambda & -3 & 5 \\ -4 & 4 - \lambda & -10 \\ 0 & 0 & 4 - \lambda \end{bmatrix} \][/tex]
To find the determinant, we can do cofactor expansion:
[tex]\[ \text{det}(\mathbf{A} - \lambda \mathbf{I}) = \begin{vmatrix} -\lambda & -3 & 5 \\ -4 & 4 - \lambda & -10 \\ 0 & 0 & 4 - \lambda \end{vmatrix} \][/tex]
Since the last row has only one non-zero element, the calculation simplifies:
[tex]\[ \text{det}(\mathbf{A} - \lambda \mathbf{I}) = (4 - \lambda) \begin{vmatrix} -\lambda & -3 \\ -4 & 4 - \lambda \end{vmatrix} \][/tex]
We now find this 2x2 determinant:
[tex]\[ \begin{vmatrix} -\lambda & -3 \\ -4 & 4 - \lambda \end{vmatrix} = \lambda(4 - \lambda) - (-3)(-4) = \lambda(4 - \lambda) - 12 \][/tex]
[tex]\[ = 4\lambda - \lambda^2 - 12 \][/tex]
Thus, the determinant becomes:
[tex]\[ (4 - \lambda)(4\lambda - \lambda^2 - 12) \][/tex]
Expanding and simplifying, we obtain the characteristic polynomial:
[tex]\[ -\lambda^3 + 8\lambda^2 - 4\lambda + 48 = 0 \][/tex]
Thus, the characteristic equation is:
[tex]\[ \boxed{\lambda^3 - (-8.0)\lambda^2 + 4.0\lambda - 48.0 = 0} \][/tex]
(b) Finding the Eigenvalues and Bases for Eigenspaces:
The roots of the characteristic equation [tex]\(\lambda^3 - (-8.0)\lambda^2 + 4.0\lambda - 48.0 = 0\)[/tex] give us the eigenvalues.
The eigenvalues are:
[tex]\[ \boxed{\lambda_1 = -2.0, \lambda_2 = 4.0, \lambda_3 = 6.0} \][/tex]
To find a basis for each eigenspace, we need to solve [tex]\((\mathbf{A} - \lambda_i \mathbf{I})\mathbf{x} = 0\)[/tex] for each eigenvalue [tex]\(\lambda_i\)[/tex].
For [tex]\(\lambda_1 = -2.0\)[/tex]:
[tex]\[ x_1 = \begin{bmatrix} -0.83205029 \\ -0.5547002 \\ 0 \end{bmatrix} \][/tex]
For [tex]\(\lambda_2 = 4.0\)[/tex]:
[tex]\[ x_2 = \begin{bmatrix} -0.44022545 \\ 0.88045091 \\ 0.17609018 \end{bmatrix} \][/tex]
For [tex]\(\lambda_3 = 6.0\)[/tex]:
[tex]\[ x_3 = \begin{bmatrix} 0.4472136 \\ -0.89442719 \\ 0 \end{bmatrix} \][/tex]
Thus, the bases for the eigenspaces associated with the eigenvalues [tex]\(\lambda_1, \lambda_2,\)[/tex] and [tex]\(\lambda_3\)[/tex] are:
[tex]\[ \boxed{ \begin{aligned} x_1 &= \begin{bmatrix} -0.83205029 \\ -0.5547002 \\ 0 \end{bmatrix},\\ x_2 &= \begin{bmatrix} -0.44022545 \\ 0.88045091 \\ 0.17609018 \end{bmatrix},\\ x_3 &= \begin{bmatrix} 0.4472136 \\ -0.89442719 \\ 0 \end{bmatrix} \end{aligned} } \][/tex]
[tex]\[ \mathbf{A} = \begin{bmatrix} 0 & -3 & 5 \\ -4 & 4 & -10 \\ 0 & 0 & 4 \end{bmatrix} \][/tex]
Let's go through the steps in detail:
(a) Finding the Characteristic Equation:
