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Let $A=(a_{i,j})_{1\leq i,j\leq n}\in M_n(\mathbb{R})$ and $trace(A)=a_{1,1}+\cdots +a_{n,n}$. If $S_n(\mathbb{R}):=\left\{A\in M_n(\mathbb{R}):A^t=A\right\}$ and $\langle \cdot,\cdot \rangle:S_n(\mathbb{R})\times S_n(\mathbb{R})\to \mathbb{R}$ given by $\langle A,B\rangle =trace(AB)$.

What is the dimension of $S_n(\mathbb{R})$?

If $n=2$, I know that $\begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix},\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix},\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$ is a basis of $S_2(\mathbb{R})$

I've seen solutions to this question but they don't use the inner product defined by the trace. How can the trace be used to prove what was requested? any hint to this? Thank you.

eraldcoil
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  • The definition of the set of symmetric matrices doesn't use the trace function at all. Are you sure the trace is not used in a later part of the problem?

    How you prove the problem is by noticing that once you have picked the entries in the upper triangular part (including the diagonal), the rest of the matrix is given if it is to be symmetric.

     – Richard Jensen Aug 18 '21 at 08:25
  • Thanks a lot. I have already tried the exercise. I tested it using that the orthogonal complement of the symmetric matrices are the antisymmetric ones together with the deduction of the dimension of the antisymmetric ones allow to find the dimension of the symmetric ones. – eraldcoil Aug 18 '21 at 15:40

1 Answers1

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Step 1: $(S_n(\mathbb{R}))^{\perp}=\left\{A\in M_n(\mathbb{R}):A^t=-A\right\}$

Step 2: $dim((S_n(\mathbb{R}))^{\perp}=(n^2-n)/2$

Step 3: $M_n(\mathbb{R})=S_n(\mathbb{R})\oplus (S_n(\mathbb{R}))^{\perp}$

Step 4: $dim(S_n(\mathbb{R}))=\frac{n(n+1)}{2}.$

eraldcoil
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