Join the IDNLearn.com community and get your questions answered by knowledgeable individuals. Discover in-depth and reliable answers to all your questions from our knowledgeable community members who are always ready to assist.
Sagot :
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]
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Trust IDNLearn.com for all your queries. We appreciate your visit and hope to assist you again soon.
