# Tensor Analysis problem

#### ssh

##### New member
Show that a symmetric tensor has n(n+1) \ 2 quantities.

In a symmetric tensor we have that Aij = Aji which means that
A12 = A21

A23 = A32 and so on. Thus these n quantites are similar. What do we do next?

#### Ackbach

##### Indicium Physicus
Staff member
A symmetric tensor must, by definition, be square. Hence, you have a situation like the following:
$$\begin{bmatrix} a_{11} &a_{21} &\dots &a_{n1}\\ a_{21} &a_{22} &\dots &a_{n2}\\ \vdots &\vdots &\ddots &\vdots\\ a_{n1} &a_{n2} &\dots &a_{nn}. \end{bmatrix}$$
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.