Determine eigenvalue pairs. Assume that $A$ is an $n\times n$ matrix. Assume that $u$ and $v$ are eigenvectors corresponding
to eigenvalues $5$ and $9$. Determine all pairs of real numbers $α$ and $β$ such that $αu+βv$ is also an eigenvector of $A$.

Could someone please advise if this works!
Since u and v are eigenvectors corresponding to eigenvalues 5 and 9, 

We have Au = 5u …(1)
      Av = 9v  …(2)
Now we want to find all pairs of α and β  lets call them  α_i  and β_i  such that (α_i u+β_i v)
 is eigen vector of A i.e.
          $$      A(α_i u+β_i v)=e_i (α_i u+β_i v)
   &lt; => Aα_{i} u+Aβ_{i} v=e_{i} (α_{i} u+β_{i} v)
   &lt; => α_{i} Au+β_{i} Av=e_{i} (α_{i} u+β_{i} v)
    &lt; => α_{i} 5u+β_{i} 9v=e_{i} (α_{i} u+β_{i} v)
    &lt; => e_{i}=((α_{i} 5u+β_{i} 9v))/((α_{i} u+β_{i} v) )   …(3)

 e_{i} $$ will be=k when  
      $$ α_{i}$$  happens to be the multiple of 9 i.e.$$α_{i}=9*k, 
      β_{i} $$ happens to be multiple of 5 i.e.5*k 
 except when k = 0

 Thus for$$ (α_{i} u+β_{i} v) $$ to be eigenvector of A $$ (α_{i},β_{i} )=[(9*k,5*k)]  ∀ k = ±1,±2,±3$$…and k!=0 


So my question is different from the possible duplicate highlighted below - as it has a specific linear combination to deal with
Suppose $\alpha u+\beta v$ is an eigenvector with respect to the eigenvalue $\lambda$:
$$
A(\alpha u+\beta v)=\lambda(\alpha u+\beta v)
$$
By assumption, this can be rewritten as
$$
5\alpha u+9\beta v=\lambda\alpha u+\lambda\beta v
$$
and so
$$
(5-\lambda)\alpha u+(9-\lambda)\beta v=0
$$
Now recall that eigenvectors relative to distinct eigenvalues are linearly independent.
you have $$Av=5v$$ and $$Au=9u$$ now you have to find all a nad b such that $$ A(au+bv)=e(au+bv)$$ where e is new eigenvalue.

you have $$A(au+bv)=(aAu+bAv)=(a*9u+b*5v)=((9a)u+(5b)v)=e(au+bv)$$
but 5 and 9 are prime related to each other, you never can find such a and b,