Compute the eigenvalues and eigenvectors of the matrix A. You’ll need to show your work. You’ll need to write out the characteristic polynomial and show your solution.
\[ A = \left[\begin{array}{cc} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right] \]
Characteristic Equation:
\[P_{A}(x) = det(A-xI_{3})\] \[=\left|\begin{array}{cc}
1-x & 2 & 3 \\
0 & 4-x & 5 \\
0 & 0 & 6-x
\end{array}\right|\]
\[=(1-x)\left|\begin{array}{cc} 4-x & 5 \\ 0 & 6-x \end{array}\right| - 2\left|\begin{array}{cc} 0 & 5 \\ 0 & 6-x \end{array}\right| + 3\left|\begin{array}{cc} 0 & 4-x \\ 0 & 0 \end{array}\right|\]
\[= (1-x)[(4-x)(6-x)-0]-2[0]+3[0]\] \[= (1-x)(x^2-10x+24)\] \[= -x^3 +11x^2-34x +24\] \[= x^3 -11x^2+34x -24\]
Eigenvalues:
To find the eigenvaules of this characteristic equation, we can use the last term in our characteristic equation. The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, and 24.
x = 1 is a solution: \[1^3 -11(1)^2+34(1) -24 = 0\] \[0 = 0\]
x = 4 is a solution:
\[4^3 -11(4)^2+34(4) -24 = 0\] \[0 = 0\] x = 6 is a solution:
\[6^3 -11(6)^2+34(6) -24 = 0\] \[0 = 0\]
These 3 numbers multiply to 24 and add to 11 (our second term in the characteristic equation), so this further confirms our findings.
Our final eigenvalues are: \[\lambda_1 = 1\] \[\lambda_2 = 4\] \[\lambda_3 = 6\]
Eigenvectors:
\(\lambda_1 = 1\)
\[(A-\lambda_1I_n)x = 0 \]
\[(\left[\begin{array}{cc} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6 \\ \end{array}\right] - \left[\begin{array}{cc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{array}\right])( \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right]) = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[\left[\begin{array}{cc} 0 & 2 & 3\\ 0 & 3 & 5\\ 0 & 0 & 5 \\ \end{array}\right] \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[R_2 = R_2-R_3: \left[\begin{array}{cc} 0 & 2 & 3\\ 0 & 3 & 0\\ 0 & 0 & 5 \\ \end{array}\right] \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[R_2 = \frac{1}{3}R_2: \left[\begin{array}{cc} 0 & 2 & 3\\ 0 & 1 & 0\\ 0 & 0 & 5 \\ \end{array}\right] \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[R_1 = R_1-2R_2: \left[\begin{array}{cc} 0 & 0 & 3\\ 0 & 1 & 0\\ 0 & 0 & 5 \\ \end{array}\right] \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[R_1 = \frac{1}{3}R_1: \left[\begin{array}{cc} 0 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 5 \\ \end{array}\right] \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[R_3 = R_3-5R_1: \left[\begin{array}{cc} 0 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 0 \\ \end{array}\right] \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[E_A(1) = \{\left[\begin{array}{cc} 1 \\ 0 \\ 0 \end{array}\right]\}\]
\(\lambda_2 = 4\)
\[(A-\lambda_2I_n)x = 0 \]
\[(\left[\begin{array}{cc} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6 \\ \end{array}\right] - \left[\begin{array}{cc} 4 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 4\\ \end{array}\right])( \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right]) = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[\left[\begin{array}{cc} -3 & 2 & 3\\ 0 & 0 & 5\\ 0 & 0 & 2 \\ \end{array}\right] \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[R_3 = \frac{1}{2}R_3: \left[\begin{array}{cc} -3 & 2 & 3\\ 0 & 0 & 5\\ 0 & 0 & 1 \\ \end{array}\right] \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[R_2 = R_2-5R_3: \left[\begin{array}{cc} -3 & 2 & 3\\ 0 & 0 & 0\\ 0 & 0 & 1 \\ \end{array}\right] \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[R_1 = -\frac{1}{3}R_1: \left[\begin{array}{cc} 1 & -\frac{2}{3} & -1\\ 0 & 0 & 0\\ 0 & 0 & 1 \\ \end{array}\right] \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[R_1 = R_1+R_3: \left[\begin{array}{cc} 1 & -\frac{2}{3} & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \\ \end{array}\right] \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[E_A(4) = \{\left[\begin{array}{cc} 2 \\ 3 \\ 0 \end{array}\right]\}\]
\(\lambda_3 = 6\)
\[(A-\lambda_6I_n)x = 0 \]
\[(\left[\begin{array}{cc} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6 \\ \end{array}\right] - \left[\begin{array}{cc} 6 & 0 & 0\\ 0 & 6 & 0\\ 0 & 0 & 6\\ \end{array}\right])( \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right]) = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[\left[\begin{array}{cc} -5 & 2 & 3\\ 0 & -2 & 5\\ 0 & 0 & 0 \\ \end{array}\right] \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[R_1 = R_1+R_2: \left[\begin{array}{cc} -5 & 0 & 0\\ 0 & -2 & 5\\ 0 & 0 & 0 \\ \end{array}\right] \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[R_1 = \frac{-1}{5}R_1, R_2 = \frac{-1}{2}R_2: \left[\begin{array}{cc} 1 & 0 & \frac{-8}{5}\\ 0 & 1 & \frac{-5}{2}\\ 0 & 0 & 0 \\ \end{array}\right] \left[\begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right] = \left[\begin{array}{cc} 0 \\ 0 \\ 0 \end{array}\right]\]
\[E_A(6) = \{\left[\begin{array}{cc} \frac{8}{5} \\ \frac{5}{2} \\ 1 \end{array}\right]\} = \{\left[\begin{array}{cc} 16 \\ 25 \\ 10 \end{array}\right]\} \]