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## Solution to Linear Algebra Done Wrong Exercise 1.2.4

Write down a basis for the space of

a) $3\times 3$ symmetric matrices;
b) $n\times n$ symmetric matrices;
c) $n\times n$ anti-symmetric ($A^T = -A$) matrices;

Solution:

a) A symmetric $3\times 3$ matrix $A$ must have the form
$$A=\begin{pmatrix} a & b & c \\ b & a’ & b’\\ c & b’ & c’\end{pmatrix}.$$ Define the following matrices
$${\bf e}_{11}=\begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{pmatrix},{\bf e}_{12}=\begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0\end{pmatrix},$$ $${\bf e}_{13}=\begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 1 & 0 & 0\end{pmatrix},{\bf e}_{22}=\begin{pmatrix} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0\end{pmatrix},$$ $${\bf e}_{23}=\begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0\end{pmatrix},{\bf e}_{33}=\begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1\end{pmatrix}.$$ Then the matrix $A$ above can be uniquely expressed as
$$A=a{\bf e}_{11}+b{\bf e}_{12}+c{\bf e}_{13}+a'{\bf e}_{22}+b'{\bf e}_{23}+c'{\bf e}_{33}.$$ Therefore ${\bf e}_{11}$, ${\bf e}_{12}$, ${\bf e}_{13}$, ${\bf e}_{22}$, ${\bf e}_{23}$, ${\bf e}_{33}$ is a basis in the space of symmetric $3\times 3$ matrices.

b) This is very similar to part a). Here we only give the answer. For $1\leqslant i<j\leqslant n$, let ${\bf e}_{ij}$ be the $n\times n$ matrix whose entries in row number $i$ and column number $j$ AND in row number $j$ and column number $i$ are 1 while all other entries are zero, see for example ${\bf e}_{12}$ in a). For $1\leqslant k\leqslant n$, let ${\bf e}_{kk}$ be the $n\times n$ matrix whose entry in row number $k$ and column number $k$ is one while all other entries are zero. Then the list ${\bf e}_{ij}$ for all $1\leqslant i\leqslant j\leqslant n$ is a basis of the space of $n\times n$ symmetric matrices.

c) We first consider the case $n=3$. An anti-symmetric $3\times 3$ matrix $A$ must have the form
$$A=\begin{pmatrix} 0 & a & b \\- a & 0 & c\\ -b & -c & 0\end{pmatrix}.$$ Define the following matrices
$${\bf e}_{12}=\begin{pmatrix} 0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0\end{pmatrix},$$ $${\bf e}_{13}=\begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ -1 & 0 & 0\end{pmatrix},$$ $${\bf e}_{23}=\begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0\end{pmatrix}.$$ Then the matrix $A$ above can be uniquely expressed as
$$A=a{\bf e}_{12}+b{\bf e}_{13}+c{\bf e}_{23}.$$ Therefore ${\bf e}_{12}$, ${\bf e}_{13}$, ${\bf e}_{23}$ is a basis in the space of anti-symmetric $3\times 3$ matrices.

For general $n$, it is quite similar. For $1\leqslant i<j\leqslant n$, let ${\bf e}_{ij}$ be the $n\times n$ matrix such that the entry in row number $i$ and column number $j$ is $1$, AND the entry in row number $j$ and column number $i$ is $-1$ while all other entries are zero, see for example ${\bf e}_{12}$ above. Then the list ${\bf e}_{ij}$ for all $1\leqslant i< j\leqslant n$ is a basis of the space of $n\times n$ anti-symmetric matrices.