Given a matrix A, I need determine the number of positive, negative and zero eigen value of it without actually computing it?. The matrix is given below
\begin{pmatrix}0&...&0 & 1\\ 0&...&0 & 2\\ .&.&. &.\\ .&.&. &.\\ 0&...&0 &n-1\\ 1&...&n-1&n \end{pmatrix}
An alternative proof could use Sylvester's law of inertia i.e. the number of positive, negative and zero eigenvalues of a given symmetric matrix $A$ does not change under a change of coordinates $A \rightarrow T'AT$ ($\det T \neq 0$).
Since you can easily use the $1$ in the first row/ column to eliminate any non-zero entries up to the $n$ in the lower right corner you are basically left with the task to determine the number of positive and negative eigenvalues of $B=\begin{pmatrix} n && 1 \\ 1 && 0 \end{pmatrix}$. By computation or the same argument that was stated by @Sasha we get that $B$ has one positive and one negative Eigenvalue and therefor A has one positive, one negative and n-2 zero eigenvalues.
In general, when $u,v\in\mathbb{R}^n$ are linearly independent, $uv^\top+vu^\top$ always have one positive eigenvalue, one negative eigenvalue and $n-2$ zero eigenvalues. See an explanation in my answer to another question a week ago. In your case, we have $u^\top = (1,2,\ldots,n-1,\frac{n}{2})$ and $v^\top=(0,0,\ldots,0,1)$.
Assume $n > 1$.
Clearly vectors $v_k = -k \mathbf{e}_1 + \mathbf{e}_k$ are zero eigenvalues one for $1 < k < n$. Thus there are $n-2$ zero eigenvalues.
The trace of the matrix equals $n$, thus the sum of all eigenvalues is positive. All eigenvalues are real, because the matrix is symmetric. Thus, there is 1 positive, one negative, and $n-2$ zero eigenvalues.
