First, under the hypotheses the vectors are linearly independent with probability 1. By induction, and it suffices to show that $a_1,...,a_m$ are linearly independent with prob 1 as the general case is a finite union of cases like that.

The matrix whose ith column is $a_i$ consists if i.i.d. entries with a continuous distribution. They are linearly independent if the determinant of the matrix is non 0. Separating out the $a_{11}$ (= first element of $a_1$) term, the determinant is of the form $a_{11}f($(other $a_{ij})$ + $g($other $a_{ij})$. f is the determinant that occurs in a problem one dimension smaller, and is therefore non zero almost surely by the induction hypothesis. Calculate the probability that this is 0 by conditioning on the other $a_{ij}$. By continuity of the distribution, this probability is 0.

For the proposer's problem, do the same, Now the determinant is a quadratic in $a_{11}$as it occurs twice in the relevant matrix. Now it must be shown that the quadratic is not 0. When that is seen to be a non zero quadratic it follows as above that $a_{11}$ satisifies it with probability 0, as it is equal to either of its roots with probability 0 by continuity.

The coefficient of the quadratic term is not the determinant of an identical problem only in a smaller dimension, as above, but it has these features:

$a_{22}, ..., a_{2n}$ occur only once, and the determinant is a linear function of them. There are no cross terms because they are in the same row.

If you substitute $a_{n2}, ..., a_{nn}$ for $a_{22}, ..., a_{2n}$ you get the matrix from an identical problem one dimension smaller, which by the induction hypothesis is almost surely non 0.

It follows that the determinant is if the form $\lambda_2 a_{22}+ ... \lambda_n a_{2n} + \lambda$ where the $\lambda_i$ do not depend on $a_{2i}$ and not all $\lambda_i$ are 0 with probability 1, because if they were, you would get 0 when you plugged in $a_{n2}, ..., a_{nn}$, which is the determinant which is non 0 by the induction hypothesis. The result follows from :

If $X_i$ are independent with continuous distributions and $\lambda_i$ are constants not all 0 the $P ( \lambda_0 + \sum \lambda_i X_i ) = 0$ .