Answer :
To solve the problem, we will follow these steps:
1. Find the characteristic equation.
2. Find the eigenvalues.
3. Find a basis for each eigenspace.
### Step 1: Find the Characteristic Equation
The characteristic equation is given by the determinant of [tex]\(A - \lambda I\)[/tex] set to zero, [tex]\( \det(A - \lambda I) = 0 \)[/tex].
Given matrix [tex]\(A\)[/tex]:
[tex]\[ A = \left[\begin{array}{rrr} 0 & -3 & 5 \\ -4 & 4 & -10 \\ 0 & 0 & 4 \end{array} \right] \][/tex]
Subtract [tex]\( \lambda \)[/tex] times the identity matrix [tex]\(I\)[/tex] from [tex]\(A\)[/tex]:
[tex]\[ A - \lambda I = \left[\begin{array}{rrr} 0-\lambda & -3 & 5 \\ -4 & 4-\lambda & -10 \\ 0 & 0 & 4-\lambda \end{array} \right] \][/tex]
The characteristic polynomial is the determinant of this matrix:
[tex]\[ \det(A - \lambda I) = \det \left[\begin{array}{rrr} -\lambda & -3 & 5 \\ -4 & 4-\lambda & -10 \\ 0 & 0 & 4-\lambda \end{array} \right] \][/tex]
To find this determinant, we expand along the third row:
[tex]\[ \det(A - \lambda I) = (-\lambda) \left[ (4-\lambda) \times (4-\lambda) - (-10) \times 0 \right] - (-3) \left[ -4 \times (4-\lambda) - (-10) \times 0 \right] + 5(0) \][/tex]
Simplify the determinant expression:
[tex]\[ \det(A - \lambda I) = (-\lambda) ( (4-\lambda)(4-\lambda) - 0 ) + 3 ( 4(4-\lambda) ) \][/tex]
[tex]\[ \det(A - \lambda I) = (-\lambda) ( (4-\lambda)^2 ) + 12(4-\lambda) \][/tex]
[tex]\[ \det(A - \lambda I) = -\lambda ( 16 - 8\lambda + \lambda^2 ) + 48 - 12\lambda \][/tex]
[tex]\[ \det(A - \lambda I) = -16\lambda + 8\lambda^2 - \lambda^3 + 48 - 12\lambda \][/tex]
Combine like terms:
[tex]\[ \det(A - \lambda I) = -\lambda^3 + 8\lambda^2 - 28\lambda + 48 \][/tex]
Set this equal to zero to obtain the characteristic equation:
[tex]\[ -\lambda^3 + 8\lambda^2 - 28\lambda + 48 = 0 \][/tex]
### Step 2: Find the Eigenvalues
Solve the characteristic equation for [tex]\( \lambda \)[/tex]:
[tex]\[ -\lambda^3 + 8\lambda^2 - 28\lambda + 48 = 0 \][/tex]
Synthetic division or an algebra solver can help here. By testing possible rational roots, we find that [tex]\( \lambda = 4 \)[/tex], [tex]\( \lambda = 2 \)[/tex], and [tex]\( \lambda = -6 \)[/tex].
### Eigenvalues
[tex]\[ \lambda_1 = -6, \lambda_2 = 2, \lambda_3 = 4 \][/tex]
### Step 3: Find Eigenspaces
For each eigenvalue, we solve [tex]\( (A - \lambda I) \mathbf{x} = 0 \)[/tex] to find a basis for each corresponding eigenspace.
