= Solution
{c}
The <transpose> is a <linear map>, so $T(A)=A+A^T$ is a <linear map> on the $n^2$-dimensional <vector space> $M_{n\times n}(F)$. Let $E_{ij}$ denote a <matrix unit>. The answer depends on the <characteristic of a field>.
If $\operatorname{char}F\ne2$, then $T(A)=0$ means $A^T=-A$. Such <skew-symmetric matrices> have zero diagonal, and their entries above the diagonal can be chosen freely. A <basis> of the <kernel of a linear map> is $E_{ij}-E_{ji}$ for $i<j$, giving <nullity> $n(n-1)/2$. The <image of a linear map> consists of all <symmetric matrices>: $T(A)$ is symmetric, and every <symmetric matrix> $S$ is $T(S/2)$. Its <basis> comprises $E_{ii}$ and $E_{ij}+E_{ji}$ for $i<j$, giving <rank of a linear map> $n(n+1)/2$.
If $\operatorname{char}F=2$, then $-A=A$, so $\ker T$ consists of all <symmetric matrices>, with <dimension> $n(n+1)/2$. Every $T(A)$ is symmetric with zero diagonal, and every symmetric zero-diagonal <matrix> $S$ arises by taking $A_{ij}=S_{ij}$ for $i<j$ and all other entries zero. The <image of a linear map> therefore has <basis> $E_{ij}+E_{ji}$ for $i<j$, and <dimension> $n(n-1)/2$. This is the <symmetrization map in characteristic two>.
In both cases the <rank-nullity theorem> is verified explicitly:
$$
\boxed{\begin{array}{c|cc}
&\dim\ker T&\dim\operatorname{im}T\\\hline
\operatorname{char}F\ne2&\dfrac{n(n-1)}2&\dfrac{n(n+1)}2\\
\operatorname{char}F=2&\dfrac{n(n+1)}2&\dfrac{n(n-1)}2
\end{array}\qquad
\dim\ker T+\dim\operatorname{im}T=n^2.}
$$
Back to article page
