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.

  • $\begingroup$ 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

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 $(E_{ij}+E_{ji})/2$.

  • $\begingroup$ +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

To see what the matrices look like, here are some...

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.