Given the matrix A to find the Eigenvalues we have the following: d e t ( A − λ I ) = 0 ⟹ ∣ − λ 1 1 3 − λ − 4 − 3 − 5 7 6 − λ ∣ = 0 Add column 2 multiplied by λ to column 1: C 1 = C 1 + λ C 2 ∣ − λ 1 1 3 − λ − 4 − 3 − 5 7 6 − λ ∣ = ∣ 0 1 1 − λ ( λ + 4 ) + 3 − λ − 4 − 3 7 λ − 5 7 6 − λ ∣ Subtract column 2 from column 3: C 3 = C 3 − C 2 ∣ 0 1 1 − λ ( λ + 4 ) + 3 − λ − 4 − 3 7 λ − 5 7 6 − λ ∣ = ∣ 0 1 0 − λ ( λ + 4 ) + 3 − λ − 4 λ + 1 7 λ − 5 7 − λ − 1 ∣ Expand along row 1: ∣ 0 1 0 − λ ( λ + 4 ) + 3 − λ − 4 λ + 1 7 λ − 5 7 − λ − 1 ∣ = ( 0 ) ( − 1 ) 1 + 1 ∣ − λ − 4 λ + 1 7 − λ − 1 ∣ + ( 1 ) ( − 1 ) 1 + 2 ∣ − λ ( λ + 4 ) + 3 λ + 1 7 λ − 5 − λ − 1 ∣ + ( 0 ) ( − 1 ) 1 + 3 ∣ − λ ( λ + 4 ) + 3 − λ − 4 7 λ − 5 7 ∣ = − ∣ − λ ( λ + 4 ) + 3 λ + 1 7 λ − 5 − λ − 1 ∣ Note that: ∣ − λ ( λ + 4 ) + 3 λ + 1 7 λ − 5 − λ − 1 ∣ = ( − λ ( λ + 4 ) + 3 ) ⋅ ( − λ − 1 ) − ( λ + 1 ) ⋅ ( 7 λ − 5 ) = λ 3 − 2 λ 2 − λ + 2 Finally we have the following: ( − 1 ) ⋅ ( λ 3 − 2 λ 2 − λ + 2 ) = − ( λ − 2 ) ( λ − 1 ) ( λ + 1 ) . ⟹ d e t ( A − λ I ) = − ( λ − 2 ) ( λ − 1 ) ( λ + 1 ) = 0 Solving this we have that: λ = − 1 , 1 , 2. ∴ The eigenvalues of A are λ = − 1 , λ = 1 , λ = 2 We now fin the eigenvectors of the matrix. 1. For λ = 2 : [ − λ 1 1 3 − λ − 4 − 3 − 5 7 6 − λ ] = [ − 2 1 1 3 − 6 − 3 − 5 7 4 ] The nullspace of this matrix is [ 1 3 − 1 3 1 ] which is the eigenvector. 2. For λ = 1 : [ − λ 1 1 3 − λ − 4 − 3 − 5 7 6 − λ ] = [ − 1 1 1 3 − 5 − 3 − 5 7 5 ] The nullspace for this matrix is: [ 1 0 1 ] and this is the eigenvector 3. For λ = − 1 : [ − λ 1 1 3 − λ − 4 − 3 − 5 7 6 − λ ] = [ 1 1 1 3 − 3 − 3 − 5 7 7 ] The nullspace of this matrix is [ 0 − 1 1 ] which is the eigenvector. A is diagonalisable since each of the eigenvectors of A are linearly independent and they are distinct. \text{Given the matrix \(A\) to find the Eigenvalues we have the following:}\\
{\rm det}(A - \lambda I) = 0\\
\implies \begin{vmatrix}- \lambda & 1 & 1\\3 & - \lambda - 4 & -3\\-5 & 7 & 6 - \lambda\end{vmatrix}\ = 0 \\
\text{Add column }2\text{ multiplied by λ to column 1:} C_{1} = C_{1} + \lambda C_{2}\\
\left|\begin{array}{ccc}- \lambda & 1 & 1\\3 & - \lambda - 4 & -3\\-5 & 7 & 6 - \lambda\end{array}\right| = \left|\begin{array}{ccc}0 & 1 & 1\\- \lambda \left(\lambda + 4\right) + 3 & - \lambda - 4 & -3\\7 \lambda - 5 & 7 & 6 - \lambda\end{array}\right| \\
\text{Subtract column }2\text{ from column 3: } C_{3} = C_{3} - C_{2} \\
