Fixed points of $2$-dimensional linear system 
Consider $$\begin{aligned} \dot{x} &= ax + by\\ \dot{y} &= bx + ay\end{aligned}$$ where $a>0$ and $b<0$. Find the fixed points and classify them.


I haven't been able to find any fixed points other than the trivial $(x,y)=(0,0)$ solution. Any help would be appreciated!

Fixed points are going to be whatever coordinates in the plane that make the derivative zero.  If you set this up as a matrix differential equation you get
$$
\frac{d}{dt} \left [ \begin{array}{c}
x\\ y\\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{cc}
a & b \\
b & a \\
\end{array} \right ]\left [ \begin{array}{c}
x\\ y\\
\end{array} \right ]
$$
and you can diagonalize this matrix as 
$$
\left [ \begin{array}{cc}
a & b \\
b & a \\
\end{array} \right ] \;\; =\;\; \underbrace{\left [ \begin{array}{cc}
1/\sqrt{2} & -1/\sqrt{2} \\
1/\sqrt{2} & 1/\sqrt{2} \\
\end{array} \right ]}_{ = P} \underbrace{\left [ \begin{array}{cc}
a + b  & 0 \\
0 & a-b \\
\end{array} \right ]}_{=D} \underbrace{\left [ \begin{array}{cc}
1/\sqrt{2} & 1/\sqrt{2} \\
-1/\sqrt{2} & 1/\sqrt{2} \\
\end{array} \right ]}_{=P^T}.
$$
Notice that since $a>0$ and $b<0$ then the only way for this matrix to be singular is if $b = -a$.  This would make the eigenvalue $a+b = 0$, and the corresponding eigenvector to this is $\left [ \begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array} \right ]^T$.  Therefore we get two cases for fixed points:
$$
\textbf{Fixed Points} \;\; =\;\; \begin{cases}
\text{Span} \left \{\left [ \begin{array}{c}
1 \\ 1 \\
\end{array} \right ] \right \}, & \text{if} \; b = -a \\
\left [ \begin{array}{c}
0 \\ 0 \\
\end{array} \right ], & \text{otherwise} \\
\end{cases}.
$$
