Question 5.1.3: Find the eigenvalues of A= [ 1 0 0 0 0 1 5 -10 1 0 2 0 1 0 0...
Find the eigenvalues of
A =\begin{bmatrix} 1&0&0&0 \\0&1&5&-10 \\ 1&0&2&0\\ 1&0&0&3 \end{bmatrix}
and find a basis for each of the corresponding eigenspaces
The characteristic equation of A is
det(A − λI) = \begin{vmatrix} 1 − λ & 0&0&0\\ 0 &1− λ & 5&-10 \\ 1&0& 2 − λ & 0\\ 1&0&0& 3 − λ \end{vmatrix} = (λ − 1)²(λ − 2)(λ − 3) = 0
Thus, the eigenvalues are
\lambda _{1} = 1 \lambda _{2} =2 and \lambda _{3} = 3
Since the exponent of the factor λ − 1 is 2, we say that the eigenvalue \lambda _{1} = 1 has algebraic multiplicity 2. To find the eigenspace for \lambda _{1} = 1, we reduce the matrix
A − (1)I = \begin{bmatrix} 0&0&0&0 \\0&0&5&-10 \\ 1&0&1&0\\ 1&0&0&2 \end{bmatrix} to \begin{bmatrix} 1&0&0&2 \\0&0&1&-2 \\ 0&0&0&0\\ 0&0&0&0 \end{bmatrix}
Hence, the eigenspace corresponding to \lambda _{1} = 1 is
V_{1} = \left\{s\begin{bmatrix} 0 \\ 1 \\0\\0 \end{bmatrix} + t \begin{bmatrix} -2 \\ 0 \\2\\1 \end{bmatrix} |s,t ∈R \right\}
Observe that the two vectors
\begin{bmatrix} 0 \\ 1 \\0\\0 \end{bmatrix} and \begin{bmatrix} -2 \\ 0 \\2\\1 \end{bmatrix}
are linearly independent and hence form a basis for V_{\lambda 1}. Since dim(V_{\lambda 1} ) = 2, we say that \lambda _{1} has geometric multiplicity equal to 2. Alternatively, we can write
V_{\lambda 1} =span \left\{\begin{bmatrix} 0 \\ 1\\0\\0 \end{bmatrix} ,\begin{bmatrix} -2 \\ 0\\2\\1 \end{bmatrix} \right\}
Similarly, the eigenspaces corresponding to \lambda _{2} = 2 and \lambda _{3} = 3 are, respectively,
V_{\lambda 2} =span \left\{\begin{bmatrix} 0 \\ 5\\1\\0 \end{bmatrix} \right\} and V_{\lambda 3} =span \left\{\begin{bmatrix} 0 \\ -5\\0\\1 \end{bmatrix} \right\}