Eigenvalue [tex]\(\lambda = -6\)[/tex]:
[tex]\[ A - (-6)I = \left[\begin{array}{rrr} 6 & -3 & 5 \\ -4 & 10 & -10 \\ 0 & 0 & 10 \end{array} \right] \][/tex]
Row reduce this matrix to obtain:
[tex]\[ \left[\begin{array}{rrr} 1 & -0.5 & 0.8333 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{array} \right] \][/tex]
This simplifies to:
[tex]\[ x_1 = t \left[ \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \right] \][/tex]
Eigenvalue [tex]\(\lambda = 2\)[/tex]:
[tex]\[ A - 2I = \left[\begin{array}{rrr} -2 & -3 & 5 \\ -4 & 2 & -10 \\ 0 & 0 & 2 \end{array} \right] \][/tex]
Row reduce this matrix:
[tex]\[ \left[\begin{array}{rrr} 1 & 3/2 & -5/2 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{array} \right] \][/tex]
This simplifies to:
[tex]\[ x_2 = s \left[ \begin{array}{r} -1 \\ 1.5 \\ 0 \end{array} \right] + t \left[ \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \right] \][/tex]
Eigenvalue [tex]\(\lambda = 4\)[/tex]:
[tex]\[ A - 4I = \left[\begin{array}{rrr} -4 & -3 & 5 \\ -4 & 0 & -10 \\ 0 & 0 & 0 \end{array} \right] \][/tex]
Row reduce this matrix:
[tex]\[ \left[\begin{array}{rrr} 1 & 0.75 & -1.25 \\ 0 & 1 & -2.5 \\ 0 & 0 & 0 \end{array} \right] \][/tex]
This simplifies to:
[tex]\[ x_3 = s \left[ \begin{array}{r} 0.25 \\ -2.5 \\ 1 \end{array} \right] \][/tex]
Thus, the final answers are:
(a) The characteristic equation:
[tex]\[ -\lambda^3 + 8\lambda^2 - 28\lambda + 48 = 0 \][/tex]
(b) The eigenvalues (from smallest to largest):
[tex]\[ (\lambda_1, \lambda_2, \lambda_3) = (-6, 2, 4) \][/tex]
A basis for each of the corresponding eigenspaces:
[tex]\[ x_1 = \left\{ \left[ \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \right] \right\} \][/tex]
[tex]\[ x_2 = \left\{ \left[ \begin{array}{r} -1 \\ 1.5 \\ 0 \end{array} \right], \left[ \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \right] \right\} \][/tex]
[tex]\[ x_3 = \left\{ \left[ \begin{array}{r} 0.25 \\ -2.5 \\ 1 \end{array} \right] \right\} \][/tex]
1. Find the characteristic equation.
2. Find the eigenvalues.
3. Find a basis for each eigenspace.
### Step 1: Find the Characteristic Equation
The characteristic equation is given by the determinant of [tex]\(A - \lambda I\)[/tex] set to zero, [tex]\( \det(A - \lambda I) = 0 \)[/tex].
Given matrix [tex]\(A\)[/tex]:
[tex]\[ A = \left[\begin{array}{rrr} 0 & -3 & 5 \\ -4 & 4 & -10 \\ 0 & 0 & 4 \end{array} \right] \][/tex]
Subtract [tex]\( \lambda \)[/tex] times the identity matrix [tex]\(I\)[/tex] from [tex]\(A\)[/tex]:
[tex]\[ A - \lambda I = \left[\begin{array}{rrr} 0-\lambda & -3 & 5 \\ -4 & 4-\lambda & -10 \\ 0 & 0 & 4-\lambda \end{array} \right] \][/tex]
The characteristic polynomial is the determinant of this matrix:
[tex]\[ \det(A - \lambda I) = \det \left[\begin{array}{rrr} -\lambda & -3 & 5 \\ -4 & 4-\lambda & -10 \\ 0 & 0 & 4-\lambda \end{array} \right] \][/tex]
To find this determinant, we expand along the third row:
[tex]\[ \det(A - \lambda I) = (-\lambda) \left[ (4-\lambda) \times (4-\lambda) - (-10) \times 0 \right] - (-3) \left[ -4 \times (4-\lambda) - (-10) \times 0 \right] + 5(0) \][/tex]
Simplify the determinant expression:
[tex]\[ \det(A - \lambda I) = (-\lambda) ( (4-\lambda)(4-\lambda) - 0 ) + 3 ( 4(4-\lambda) ) \][/tex]
[tex]\[ \det(A - \lambda I) = (-\lambda) ( (4-\lambda)^2 ) + 12(4-\lambda) \][/tex]
[tex]\[ \det(A - \lambda I) = -\lambda ( 16 - 8\lambda + \lambda^2 ) + 48 - 12\lambda \][/tex]
[tex]\[ \det(A - \lambda I) = -16\lambda + 8\lambda^2 - \lambda^3 + 48 - 12\lambda \][/tex]
Combine like terms:
[tex]\[ \det(A - \lambda I) = -\lambda^3 + 8\lambda^2 - 28\lambda + 48 \][/tex]
Set this equal to zero to obtain the characteristic equation:
[tex]\[ -\lambda^3 + 8\lambda^2 - 28\lambda + 48 = 0 \][/tex]
### Step 2: Find the Eigenvalues
Solve the characteristic equation for [tex]\( \lambda \)[/tex]:
[tex]\[ -\lambda^3 + 8\lambda^2 - 28\lambda + 48 = 0 \][/tex]
Synthetic division or an algebra solver can help here. By testing possible rational roots, we find that [tex]\( \lambda = 4 \)[/tex], [tex]\( \lambda = 2 \)[/tex], and [tex]\( \lambda = -6 \)[/tex].
