Answer :
To find the characteristic equation, eigenvalues, and eigenvectors of the matrix
[tex]\[ \begin{pmatrix} 2 & -2 & 9 \\ 0 & 3 & -2 \\ 0 & -1 & 2 \end{pmatrix}, \][/tex]
we can follow the steps below:
#### (a) Characteristic Equation
1. The characteristic equation of a matrix [tex]\( A \)[/tex] is derived from the determinant of [tex]\( A - \lambda I \)[/tex], where [tex]\( \lambda \)[/tex] is an eigenvalue and [tex]\( I \)[/tex] is the identity matrix.
2. For the given matrix [tex]\( A \)[/tex]:
[tex]\[ A - \lambda I = \begin{pmatrix} 2-\lambda & -2 & 9 \\ 0 & 3-\lambda & -2 \\ 0 & -1 & 2-\lambda \end{pmatrix} \][/tex]
3. Calculate the determinant of this matrix to get the characteristic equation:
[tex]\[ \text{det}(A - \lambda I) = \left|\begin{array}{ccc} 2-\lambda & -2 & 9 \\ 0 & 3-\lambda & -2 \\ 0 & -1 & 2-\lambda \end{array}\right| \][/tex]
4. Expanding along the first row:
[tex]\[ \text{det}(A - \lambda I) = (2-\lambda) \left|\begin{array}{cc} 3-\lambda & -2 \\ -1 & 2-\lambda \end{array}\right| \][/tex]
[tex]\[ = (2-\lambda) \left[ (3-\lambda)(2-\lambda) - (-2)(-1) \right] \\ = (2-\lambda) \left[ (3-\lambda)(2-\lambda) - 2 \right] \][/tex]
[tex]\[ = (2-\lambda) \left(6 - 5\lambda + \lambda^2 - 2\right) \\ = (2-\lambda) (\lambda^2 - 5\lambda + 4) \][/tex]
5. Solve for the roots of the polynomial [tex]\(\lambda^3 - 7\lambda^2 + 14\lambda - 8 = 0\)[/tex].
Thus, the characteristic equation is:
[tex]\[ \lambda^3 - 7\lambda^2 + 14\lambda - 8 = 0 \][/tex]
#### (b) Eigenvalues and Eigenspaces
1. Find the roots of the characteristic equation to determine the eigenvalues:
[tex]\[ (\lambda - 1)(\lambda - 2)(\lambda - 4) = 0 \][/tex]
2. The eigenvalues are [tex]\(\lambda_1 = 1\)[/tex], [tex]\(\lambda_2 = 2\)[/tex], and [tex]\(\lambda_3 = 4\)[/tex].
3. Now, for each eigenvalue, we find the corresponding eigenvectors (basis for the eigenspaces).
##### Eigenvalue [tex]\(\lambda = 1\)[/tex]:
Solve [tex]\((A - I)x = 0\)[/tex]:
[tex]\[ \begin{pmatrix} 1 & -2 & 9 \\ 0 & 2 & -2 \\ 0 & -1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \][/tex]
[tex]\[ \Rightarrow \begin{cases} x_1 - 2x_2 + 9x_3 = 0 \\ 2x_2 - 2x_3 = 0 \\ -x_2 + x_3 = 0 \end{cases} \][/tex]
[tex]\[ \Rightarrow x_2 = x_3, x_1 = -7x_2 \\ \Rightarrow x = k \begin{pmatrix} -7 \\ 1 \\ 1 \end{pmatrix} \][/tex]
A basis for the eigenspace corresponding to [tex]\(\lambda = 1\)[/tex] is:
[tex]\[ \begin{pmatrix} -7 \\ 1 \\ 1 \end{pmatrix} \][/tex]
##### Eigenvalue [tex]\(\lambda = 2\)[/tex]:
Solve [tex]\((A - 2I)x = 0\)[/tex]:
[tex]\[ \begin{pmatrix} 0 & -2 & 9 \\ 0 & 1 & -2 \\ 0 & -1 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \][/tex]
