Exercise 3.2.11

We may write $B_0^{\prime }=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill \\ & & & & & & & & & \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill B^{\prime }\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill \end{array}\right)$. And thus we have that

Let

be a subspace in ${𝔽}^{m+1}$. So we have that $R(L_B)=L+R(B_0^{\prime })$. And it’s easy to observe that all element in $L$ has its first entry nonzero except $0$. But all element in $R(B_0^{\prime })$ has it first entry zero. So we know that $L\cap R(B_0^{\prime })=\{0\}$ and hence $R(L_B)=L\oplus R(B_0^{\prime })$. By Exercise 1.6.29(b) we know that dim$(R(L_B))=$dim$(L)+$dim$(R(B_0^{\prime }))=1+$dim$(R(B_0^{\prime }))$.

Next we want to prove that dim$(R(B_0^{\prime }))=$dim$(B^{\prime })$ by showing $N(L_{B_0^{\prime }})=N(L_{B^{\prime }})$. We may let

and scrutinize the fact

is true. So $N(L_{B_0^{\prime }})=N(L_{B^{\prime }})$ is an easy result of above equalitiies. Finally since $L_{B_0^{\prime }}$ and $L_B^{\prime }$ has the same domain, by Dimension Theorem we get the desired conclusion