I am new to matrix algebra, and trying to understand why
$\operatorname{tr} A = \det A$, where
$$A = \begin{pmatrix}
1 & -1 & \ldots & -1\\
1 & 1 & 0 &0\\
\vdots & 0 & \ddots & 0 \\
1&0&\ldots&1
\end{pmatrix}$$
To be clear about my notation, all off diagonal elements are zeroes except the elements in the first column and first row. The proof of this could be done by using mathematical induction, yet I am interested in knowing why it is true from the matrix algebra perspective.
Best Answer
We can write $$A = \begin{bmatrix}1 & -\vec{1}_n^T \\ \vec{1}_n & I_n\end{bmatrix}$$ where $\vec{1}_n$ is a column vector of $n$ ones and $I_n$ is the $n \times n$ identity matrix. Trivially, $\text{tr}(A) = n+1$. The block matrix determinant formula states that $$\det\left(\begin{bmatrix}A_{11} & A_{12} \\ A_{21} & A_{22}\end{bmatrix}\right) = \det(A_{22})\det(A_{11}-A_{12}A_{22}^{-1}A_{21})$$ provided that $A_{22}$ is invertible. Applying that formula here gives us $$\det(A) = \det(I_n) \det(1+\vec{1}_n^TI_n^{-1}\vec{1}_n) = \det(I_n)\det(1+\vec{1}_n^T\vec{1}_n) = \det(I_n)\det(n+1) = n+1.$$ So $\text{tr}(A) = \det(A) = n+1$.