= 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 $a_1=a_3+2a_4+3a_5$. Consequently
$$
(a_1,a_2,a_3,a_4,a_5)=a_3(1,-2,1,0,0)+a_4(2,-3,0,1,0)+a_5(3,-4,0,0,1).
$$
Each 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>
$$
\boxed{\{(1,-2,1,0,0),\ (2,-3,0,1,0),\ (3,-4,0,0,1)\},\qquad\dim V=3.}
$$
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