\left|\begin{array}{ccc}0 & 1 & 1\\- \lambda \left(\lambda + 4\right) + 3 & - \lambda - 4 & -3\\7 \lambda - 5 & 7 & 6 - \lambda\end{array}\right| = \left|\begin{array}{ccc}0 & 1 & 0\\- \lambda \left(\lambda + 4\right) + 3 & - \lambda - 4 & \lambda + 1\\7 \lambda - 5 & 7 & - \lambda - 1\end{array}\right| \\
\text{Expand along row 1:} \\
\left|\begin{array}{ccc}0 & 1 & 0\\- \lambda \left(\lambda + 4\right) + 3 & - \lambda - 4 & \lambda + 1\\7 \lambda - 5 & 7 & - \lambda - 1\end{array}\right| = \left(0\right) \left(-1\right)^{1 + 1} \left|\begin{array}{cc}- \lambda - 4 & \lambda + 1\\7 & - \lambda - 1\end{array}\right| + \left(1\right) \left(-1\right)^{1 + 2} \left|\begin{array}{cc}- \lambda \left(\lambda + 4\right) + 3 & \lambda + 1\\7 \lambda - 5 & - \lambda - 1\end{array}\right| + \left(0\right) \left(-1\right)^{1 + 3} \left|\begin{array}{cc}- \lambda \left(\lambda + 4\right) + 3 & - \lambda - 4\\7 \lambda - 5 & 7\end{array}\right| = - \left|\begin{array}{cc}- \lambda \left(\lambda + 4\right) + 3 & \lambda + 1\\7 \lambda - 5 & - \lambda - 1\end{array}\right| \\
\text{Note that: } \\
\left|\begin{array}{cc}- \lambda \left(\lambda + 4\right) + 3 & \lambda + 1\\7 \lambda - 5 & - \lambda - 1\end{array}\right| = \left(- \lambda \left(\lambda + 4\right) + 3\right)\cdot \left(- \lambda - 1\right) - \left(\lambda + 1\right)\cdot \left(7 \lambda - 5\right) = \lambda^{3} - 2 \lambda^{2} - \lambda + 2 \\
\text{Finally we have the following: } \\
\left(-1\right)\cdot \left(\lambda^{3} - 2 \lambda^{2} - \lambda + 2\right) = - \left(\lambda - 2\right) \left(\lambda - 1\right) \left(\lambda + 1\right).\\
\implies {\rm det}(A - \lambda I) = - \left(\lambda - 2\right) \left(\lambda - 1\right) \left(\lambda + 1\right) =0 \\
\text{Solving this we have that: } \lambda = -1, 1,2.\\
\therefore \text{The eigenvalues of \(A\) are } \lambda = -1, \lambda = 1, \lambda = 2\\
\text{We now fin the eigenvectors of the matrix. }\\
1. \text{For } \lambda = 2:\\
\left[\begin{array}{ccc}- \lambda & 1 & 1\\3 & - \lambda - 4 & -3\\-5 & 7 & 6 - \lambda\end{array}\right] = \left[\begin{array}{ccc}-2 & 1 & 1\\3 & -6 & -3\\-5 & 7 & 4\end{array}\right]\\
\text{The nullspace of this matrix is }
\left[\begin{array}{c}\frac{1}{3}\\- \frac{1}{3}\\1\end{array}\right] \text{which is the eigenvector.} \\
2. \text{For } \lambda = 1:\\
\left[\begin{array}{ccc}- \lambda & 1 & 1\\3 & - \lambda - 4 & -3\\-5 & 7 & 6 - \lambda\end{array}\right] = \left[\begin{array}{ccc}-1 & 1 & 1\\3 & -5 & -3\\-5 & 7 & 5\end{array}\right]\\
