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The full problem statement is:

$A \in M_{nn}$ is called symmetric if $A^T = A$ and $B^T=B$. Let $V = \{ A \in M_{nn} | A$ is symmetric$\}$. Determine a basis for $V$.

I am having difficult starting with this problem. If I understand correctly, $V$ is a set of symmetric matrices. Meaning the only restriction is that for all matrices in $V$, every elements $e_{ij}$ for $0 < i \le j \le n$ must equal their counterparts $e_{ji}$.

The identity matrix $I_n$ seems like it is a basis but I don't know how to determine that. I could verify that $I_n$ spans $V$ and is linearly independent proving it is a basis of some kind, but not necessarily of $V$.

I feel like I'm close to the solution but am missing something.

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  • $\begingroup$ You can take all ${n \choose 2}$ matrices for which $A_{rs}=(a)_{ij}$ such that $a_{ij}=a_{ji}=1$ for $\{i,j\}=\{r,s\}$ and $0$ otherwise. $\endgroup$ – Jlamprong Jan 16 '14 at 5:42
  • $\begingroup$ HOw can identity matrix be basis of this space? Do you mean to say that $V$ is one dimensional? $\endgroup$ – Vishal Gupta Jan 16 '14 at 6:52
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Note that the matrices $E_{ij}$ where the only non zero entry is $a_{ij}=1$ form a basis of $M_{nn}$. Now, $E_{ii}'s$ are present in your basis for $i=1..n$. The other basis members are $(E_{ij}+E_{ji})/2$.

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  • $\begingroup$ +1 -- but, why do you divide by 2? That's not neccessary, is it? $\endgroup$ – user127.0.0.1 Jan 16 '14 at 5:43
  • $\begingroup$ no- it's not. I just did that out of habit I suppose. $\endgroup$ – voldemort Jan 16 '14 at 5:44
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To see what the matrices look like, here are some...

$$ \pmatrix{1 & 0 & \cdots \\ 0 & 0 &\cdots \\\vdots & \vdots & \ddots}, \pmatrix{0& 1 & \cdots \\ 1 &0 &\cdots \\\vdots & \vdots & \ddots}, \pmatrix{0& 0&1 & \cdots \\0&0&0&\cdots \\ 1 &0 &0&\cdots \\\vdots & \vdots & \vdots&\ddots},\cdots,\\ \pmatrix{0 & 0 & \cdots \\ 0 & 1 &\cdots \\\vdots & \vdots & \ddots}, \pmatrix{0& 0&0 & \cdots \\0&0&1&\cdots \\ 0 &1 &0&\cdots \\\vdots & \vdots & \vdots&\ddots},\cdots \\ \text{ etc. } $$

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