[tex]\[ \Rightarrow \begin{cases} -x_2 + 9x_3 = 0 \\ x_2 = 0 \\ -x_2 = 0 \end{cases} \][/tex]
[tex]\[ \Rightarrow x_1 \text{ is free}, x_2 = 0, x_3 = 0 \\ \Rightarrow x = k \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \][/tex]
A basis for the eigenspace corresponding to [tex]\(\lambda = 2\)[/tex] is:
[tex]\[ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \][/tex]
##### Eigenvalue [tex]\(\lambda = 4\)[/tex]:
Solve [tex]\((A - 4I)x = 0\)[/tex]:
[tex]\[ \begin{pmatrix} -2 & -2 & 9 \\ 0 & -1 & -2 \\ 0 & -1 & -2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \][/tex]
[tex]\[ \Rightarrow \begin{cases} -2x_1 - 2x_2 + 9x_3 = 0 \\ -x_2 - 2x_3 = 0 \\ -x_2 - 2x_3 = 0 \end{cases} \][/tex]
[tex]\[ \Rightarrow x_2 = -2x_3, x_1 = \frac{13}{2} x_3 \\ \Rightarrow x = k \begin{pmatrix} \frac{13}{2}\\ -2 \\ 1 \end{pmatrix} \][/tex]
A basis for the eigenspace corresponding to [tex]\(\lambda = 4\)[/tex] is:
[tex]\[ \begin{pmatrix} \frac{13}{2} \\ -2 \\ 1 \end{pmatrix} \][/tex]
#### Summary:
1. The characteristic equation is:
[tex]\[ \lambda^3 - 7\lambda^2 + 14\lambda - 8 = 0 \][/tex]
2. The eigenvalues are:
[tex]\[ \lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 4 \][/tex]
3. A basis for each of the corresponding eigenspaces is:
[tex]\[ x_1 = \begin{pmatrix} -7 \\ 1 \\ 1 \end{pmatrix}, x_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, x_3 = \begin{pmatrix} \frac{13}{2} \\ -2 \\ 1 \end{pmatrix} \][/tex]
[tex]\[ \begin{pmatrix} 2 & -2 & 9 \\ 0 & 3 & -2 \\ 0 & -1 & 2 \end{pmatrix}, \][/tex]
we can follow the steps below:
#### (a) Characteristic Equation
1. The characteristic equation of a matrix [tex]\( A \)[/tex] is derived from the determinant of [tex]\( A - \lambda I \)[/tex], where [tex]\( \lambda \)[/tex] is an eigenvalue and [tex]\( I \)[/tex] is the identity matrix.
2. For the given matrix [tex]\( A \)[/tex]:
[tex]\[ A - \lambda I = \begin{pmatrix} 2-\lambda & -2 & 9 \\ 0 & 3-\lambda & -2 \\ 0 & -1 & 2-\lambda \end{pmatrix} \][/tex]
3. Calculate the determinant of this matrix to get the characteristic equation:
[tex]\[ \text{det}(A - \lambda I) = \left|\begin{array}{ccc} 2-\lambda & -2 & 9 \\ 0 & 3-\lambda & -2 \\ 0 & -1 & 2-\lambda \end{array}\right| \][/tex]
4. Expanding along the first row:
[tex]\[ \text{det}(A - \lambda I) = (2-\lambda) \left|\begin{array}{cc} 3-\lambda & -2 \\ -1 & 2-\lambda \end{array}\right| \][/tex]
[tex]\[ = (2-\lambda) \left[ (3-\lambda)(2-\lambda) - (-2)(-1) \right] \\ = (2-\lambda) \left[ (3-\lambda)(2-\lambda) - 2 \right] \][/tex]
[tex]\[ = (2-\lambda) \left(6 - 5\lambda + \lambda^2 - 2\right) \\ = (2-\lambda) (\lambda^2 - 5\lambda + 4) \][/tex]
