MHB How many possibly unique entries are there in a symmetric tensor?

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A symmetric tensor has n(n+1)/2 unique quantities due to the property that Aij = Aji, meaning certain entries are equivalent. The tensor is square, leading to a matrix structure where the unique entries consist of those on the main diagonal and either above or below it. To calculate the total unique entries, one can sum the counts of entries in each row, resulting in the formula 1 + 2 + ... + n. This summation simplifies to n(n+1)/2, confirming the total number of unique entries in a symmetric tensor. Thus, the discussion effectively demonstrates the derivation of this formula.
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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?
 
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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.
 
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