Solution B.C12 Chris Black

Let $A$ be an arbitrary element of the specified vector space $T$. Then there exist $a$, $b$, $c$, $d$, $e$ and $f$ so that $A = \begin{bmatrix} a & 0 & 0\\ b & c & 0\\ d & e & f \end{bmatrix}$. Then \begin{align*} A &= a\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} + b\begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} + c\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix} + d\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0\\ 1 & 0 & 0 \end{bmatrix} + e\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 1 & 0 \end{bmatrix} + f\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 0 & 1 \end{bmatrix} \end{align*} Consider the set \begin{align*} B &= \set{ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0\\ 1 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 0 & 1 \end{bmatrix} } \end{align*} The six vectors in $B$ span the vector space $T$, and we can check rather simply that they are also linearly independent. Thus, $B$ is a basis of $T$.