Show $A^T$ has an eigenvector with all components rational 
Matrix $A$ is a $5 \times 5$ matrix with rational entries such that $(1, \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5})^T$ is an eigenvector of A. Show that $A^T$ has eigenvector with all components rational.

My idea is: let the eigenvalue associated with the above eigenvector be $λ$. Since all matrix entries are rational numbers so an irrational number will be linearly independent. Use this
$2(a_{11} + 2a_{14}) = a_{41} + 2a_{44}$
$a_{21}+2a_{24} = 0$
$a_{31}+2a_{34} = 0$
$a_{51}+2a_{54} = 0$
but I can't find transposed matrix's eigenvector.

Let $v = (1, \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5})^T$ and assume $A v = \lambda v$. From the first row we get
$$a_{1,1} + 2 a_{1,4} + a_{1,2} \sqrt{2} + a_{1,3} \sqrt{3} + a_{1,5} \sqrt{5} = \lambda$$
From the second row we get:
$$a_{2,1} + 2 a_{2,4} + a_{2,2} \sqrt{2} + a_{2,3} \sqrt{3} + a_{2,5} \sqrt{5} = \lambda \sqrt{2}$$
Now substitute $\lambda$:
$$a_{2,1} + 2 a_{2,4} + a_{2,2} \sqrt{2} + a_{2,3} \sqrt{3} + a_{2,5} \sqrt{5} = (a_{1,1} + 2 a_{1,4} + a_{1,2} \sqrt{2} + a_{1,3} \sqrt{3} + a_{1,5} \sqrt{5}) \sqrt{2}$$
Multiply and rearrange the terms:
$$a_{2,1} + 2 a_{2,4} - 2a_{1,2} + (a_{2,2}-a_{1,1}-2a_{1,4}) \sqrt{2} + a_{2,3} \sqrt{3} + a_{2,5} \sqrt{5} - a_{1,3} \sqrt{6} - a_{1,5} \sqrt{10} = 0$$
Since the roots of the squarefree positive integers are linearly independent over $\mathbb{Q}$, we obtain in particular $a_{1,3} = 0$ and $a_{1,5} = 0$,
so
$$\lambda = a_{1,1} + 2a_{1,4} + a_{1,2} \sqrt{2}.$$
From the third row we get:
$$a_{3,1} + 2 a_{3,4} + a_{3,2} \sqrt{2} + a_{3,3} \sqrt{3} + a_{3,5} \sqrt{5} = \lambda \sqrt{3}$$
Again, substitute $\lambda$:
$$a_{3,1} + 2 a_{3,4} + a_{3,2} \sqrt{2} + a_{3,3} \sqrt{3} + a_{3,5} \sqrt{5} = (a_{1,1} + 2a_{1,4} + a_{1,2} \sqrt{2}) \sqrt{3}$$
Multiply and rearrange the terms:
$$a_{3,1} + 2 a_{3,4} + a_{3,2} \sqrt{2} + (a_{3,3}-a_{1,1}-2a_{1,4}) \sqrt{3} + a_{3,5} \sqrt{5} - a_{1,2} \sqrt{6} = 0$$
So we obtain $a_{1,2} = 0$ and
$$\lambda = a_{1,1} + 2 a_{1,4}$$
Thus $\lambda \in \mathbb{Q}$. Since $A^T$ and $A$ have the same set of eigenvalues, $A^T$ is a matrix with rational components and a rational eigenvalue. Therefore $A^T$ has an eigenvector with rational components.
