How to write basis for symmetric nxn matrices

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SUMMARY

The basis for the space of nxn symmetric matrices consists of n(n+1)/2 matrices. For a given n, the basis includes n matrices with 1s along the main diagonal and additional matrices formed by placing 1s in positions above the diagonal, with corresponding 1s below. For example, for n=2, the basis matrices are [1, 0; 0, 0], [0, 0; 0, 1], and [0, 1; 1, 0]. For n=3, the basis expands to 6 matrices, demonstrating the systematic approach to constructing symmetric matrix bases.

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Homework Statement



Write down a basis for the space of nxn symmetric matrices.



The Attempt at a Solution



I just need to know what the notation for this sort of thing is. I understand what the basis looks like, and I was even able to calculate that it would have dimension (n/2)(n+1), but I can't think of how you're supposed to compactly write down an answer for this that isn't simply a rambling paragraph. For a given n, I could simply write down a whole bunch of matrices with ones and zeros in appropriate places. But for a general n, I don't know what the format would be.
 
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You could start by defining the canonical basis for the space of nx1 vectors, say [itex]e_i[/itex] = the column vector with a 1 in the i'th position and 0 everywhere else. You can use this to succinctly write the matrix that has a 1 in the (i,j) position and 0 everywhere else, and from there it's easy enough to write a basis for the space of nxn symmetric matrices.
 
First the are the n matrices with 1 at a single entry along the main diagonal and all other entries 0. Now there are [itex]n^2- n[/itex] entries NOT on the main diagonal and so n(n-1)/2 entries above the diagonal. Put a 1 into anyone of those and a 1 in the corresponding position below the main diagonal. That will give you the [itex]n+ \frac{n(n-1)}{2}= \frac{2n+ n^2- n}{2}= \frac{n^2+ n}{2}= \frac{n(n+1)}{n}[/itex] basis matrices.

If n= 2, 2(2+1)/2= 3 and the three basis matrices are
[tex]\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix}0 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}[/tex]

Now, you try to write out the 3(3+1)/2= 6 basis matrices for the 3 by 3 symmetric matrices.
 

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