so we are given a parameter $p$ and our new basis to test whish is a subspace for $\mathbb{R}^n$ is given with: $$ a_i=(p,...,p)+\frac{1}{i}e_i ,1\leq i \leq n$$ where $e_i$ is the canonical basis vector for $\mathbb{R}^n$ . So this problem boils down to findig the rank of a matrix depending on a parameter. $$ \left|\begin{matrix} p+1&p&\cdots&p\\ \vdots&\vdots&\ddots&\vdots\\ p&p&\cdots&p+\frac{1}{n} \end{matrix}\right| $$
MY SOLUTION : denote triangular number sequence as $t_n=\frac{n(n+1)}{2}$
so after we row reduce we get the main diagonal with this general term $d_n=\frac{\frac{n(n+1)}{2}p+1}{n(\frac{n(n-1)}{2}p+1)}$ where top half is $u_n=\frac{n(n+1)}{2}p+1$ and before getting a zero there is an assumption of $p$ will not be for example we suppose wether or not $p$ is equal or not to $-1$ and so on for consecutive members.
so i want to prove that the dimension will be $n-1$ when $p=\frac{-2}{n(n+1)} $ and $n$ otherwise. and in my proof i go about my way with induction proving $n=2$ and supposing $n$ then adding $(n+1)$-th vector but then i have no idea other than simply calculating or saying that each member to the right of the diagonal will be of the form $\frac{p}{u_i}$ for the i-th row and then just calculating that after adding $a_{n+1}$ the member under $d_{n-1}$ is equal to $\frac{p}{u_{n-1}}$ and the $d_n=\frac{(t_{n-1}+1)p+1}{nu_{n-2}}$ before adding the $(n-1)$-th row to get the upper triangular form.
and the case where $p=\frac{-1}{t_{n-k}}$ for $1\leq k \leq n-2$ we just switch it with the row after and still get dimension number $n$ as the last row wont be all zeroes
thank you for any help in advance