A $5 \times 5$ real matrix has an eigenvector in $\mathbb{R}^5$.

Let $A : \mathbb { R } ^ { 2 } \rightarrow \mathbb { R } ^ { 2 }$ be a linear transformation with eigenvalues $\frac { 2 } { 3 }$ and $\frac { 9 } { 5 }$. Then, there exists a non-zero vector $v \in \mathbb { R } ^ { 2 }$ such that
(a) $\| A v \| > 2 \| v \|$;
(b) $\| A v \| < \frac { 1 } { 2 } \| v \|$;
(c) $\| A v \| = \| v \|$;
(d) $A v = 0$;