# Inverse of a symmetric matrix in which upper triangular elements for a given column are equal to the corresponding diagonal element

I've been studying a symmetric matrix with the following form:

$$\begin{bmatrix} a_1 & a_2 & a_3 & a_4 & \cdots & a_n \\ a_2 & a_2 & a_3 & a_4 & \cdots & a_n \\ a_3 & a_3 & a_3 & a_4 & \cdots & a_n \\ a_4 & a_4 & a_4 & a_4 & \cdots & a_n \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ a_n & a_n & a_n & a_n & \cdots & a_n \end{bmatrix}$$

where $$a_i > 0$$ for $$i = 1, \dots, n$$.

I was curious if a matrix with this form has been described previously and, if so, what properties it has, particularly its inverse.

• Note the $(i,j)$ entry is equal to $a_{\max(i,j)}$. Dec 3, 2019 at 10:22
• Did you try for the $2 \times 2$ and $3 \times 3$ cases? Dec 3, 2019 at 10:26
• Hint: Try Guassian Elimination Dec 3, 2019 at 10:51

Denote the given matrix by $$S_n$$.

$$S_n^{-1}$$ exists if and only if $$a_n \neq 0$$ and $$a_i \neq a_{i-1} ~\forall~ i=2,...,n$$.

To see this perform elementary row operations to transform the given matrix to

$$\left[ \begin{matrix} a_1 & a_2 & a_3 &\cdots & a_n \\ a_2-a_1 & 0 & 0& \cdots & 0 \\ a_3-a_1 & a_3-a_2 & 0 &\cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a_n-a_1 & a_n-a_2 & a_n-a_3 & \cdots & 0 \\ \end{matrix} \right]$$

Now expanding the determinant along the last column we get

$$\text{det}(S_n)=(-1)^{1+n}a_n\prod_{i=2}^{n} (a_i-a_{i-1})$$ from which the above property is apparent.

Also, note that the leading principal submatrix of $$S_n$$ could be thought of as $$S_{n-1}$$ which could lead to nice proofs by induction.