Proof of Theorem SUVB

We must show that the set $B$ is both linearly independent and a spanning set for $\complex{m}$. First, the vectors in $B$ are, by Definition SUV, the columns of the identity matrix, which we know is nonsingular (since it row-reduces to the identity matrix, Theorem NMRRI). And the columns of a nonsingular matrix are linearly independent by Theorem NMLIC.

Suppose we grab an arbitrary vector from $\complex{m}$, say \begin{equation*} \vect{v}=\colvector{v_1\\v_2\\v_3\\\vdots\\v_m}. \end{equation*}

Can we write $\vect{v}$ as a linear combination of the vectors in $B$? Yes, and quite simply. \begin{align*} \colvector{v_1\\v_2\\v_3\\\vdots\\v_m}&= v_1\colvector{1\\0\\0\\\vdots\\0}+ v_2\colvector{0\\1\\0\\\vdots\\0}+ v_3\colvector{0\\0\\1\\\vdots\\0}+ \cdots+ v_m\colvector{0\\0\\0\\\vdots\\1}\\ \vect{v}&=v_1\vect{e}_1+v_2\vect{e}_2+v_3\vect{e}_3+\cdots+v_m\vect{e}_m \end{align*}

This shows that $\complex{m}\subseteq\spn{B}$, which is sufficient to show that $B$ is a spanning set for $\complex{m}$.