The characteristic equation of a matrix is found by solving the determinant of [tex]\(\mathbf{A} - \lambda \mathbf{I} = 0\)[/tex], where [tex]\(\mathbf{I}\)[/tex] is the identity matrix and [tex]\(\lambda\)[/tex] represents the eigenvalues. For our given matrix [tex]\(\mathbf{A}\)[/tex],
[tex]\[ \mathbf{A} - \lambda \mathbf{I} = \begin{bmatrix} 0 - \lambda & -3 & 5 \\ -4 & 4 - \lambda & -10 \\ 0 & 0 & 4 - \lambda \end{bmatrix} \][/tex]
To find the determinant, we can do cofactor expansion:
[tex]\[ \text{det}(\mathbf{A} - \lambda \mathbf{I}) = \begin{vmatrix} -\lambda & -3 & 5 \\ -4 & 4 - \lambda & -10 \\ 0 & 0 & 4 - \lambda \end{vmatrix} \][/tex]
Since the last row has only one non-zero element, the calculation simplifies:
[tex]\[ \text{det}(\mathbf{A} - \lambda \mathbf{I}) = (4 - \lambda) \begin{vmatrix} -\lambda & -3 \\ -4 & 4 - \lambda \end{vmatrix} \][/tex]
We now find this 2x2 determinant:
[tex]\[ \begin{vmatrix} -\lambda & -3 \\ -4 & 4 - \lambda \end{vmatrix} = \lambda(4 - \lambda) - (-3)(-4) = \lambda(4 - \lambda) - 12 \][/tex]
[tex]\[ = 4\lambda - \lambda^2 - 12 \][/tex]
Thus, the determinant becomes:
[tex]\[ (4 - \lambda)(4\lambda - \lambda^2 - 12) \][/tex]
Expanding and simplifying, we obtain the characteristic polynomial:
[tex]\[ -\lambda^3 + 8\lambda^2 - 4\lambda + 48 = 0 \][/tex]
Thus, the characteristic equation is:
[tex]\[ \boxed{\lambda^3 - (-8.0)\lambda^2 + 4.0\lambda - 48.0 = 0} \][/tex]
(b) Finding the Eigenvalues and Bases for Eigenspaces:
The roots of the characteristic equation [tex]\(\lambda^3 - (-8.0)\lambda^2 + 4.0\lambda - 48.0 = 0\)[/tex] give us the eigenvalues.
The eigenvalues are:
[tex]\[ \boxed{\lambda_1 = -2.0, \lambda_2 = 4.0, \lambda_3 = 6.0} \][/tex]
To find a basis for each eigenspace, we need to solve [tex]\((\mathbf{A} - \lambda_i \mathbf{I})\mathbf{x} = 0\)[/tex] for each eigenvalue [tex]\(\lambda_i\)[/tex].
For [tex]\(\lambda_1 = -2.0\)[/tex]:
[tex]\[ x_1 = \begin{bmatrix} -0.83205029 \\ -0.5547002 \\ 0 \end{bmatrix} \][/tex]
For [tex]\(\lambda_2 = 4.0\)[/tex]:
[tex]\[ x_2 = \begin{bmatrix} -0.44022545 \\ 0.88045091 \\ 0.17609018 \end{bmatrix} \][/tex]
For [tex]\(\lambda_3 = 6.0\)[/tex]:
[tex]\[ x_3 = \begin{bmatrix} 0.4472136 \\ -0.89442719 \\ 0 \end{bmatrix} \][/tex]
Thus, the bases for the eigenspaces associated with the eigenvalues [tex]\(\lambda_1, \lambda_2,\)[/tex] and [tex]\(\lambda_3\)[/tex] are:
[tex]\[ \boxed{ \begin{aligned} x_1 &= \begin{bmatrix} -0.83205029 \\ -0.5547002 \\ 0 \end{bmatrix},\\ x_2 &= \begin{bmatrix} -0.44022545 \\ 0.88045091 \\ 0.17609018 \end{bmatrix},\\ x_3 &= \begin{bmatrix} 0.4472136 \\ -0.89442719 \\ 0 \end{bmatrix} \end{aligned} } \][/tex]
