Proof of Theorem RPNDD

Let $r=\rank{T}$ and $s=\nullity{T}$. Suppose that $R=\set{\vectorlist{v}{r}}\subseteq V$ is a basis of the range of $T$, $\rng{T}$, and $S=\set{\vectorlist{u}{s}}\subseteq U$ is a basis of the kernel of $T$, $\krn{T}$. Note that $R$ and $S$ are possibly empty, which means that some of the sums in this proof are “empty” and are equal to the zero vector.

Because the elements of $R$ are all in the range of $T$, each must have a nonempty pre-image by Theorem RPI. Choose vectors $\vect{w}_i\in U$, $1\leq i\leq r$ such that $\vect{w}_i\in\preimage{T}{\vect{v}_i}$. So $\lt{T}{\vect{w}_i}=\vect{v}_i$, $1\leq i\leq r$. Consider the set \begin{equation*} B=\set{\vectorlist{u}{s},\,\vectorlist{w}{r}} \end{equation*} We claim that $B$ is a basis for $U$.

To establish linear independence for $B$, begin with a relation of linear dependence on $B$. So suppose there are scalars $\scalarlist{a}{s}$ and $\scalarlist{b}{r}$ \begin{equation*} \zerovector=\lincombo{a}{u}{s}+\lincombo{b}{w}{r} \end{equation*}

Then \begin{align*} \zerovector&=\lt{T}{\zerovector}&&\knowl{./knowls/theorem.LTTZZ.knowl}{\text{Theorem LTTZZ}}\\ &=T\left(\lincombo{a}{u}{s}+\right.\\ &\quad\quad\quad\quad\left.\lincombo{b}{w}{r}\right)&&\knowl{./knowls/definition.LI.knowl}{\text{Definition LI}}\\ &=a_1\lt{T}{\vect{u}_1}+a_2\lt{T}{\vect{u}_2}+a_3\lt{T}{\vect{u}_3}+\cdots+a_s\lt{T}{\vect{u}_s}+\\ &\quad\quad b_1\lt{T}{\vect{w}_1}+b_2\lt{T}{\vect{w}_2}+b_3\lt{T}{\vect{w}_3}+\cdots+b_r\lt{T}{\vect{w}_r} &&\knowl{./knowls/theorem.LTLC.knowl}{\text{Theorem LTLC}}\\ &=a_1\zerovector+a_2\zerovector+a_3\zerovector+\cdots+a_s\zerovector+\\ &\quad\quad b_1\lt{T}{\vect{w}_1}+b_2\lt{T}{\vect{w}_2}+b_3\lt{T}{\vect{w}_3}+\cdots+b_r\lt{T}{\vect{w}_r} &&\knowl{./knowls/definition.KLT.knowl}{\text{Definition KLT}}\\ &=\zerovector+\zerovector+\zerovector+\cdots+\zerovector+\\ &\quad\quad b_1\lt{T}{\vect{w}_1}+b_2\lt{T}{\vect{w}_2}+b_3\lt{T}{\vect{w}_3}+\cdots+b_r\lt{T}{\vect{w}_r} &&\knowl{./knowls/theorem.ZVSM.knowl}{\text{Theorem ZVSM}}\\ &=b_1\lt{T}{\vect{w}_1}+b_2\lt{T}{\vect{w}_2}+b_3\lt{T}{\vect{w}_3}+\cdots+b_r\lt{T}{\vect{w}_r}&&\knowl{./knowls/property.Z.knowl}{\text{Property Z}}\\ &=b_1\vect{v}_1+b_2\vect{v}_2+b_3\vect{v}_3+\cdots+b_r\vect{v}_r&&\knowl{./knowls/definition.PI.knowl}{\text{Definition PI}} \end{align*}

This is a relation of linear dependence on $R$ (Definition RLD), and since $R$ is a linearly independent set (Definition LI), we see that $b_1=b_2=b_3=\ldots=b_r=0$. Then the original relation of linear dependence on $B$ becomes \begin{align*} \zerovector&=\lincombo{a}{u}{s}+0\vect{w}_1+0\vect{w}_2+\ldots+0\vect{w}_r\\ &=\lincombo{a}{u}{s}+\zerovector+\zerovector+\ldots+\zerovector&&\knowl{./knowls/theorem.ZSSM.knowl}{\text{Theorem ZSSM}}\\ &=\lincombo{a}{u}{s}&&\knowl{./knowls/property.Z.knowl}{\text{Property Z}} \end{align*}

But this is again a relation of linear independence (Definition RLD), now on the set $S$. Since $S$ is linearly independent (Definition LI), we have $a_1=a_2=a_3=\ldots=a_r=0$. Since we now know that all the scalars in the relation of linear dependence on $B$ must be zero, we have established the linear independence of $S$ through Definition LI.

To now establish that $B$ spans $U$, choose an arbitrary vector $\vect{u}\in U$. Then $\lt{T}{\vect{u}}\in R(T)$, so there are scalars $\scalarlist{c}{r}$ such that \begin{equation*} \lt{T}{\vect{u}}=\lincombo{c}{v}{r} \end{equation*}

Use the scalars $\scalarlist{c}{r}$ to define a vector $\vect{y}\in U$, \begin{equation*} \vect{y}=\lincombo{c}{w}{r} \end{equation*}

Then \begin{align*} \lt{T}{\vect{u}-\vect{y}}&=\lt{T}{\vect{u}}-\lt{T}{\vect{y}}&&\knowl{./knowls/theorem.LTLC.knowl}{\text{Theorem LTLC}}\\ &=\lt{T}{\vect{u}}-\lt{T}{\lincombo{c}{w}{r}}&&\text{Substitution}\\ &=\lt{T}{\vect{u}}-\left(c_1\lt{T}{\vect{w}_1}+c_2\lt{T}{\vect{w}_2}+\cdots+c_r\lt{T}{\vect{w}_r}\right)&&\knowl{./knowls/theorem.LTLC.knowl}{\text{Theorem LTLC}}\\ &=\lt{T}{\vect{u}}-\left(\lincombo{c}{v}{r}\right)&&\vect{w}_i\in\preimage{T}{\vect{v}_i}\\ &=\lt{T}{\vect{u}}-\lt{T}{\vect{u}}&&\text{Substitution}\\ &=\zerovector&&\knowl{./knowls/property.AI.knowl}{\text{Property AI}} \end{align*}

So the vector $\vect{u}-\vect{y}$ is sent to the zero vector by $T$ and hence is an element of the kernel of $T$. As such it can be written as a linear combination of the basis vectors for $\krn{T}$, the elements of the set $S$. So there are scalars $\scalarlist{d}{s}$ such that \begin{equation*} \vect{u}-\vect{y}=\lincombo{d}{u}{s} \end{equation*}

Then \begin{align*} \vect{u}&=\left(\vect{u}-\vect{y}\right)+\vect{y}\\ &=\lincombo{d}{u}{s}+\lincombo{c}{w}{r} \end{align*}

This says that for any vector, $\vect{u}$, from $U$, there exist scalars ($\scalarlist{d}{s}$, $\scalarlist{c}{r}$) that form $\vect{u}$ as a linear combination of the vectors in the set $B$. In other words, $B$ spans $U$ (Definition SS).

So $B$ is a basis (Definition B) of $U$ with $s+r$ vectors, and thus \begin{equation*} \dimension{U}=s+r=\nullity{T}+\rank{T} \end{equation*} as desired.