Question:
(1) Find the eigenvalues and eigenvectors of a matrix,
\[ A=\left(\begin{array}{c}-5 & 6\\-4 & 5\end{array}\right) \]
(2) If \(C\) is constructed by the eigen vectors obtained above, find whether \(C\) is a regular function and whether \(C^{-1}AC\) can be a diagonal matrix.
Answer:
(1) The eigenvalues can be obtained from \(|\lambda E – A|=0\) where \(E\) is the unit matrix.
\[|\lambda E – A| =\left|\begin{array}{c}\lambda+5 & -6\\4 & \lambda-5\end{array}\right|=(\lambda+5)(\lambda-5)+24\\=\lambda^2-25 +24\\= \lambda^2-1\]
Therefore, the eigenvalues are \(\lambda = \pm 1\). The eigen equation can be expressed as
\[A{\bf{x}} = \lambda{\bf{x}} \quad \mathrm{where} {\bf{x}}=\left(\begin{array}{c}x\\y\end{array}\right) \neq 0\]
Thus,
\[\left(\begin{array}{c}-5-\lambda & 6\\-4 & 5-\lambda\end{array}\right)\left( \begin{array}{c}x\\y\end{array} \right)=0\\\left(\begin{array}{c}-5-1 & \\-4 & 5-1\end{array}\right)\left( \begin{array}{c}x\\y\end{array} \right)=\left(\begin{array}{c}-6 & 6\\-4 & 4\end{array}\right)\left( \begin{array}{c}x \\y\end{array} \right)=0\]
Since the vector is not zero, in order to hold the equation, the ratio must be \(x:y=1:1\). Likewise, plug in \(\lambda=-1\); then, we have \(x:y=3:2\). Therefore, the eigen vectors are
\[{\bf p}=\left( \begin{array}{c}x\\y\end{array} \right)=\left(\begin{array}{c}1\\1\end{array}\right),\quad{\bf q}=\left( \begin{array}{c}x\\y\end{array} \right)=\left(\begin{array}{c}3\\2\end{array}\right)\]
(2) A matrix \(C\) is combined by two eigen vectors obtained in question (1). Namely, \(C=({\bf p},{\bf q})\).
\[C = \left(\begin{array}{c}1 & 3\\1 & 2\end{array}\right)\]
Calculate the determinant.
\[\det C = 1 \cdot 2 – 1 \cdot 3 = -1\]
Since \(\det \neq 0\), \(C\) is regular. Now, calculate the following:
\[C^{-1}AC =\left(\begin{array}{c}-2 & 3\\1 & -1\end{array}\right)\left(\begin{array}{c}-5 & 6\\-4 & 5\end{array}\right)\left(\begin{array}{c}1 & 3\\1 & 2\end{array}\right)=\left(\begin{array}{c}1 & 0\\0 & -1\end{array}\right)\]
This is a diagonal matrix and notice that the values correspond to the eigenvalues of matrix \(A\).
Reference,
The inverse matrix for a 2\(\times\)2 matix is
\[M = \left(\begin{array}{c}a & b\\c & d\end{array}\right),\quad M^{-1} = \frac{1}{\det M}\left(\begin{array}{c}d & -b\\-c & a\end{array}\right)\]