\text{The nullspace for this matrix is: } \left[\begin{array}{c}1\\0\\1\end{array}\right] \text{and this is the eigenvector} \\
3. \text{For }\lambda = -1: \\
\left[\begin{array}{ccc}- \lambda & 1 & 1\\3 & - \lambda - 4 & -3\\-5 & 7 & 6 - \lambda\end{array}\right] = \left[\begin{array}{ccc}1 & 1 & 1\\3 & -3 & -3\\-5 & 7 & 7\end{array}\right] \\
\text{The nullspace of this matrix is } \left[\begin{array}{c}0\\-1\\1\end{array}\right] \text{which is the eigenvector.}\\
\text{\(A\) is diagonalisable since each of the eigenvectors of \(A\) are linearly independent and they are distinct.} Given the matrix A to find the Eigenvalues we have the following: det ( A − λ I ) = 0 ⟹ ∣ ∣ − λ 3 − 5 1 − λ − 4 7 1 − 3 6 − λ ∣ ∣ = 0 Add column 2 multiplied by λ to column 1: C 1 = C 1 + λ C 2 ∣ ∣ − λ 3 − 5 1 − λ − 4 7 1 − 3 6 − λ ∣ ∣ = ∣ ∣ 0 − λ ( λ + 4 ) + 3 7 λ − 5 1 − λ − 4 7 1 − 3 6 − λ ∣ ∣ Subtract column 2 from column 3: C 3 = C 3 − C 2 ∣ ∣ 0 − λ ( λ + 4 ) + 3 7 λ − 5 1 − λ − 4 7 1 − 3 6 − λ ∣ ∣ = ∣ ∣ 0 − λ ( λ + 4 ) + 3 7 λ − 5 1 − λ − 4 7 0 λ + 1 − λ − 1 ∣ ∣ Expand along row 1: ∣ ∣ 0 − λ ( λ + 4 ) + 3 7 λ − 5 1 − λ − 4 7 0 λ + 1 − λ − 1 ∣ ∣ = ( 0 ) ( − 1 ) 1 + 1 ∣ ∣ − λ − 4 7 λ + 1 − λ − 1 ∣ ∣ + ( 1 ) ( − 1 ) 1 + 2 ∣ ∣ − λ ( λ + 4 ) + 3 7 λ − 5 λ + 1 − λ − 1 ∣ ∣ + ( 0 ) ( − 1 ) 1 + 3 ∣ ∣ − λ ( λ + 4 ) + 3 7 λ − 5 − λ − 4 7 ∣ ∣ = − ∣ ∣ − λ ( λ + 4 ) + 3 7 λ − 5 λ + 1 − λ − 1 ∣ ∣ Note that: ∣ ∣ − λ ( λ + 4 ) + 3 7 λ − 5 λ + 1 − λ − 1 ∣ ∣ = ( − λ ( λ + 4 ) + 3 ) ⋅ ( − λ − 1 ) − ( λ + 1 ) ⋅ ( 7 λ − 5 ) = λ 3 − 2 λ 2 − λ + 2 Finally we have the following: ( − 1 ) ⋅ ( λ 3 − 2 λ 2 − λ + 2 ) = − ( λ − 2 ) ( λ − 1 ) ( λ + 1 ) . ⟹ det ( A − λ I ) = − ( λ − 2 ) ( λ − 1 ) ( λ + 1 ) = 0 Solving this we have that: λ = − 1 , 1 , 2. ∴ The eigenvalues of A are λ = − 1 , λ = 1 , λ = 2 We now fin the eigenvectors of the matrix. 1. For λ = 2 : ⎣ ⎡ − λ 3 − 5 1 − λ − 4 7 1 − 3 6 − λ ⎦ ⎤ = ⎣ ⎡ − 2 3 − 5 1 − 6 7 1 − 3 4 ⎦ ⎤ The nullspace of this matrix is ⎣ ⎡ 3 1 − 3 1 1 ⎦ ⎤ which is the eigenvector. 2. For λ = 1 : ⎣ ⎡ − λ 3 − 5 1 − λ − 4 7 1 − 3 6 − λ ⎦ ⎤ = ⎣ ⎡ − 1 3 − 5 1 − 5 7 1 − 3 5 ⎦ ⎤ The nullspace for this matrix is: ⎣ ⎡ 1 0 1 ⎦ ⎤ and this is the eigenvector 3. For λ = − 1 : ⎣ ⎡ − λ 3 − 5 1 − λ − 4 7 1 − 3 6 − λ ⎦ ⎤ = ⎣ ⎡ 1 3 − 5 1 − 3 7 1 − 3 7 ⎦ ⎤ The nullspace of this matrix is ⎣ ⎡ 0 − 1 1 ⎦ ⎤ which is the eigenvector. A is diagonalisable since each of the eigenvectors of A are linearly independent and they are distinct.
#333702 on Dec 2022
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