Question 5.1.2: Find the eigenvalues and corresponding eigenvectors of A = [...
Find the eigenvalues and corresponding eigenvectors of
A = \begin{bmatrix} 2&-12 \\1&-5 \end{bmatrix}
Give a description of the eigenspace corresponding to each eigenvalue
By Theorem 1 to find the eigenvalues, we solve the characteristic equation
det(A − λI) = \begin{vmatrix} 2 − λ & −12\\ 1 &−5 − λ \end{vmatrix}
= (2 − λ)(−5 − λ) − (1)(−12)
= λ² + 3λ + 2
= (λ + 1)(λ + 2) = 0
Thus, the eigenvalues are \lambda _{1} = −1 and \lambda _{2} = −2 . To find the eigenvectors, we need to find all nonzero vectors in the null spaces of A − \lambda _{1} I and A − \lambda _{2} I. First, for \lambda _{1} = −1,
A − \lambda _{1} I = A + I = \begin{bmatrix} 2&-12 \\1&-5 \end{bmatrix}+ \begin{bmatrix} 1&0 \\0&1 \end{bmatrix}=\begin{bmatrix} 3&-12 \\1&-4 \end{bmatrix}
The null space of A + I is found by row-reducing the augmented matrix
\left[\begin{array}{r | c}\begin{matrix} 3&-12 \\ 1&-4 \end{matrix}& \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\right] \quad to \quad \left[\begin{array}{r | c}\begin{matrix} 1&-4 \\ 0&0 \end{matrix}& \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\right]The solution set for this linear system is given by S= \left\{\left.\begin{matrix}\begin{bmatrix}4t\\t\end{bmatrix}\end{matrix}\right| t ∈ R\right \}. Choosing t = 1, we obtain the eigenvector v_{ 1} = \begin{bmatrix} 4\\ 1 \end{bmatrix} Hence, the eigenspace corresponding to \lambda _{1} = −1 is
V_{\lambda 1} = \left\{ t \left.\begin{matrix}\begin{bmatrix}4\\t\end{bmatrix}\end{matrix}\right| t \text{ is any real number}\right \}For \lambda _{2} = −2,
A − \lambda _{2} I = \begin{bmatrix} 4&-12 \\1&-3 \end{bmatrix}In a similar way we find that the vector v_{ 2} = \begin{bmatrix} 3 \\1 \end{bmatrix}is an eigenvector corresponding to \lambda _{2} = −2. The corresponding eigenspace is
V_{\lambda 2} = \left\{ t \left.\begin{matrix}\begin{bmatrix}3\\1\end{bmatrix}\end{matrix}\right| t \text{ is any real number}\right \}The eigenspaces V_{\lambda 1} and V_{\lambda 2} are lines in the direction of the eigenvectors \begin{bmatrix} 4 \\ 1 \end{bmatrix} and \begin{bmatrix} 3 \\ 1 \end{bmatrix} respectively. The images of the eigenspaces, after multiplication by A, are the same lines, since the direction vectors A\begin{bmatrix} 4 \\ 1 \end{bmatrix} and A \begin{bmatrix} 3 \\ 1 \end{bmatrix} are scalar multiples of \begin{bmatrix} 4 \\ 1 \end{bmatrix} and\begin{bmatrix} 3 \\ 1 \end{bmatrix} respectively.