 Solution

Define the linear map $A:{\mathbb{R}}^5\to {\mathbb{R}}^2$ by the two prescribed left-hand sides. Then $V=\ker A$, so it is a vector subspace: it contains zero, and linear combinations of vectors with zero image still have zero image.

Subtract the first equation from the second to obtain $a_2=-2a_3-3a_4-4a_5$. The first equation then gives $a_1=a_3+2a_4+3a_5$. ConsequentlyEach of the three displayed vectors satisfies both equations, and the decomposition proves they span $V$. If their linear combination is zero, its third, fourth and fifth coordinates respectively force all three coefficients to vanish. They are therefore linearly independent, giving the basis