1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Note the $(i,j)$ entry is equal to $a_{\max(i,j)}$. $\endgroup$ – pre-kidney Dec 3 '19 at 10:22
  • $\begingroup$ Did you try for the $2 \times 2$ and $3 \times 3$ cases? $\endgroup$ – Rodrigo de Azevedo Dec 3 '19 at 10:26
  • $\begingroup$ Hint: Try Guassian Elimination $\endgroup$ – Dhanvi Sreenivasan Dec 3 '19 at 10:51
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.