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Find a basis for the vector space of symmetric matrices with an order of $n \times n$

This is my thought:
by definition of symmetry, $a_{i,j}=a_{j,i}$.
Therefore, the basis should consist ${n^2-n} \over 2$ matrices to determine each symmetric pair.
In addition, it should also consist $n$ matrices to determine each term in the diagonal.

Therefore, the dimension of the vector space is ${n^2+n} \over 2$.

It's not hard to write down the above mathematically (in case it's true).

Two questions:

  1. Am I right? Is that the desired basis?
  2. Is there a more efficent alternative to reprsent the basis?

Thanks!

AndrePoole
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  • Are there only $n$ entries above the main diagonal? – Casteels Dec 28 '13 at 20:36
  • OK, I think I get it now. $n$ for the diagonal and ${n^2-n} \over 2$ for the symmetric pairs – AndrePoole Dec 28 '13 at 20:48
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    So far you have not given a basis. Also, note that a basis does not have a dimension. The number of elements of the basis (its cardinality) is the dimension of the vector space. – Carsten S Dec 28 '13 at 21:27

3 Answers3

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Let $E_{ij}$ be the matrix with all its elements equal to zero except for the $(i,j)$-element which is equal to one.

Then a desired basis is $$ \frac{1}{2}\big(E_{ij}+E_{ji}\big), \quad 1\le i\le j\le n. $$

Yiorgos S. Smyrlis
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Let $\phi(A) = {A+A^T \over 2}$. Note that $\phi$ is a surjective map onto the space of symmetric matrices. Now choose a basis for the $n \times n$ matrices, then $\phi$ will map these into a spanning set. Now choose a maximal, linearly independent subset.

copper.hat
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Hint: a symmetric matrix is determined by the coefficients on and above the diagonal. The matrix having $1$ at the place $(1,2)$ and $(2,1)$ and $0$ elsewhere is symmetric, for instance. Can you go on? Just take as model the standard basis for the space of all matrices (those with only one $1$ and all other entries $0$).

egreg
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  • Yeah, I think that's what I described. Can you test my explanation? – AndrePoole Dec 28 '13 at 21:07
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    @AndrePoole You didn't mention any specific matrix; I guess your idea was the same as mine, but it wasn't made explicit. The dimension is exactly $(n^2+n)/2$: $n$ for the first row, $n-1$ for the second row, and so on; so $n+(n-1)+\dots+2+1=(n^2+n)/2$. – egreg Dec 28 '13 at 21:09