Problem related to linear transformation

Question

Problem

Assume that a linear transformation $T$ has two eigenvectors $x$ and $y$ belonging to two distinct eigenvalue $\lambda$ and $\mu$ . If $ax+by$ is an eigenvector of$T$,prove that $a=0$ or $b=0$.

Attempt

$T(x)=\lambda x$ and $T(y) = \mu y$. $T(ax+by) = \omega(ax+by)$,where $\omega$ is an eigenvalue for $ax+by$.

$a(\lambda -\omega)=0$ and $b(\mu-\omega) =0$ since $x$ and$ y$ are independent( distinct eigenvalues) .

Problem

How to arrive at result after that ?

Answer 1

$T(ax+by) = \omega(ax+by)$, where $\omega$ is an eigenvalue for $ax+by$.

As noted, by linearity $\,\omega(ax+by) = T(ax+by) = a T(x) + bT(y) = a\lambda x + b \mu y\,$, and by linear independence $\,a \lambda = a \omega\,$ and $\,b\mu = b \omega\,$. Multiplying the first equality by $b$ and the second one by $a$ then subtracting gives $\require{cancel}\,ba\lambda-ab\mu=ba\omega-ab\omega=0 \iff ab(\lambda-\mu)=0\,$, and since $\,\lambda \ne \mu\;\ldots$

Answer 2

Your approach is correct. Now that $\lambda \ne \mu$, there can be no $\omega$ that simultaneously satisfies both the equations (assuming a and b are not equal to zero). So the possible solutions are,

• a or b = 0 and $\omega$ taking the appropriate value to satisfy the other equation.

• a, b both are 0, which means the vector becomes a 0 vector and its a trivial case.

So if ax+by is an eigenvector of T, then either a=0 or b=0.