Dimension reduction over finite fields

Question

Let $\mathbf b_1, \cdots, \mathbf b_n$ be a basis for $\mathbb F^n_q$ (where $\mathbb F_q$ is a finite field of size $q$). Assume $c \in \mathbb F_q$ is a uniformly randomly chosen number. For a given $\mathbf v \in \mathbb F_q^n$, what is the probability that $\mathbf b_1, \cdots, c\mathbf v+ \mathbf b_i, \cdots, \mathbf b_n$ is not a basis for $\mathbb F^n_q$? Isn't it $\frac{1}{|\mathbb F_q|}$? i.e., there is at most one $c\in\mathbb F_q$ that may reduce the dimension of the basis?

Answer 1

It depends on $v$. WLOG, $i=1$. If $v$ is a combination of the other basis vectors $\mathbf{b}_j$, $j\neq i$, then it will always be a basis. If not, then your vectors are linearly dependent iff $$ \det(\mathbf{b}_1+cv, \mathbf{b}_2,\ldots, \mathbf{b}_n)=\det(\mathbf{b}_1,\ldots, \mathbf{b}_n)+c\det(v,\mathbf{b}_2,\ldots, \mathbf{b}_n)=0 $$ which is a linear equation with one solution for $c$, so yes, in this case, the probability is $\frac{1}{q}$.

(Here I assumed that $v$ is fixed and $c$ randomly chosen.)