### Eigenvalues
[tex]\[ \lambda_1 = -6, \lambda_2 = 2, \lambda_3 = 4 \][/tex]
### Step 3: Find Eigenspaces
For each eigenvalue, we solve [tex]\( (A - \lambda I) \mathbf{x} = 0 \)[/tex] to find a basis for each corresponding eigenspace.
Eigenvalue [tex]\(\lambda = -6\)[/tex]:
[tex]\[ A - (-6)I = \left[\begin{array}{rrr} 6 & -3 & 5 \\ -4 & 10 & -10 \\ 0 & 0 & 10 \end{array} \right] \][/tex]
Row reduce this matrix to obtain:
[tex]\[ \left[\begin{array}{rrr} 1 & -0.5 & 0.8333 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{array} \right] \][/tex]
This simplifies to:
[tex]\[ x_1 = t \left[ \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \right] \][/tex]
Eigenvalue [tex]\(\lambda = 2\)[/tex]:
[tex]\[ A - 2I = \left[\begin{array}{rrr} -2 & -3 & 5 \\ -4 & 2 & -10 \\ 0 & 0 & 2 \end{array} \right] \][/tex]
Row reduce this matrix:
[tex]\[ \left[\begin{array}{rrr} 1 & 3/2 & -5/2 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{array} \right] \][/tex]
This simplifies to:
[tex]\[ x_2 = s \left[ \begin{array}{r} -1 \\ 1.5 \\ 0 \end{array} \right] + t \left[ \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \right] \][/tex]
Eigenvalue [tex]\(\lambda = 4\)[/tex]:
[tex]\[ A - 4I = \left[\begin{array}{rrr} -4 & -3 & 5 \\ -4 & 0 & -10 \\ 0 & 0 & 0 \end{array} \right] \][/tex]
Row reduce this matrix:
[tex]\[ \left[\begin{array}{rrr} 1 & 0.75 & -1.25 \\ 0 & 1 & -2.5 \\ 0 & 0 & 0 \end{array} \right] \][/tex]
This simplifies to:
[tex]\[ x_3 = s \left[ \begin{array}{r} 0.25 \\ -2.5 \\ 1 \end{array} \right] \][/tex]
Thus, the final answers are:
(a) The characteristic equation:
[tex]\[ -\lambda^3 + 8\lambda^2 - 28\lambda + 48 = 0 \][/tex]
(b) The eigenvalues (from smallest to largest):
[tex]\[ (\lambda_1, \lambda_2, \lambda_3) = (-6, 2, 4) \][/tex]
A basis for each of the corresponding eigenspaces:
[tex]\[ x_1 = \left\{ \left[ \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \right] \right\} \][/tex]
[tex]\[ x_2 = \left\{ \left[ \begin{array}{r} -1 \\ 1.5 \\ 0 \end{array} \right], \left[ \begin{array}{r} 0 \\ 0 \\ 1 \end{array} \right] \right\} \][/tex]
[tex]\[ x_3 = \left\{ \left[ \begin{array}{r} 0.25 \\ -2.5 \\ 1 \end{array} \right] \right\} \][/tex]
