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Let $~M_n(\mathbb R)~$ be the vector space of $~n\times n~$ matrices with real entries.

Define $~F: M_n(\mathbb R)\to M_n(\mathbb R)~$ by $~F(A) = A-A^\text{T}.~$

The kernel of $~F~$ is defined to be the subspace of $~M_n(\mathbb R)~$ consisting of all $~X~$ with $~F(X)=0~$. What is the dimension of the kernel of $~F~$?

  • https://math.stackexchange.com/questions/185802/dimensions-of-symmetric-and-skew-symmetric-matrices

    Perhaps this question will help you, since you actually seek the dimension of subspace consisting of all symmetric matrices.

     – I_Really_Want_To_Heal_Myself Aug 02 '19 at 03:49

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Well, by definition $$\ker (F) = \{ A\in \textsf{M}_{n\times n}(\mathbb R) : \, A-A^t = O \}$$ that is, the kernel of $F$ consists of all those matrices that match their own transpose, that is, all the symmetric matrices.

A basis for this subspace is $$\{A^{ij} : \, 1\leq i\leq j\leq n \}$$ where $A^{ij}$ is the $n\times n$ matrix having $1$ in the $i$-th row and $j$-th column, $1$ in the $j$-th row and $i$-th column, and $0$ elsewhere.

It follows that $$\dim \ker (F) = n+(n-1)+\cdots+2+1 = \frac{n(n+1)}{2}$$

azif00
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