- Thread starter
- #1

- Thread starter ssh
- Start date

- Thread starter
- #1

- Admin
- #2

- Jan 26, 2012

- 4,198

$$

\begin{bmatrix}

a_{11} &a_{21} &\dots &a_{n1}\\

a_{21} &a_{22} &\dots &a_{n2}\\

\vdots &\vdots &\ddots &\vdots\\

a_{n1} &a_{n2} &\dots &a_{nn}.

\end{bmatrix}

$$

So the possibly unique entries are all the ones on the main diagonal plus all the ones either above the main diagonal, or all the ones below the main diagonal. So to find out how many there are, let's say you take all the entries on and below the main diagonal. How many in the first row? $1$. How many in the second? $2$. And so on. So the number of possibly unique entries is

$$1+2+\dots+n=\sum_{j=1}^{n}j=\frac{n(n+1)}{2},$$

by a well-known formula.