Answer to Question #218166 in Linear Algebra for Dhruv bartwal

Question

Find the eigenvalues and an eigenvector per eigenvalue of the matrix
A= [ 0 1 1]
[ 3 -4 -3]
[ -5 7 6]
Is A is diagonalisable ? Give reason your answer.

Answer 1

$\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.}$