One vector $v=a_1^2+1$. The probability that it forms a spanning set is $1$ or less than $1$ depending on if $-1$ is a quadratic residue mod $p$.

Edit: In the linear case, you can just set $v_1=(a_1,a_2)$, $v_2=(a_2,-a_1)$, determinant of the matrix $=a_1^2+a_2^2$. They span if and only if the determinant is nonzero, which happens if and only if $(a_1/a_2)^2=-1$.

One vector $v=a_1^2+1$. The probability that it forms a spanning set is $1$ or less than $1$ depending on if $-1$ is a quadratic residue mod $p$.

Edit: In the linear case, you can just set $v_1=(a_1,a_2)$, $v_2=(a_2,-a_1)$, determinant of the matrix $=a_1^2+a_2^2$. They span if and only if the determinant is nonzero, which happens if and only if $(a_1/a_2)^2\neq-1$.

One vector $v=a_1^2+1$. The probability that it forms a spanning set is $1$ or less than $1$ depending on if $-1$ is a quadratic residue mod $p$.

One vector $v=a_1^2+1$. The probability that it forms a spanning set is $1$ or less than $1$ depending on if $-1$ is a quadratic residue mod $p$.

Edit: In the linear case, you can just set $v_1=(a_1,a_2)$, $v_2=(a_2,-a_1)$, determinant of the matrix $=a_1^2+a_2^2$. They span if and only if the determinant is nonzero, which happens if and only if $(a_1/a_2)^2=-1$.