5. Solve for the roots of the polynomial [tex]\(\lambda^3 - 7\lambda^2 + 14\lambda - 8 = 0\)[/tex].
Thus, the characteristic equation is:
[tex]\[ \lambda^3 - 7\lambda^2 + 14\lambda - 8 = 0 \][/tex]
#### (b) Eigenvalues and Eigenspaces
1. Find the roots of the characteristic equation to determine the eigenvalues:
[tex]\[ (\lambda - 1)(\lambda - 2)(\lambda - 4) = 0 \][/tex]
2. The eigenvalues are [tex]\(\lambda_1 = 1\)[/tex], [tex]\(\lambda_2 = 2\)[/tex], and [tex]\(\lambda_3 = 4\)[/tex].
3. Now, for each eigenvalue, we find the corresponding eigenvectors (basis for the eigenspaces).
##### Eigenvalue [tex]\(\lambda = 1\)[/tex]:
Solve [tex]\((A - I)x = 0\)[/tex]:
[tex]\[ \begin{pmatrix} 1 & -2 & 9 \\ 0 & 2 & -2 \\ 0 & -1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \][/tex]
[tex]\[ \Rightarrow \begin{cases} x_1 - 2x_2 + 9x_3 = 0 \\ 2x_2 - 2x_3 = 0 \\ -x_2 + x_3 = 0 \end{cases} \][/tex]
[tex]\[ \Rightarrow x_2 = x_3, x_1 = -7x_2 \\ \Rightarrow x = k \begin{pmatrix} -7 \\ 1 \\ 1 \end{pmatrix} \][/tex]
A basis for the eigenspace corresponding to [tex]\(\lambda = 1\)[/tex] is:
[tex]\[ \begin{pmatrix} -7 \\ 1 \\ 1 \end{pmatrix} \][/tex]
##### Eigenvalue [tex]\(\lambda = 2\)[/tex]:
Solve [tex]\((A - 2I)x = 0\)[/tex]:
[tex]\[ \begin{pmatrix} 0 & -2 & 9 \\ 0 & 1 & -2 \\ 0 & -1 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \][/tex]
[tex]\[ \Rightarrow \begin{cases} -x_2 + 9x_3 = 0 \\ x_2 = 0 \\ -x_2 = 0 \end{cases} \][/tex]
[tex]\[ \Rightarrow x_1 \text{ is free}, x_2 = 0, x_3 = 0 \\ \Rightarrow x = k \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \][/tex]
A basis for the eigenspace corresponding to [tex]\(\lambda = 2\)[/tex] is:
[tex]\[ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \][/tex]
##### Eigenvalue [tex]\(\lambda = 4\)[/tex]:
Solve [tex]\((A - 4I)x = 0\)[/tex]:
[tex]\[ \begin{pmatrix} -2 & -2 & 9 \\ 0 & -1 & -2 \\ 0 & -1 & -2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \][/tex]
[tex]\[ \Rightarrow \begin{cases} -2x_1 - 2x_2 + 9x_3 = 0 \\ -x_2 - 2x_3 = 0 \\ -x_2 - 2x_3 = 0 \end{cases} \][/tex]
[tex]\[ \Rightarrow x_2 = -2x_3, x_1 = \frac{13}{2} x_3 \\ \Rightarrow x = k \begin{pmatrix} \frac{13}{2}\\ -2 \\ 1 \end{pmatrix} \][/tex]
A basis for the eigenspace corresponding to [tex]\(\lambda = 4\)[/tex] is:
[tex]\[ \begin{pmatrix} \frac{13}{2} \\ -2 \\ 1 \end{pmatrix} \][/tex]
#### Summary:
1. The characteristic equation is:
[tex]\[ \lambda^3 - 7\lambda^2 + 14\lambda - 8 = 0 \][/tex]
2. The eigenvalues are:
[tex]\[ \lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 4 \][/tex]
3. A basis for each of the corresponding eigenspaces is:
[tex]\[ x_1 = \begin{pmatrix} -7 \\ 1 \\ 1 \end{pmatrix}, x_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, x_3 = \begin{pmatrix} \frac{13}{2} \\ -2 \\ 1 \end{pmatrix} \][/tex]
