I have this matrix that represents a $2 \times 2$ linear system and I am supposed to solve to find what $x(t)$ and $y(t)$ are.
$\left[ {\begin{array}{cc} 1 & 5 \\ -1 & -3 \\ \end{array} } \right] $
I got that my eigenvalues were $-1+i$ and $-1-i$. But how do I use these to find my eigenvectors?
$\endgroup$ 12 Answers
$\begingroup$Use the definition of "eigenvalue" and "eigenvector". If $-1+ i$ is an eigenvalue then there exists a vector, $\begin{bmatrix}x \\ y \end{bmatrix}$, such that $\begin{bmatrix}1 & 5 \\ -1 & -3 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}(-1+ i)x \\ (-1+ i)y\end{bmatrix}$. Solve for $x$ and $y$. Of course, since the set of eigenvectors corresponding to a given eigenvalue form a subspace, there will be an infinite number of possible $(x, y)$ values.
$\endgroup$ $\begingroup$$$\left[ {\begin{array}{cc} 1-(-1+i) & 5 \\ -1 & -3-(-1+i) \\ \end{array} } \right] = $$
$$ \left[ {\begin{array}{cc} 2-i & 5 \\ -1 & -2-i \\ \end{array} } \right]$$
One eigenvector is $$ \begin{pmatrix} 2+i\\-1\end{pmatrix}$$
and the other one is $$ \begin{pmatrix} 2-i\\-1\end{pmatrix}$$
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