Rank and nullity theorem proof question I was looking at my professor's proof that the dimension of the image is given, following the rank and nullity theorem, by the difference of the dimension of the domain and the dimension of the nullspace. He proves it by showing that if you extend a basis of the kernel using $n-k$ vectors now that basis spans the image. He goes on by showing that those $n-k$ vectors are linearly independent and since the first $k$ vectors go to $0$ since they belong to the nullspace the remaining vectors span the image. I do not see clearly why those vectors should span the whole image rather than just a subspace of it: them being independent does not necessarily entail that they span the whole image right? How can it be shown that they span the entire image?
 A: I will answer in the language of linear transformations. Consider a linear transformation $T:V\rightarrow W$ and assume $\{v_1,\ldots ,v_k\}$ is a basis for ker$T$. Complete it to a basis of $V$ by $\{v_1,\ldots ,v_k,\ldots ,v_n\}$. Then, the set if images $\{Tv_1,\ldots ,Tv_k,\ldots ,Tv_n\}$ span Im$(T)$. Indeed, for $w\in $Im$(T)$ there is a $v\in V$ s.t. $Tv=w$, then write $v$ as a linear combination of your basis and applying $T$ you get the desired result. Consequently, since $v_1,\ldots ,v_k\in $ker$T$ we get that $\{Tv_{k+1},\ldots ,Tv_n\}$ span Im$(T)$. Can you take it from here to show they are linearly independent?
A: In general you have that if $v_1,\dots,v_n$ spans $V$ and $T : V\to W$ is a linear map, then $Tv_1,\dots,Tv_n$ spans range $T$. To prove this, note that for $v\in V$ we have that $v=a_1v_1+\dots+a_nv_n$. Then applying $T$ to both sides gives $Tv=T(a_1v_1+\dots+a_nv_n)$. Using the linearity of $T$ we have that $Tv=a_1Tv_1+\dots+a_nTv_n$. This shows that every $Tv\in $ range $T$ can be written as a linear combination of $Tv_1,\dots,Tv_n$.
Now for rank nullity, let $v_1,\dots,v_m$ be a basis of null $T$. Extend this to a basis of $V$. Let $v_1,\dots,v_m,\dots,v_n$ be that extended basis of $V$. Then for $v\in V$ we have $v=a_1v_1+\dots+a_mv_m+\dots+a_nv_n$. Applying $T$ to both sides gives us $Tv=a_1Tv_1+\dots+a_mTv_m+\dots+a_nTv_n$. Because the vectors $v_1,\dots,v_m$ were in null $T$, we have that $Tv=a_{m+1}Tv_{m+1}+\dots+a_nTv_n$. To show that this list is linearly independent, let $c_{m+1},\dots,c_n$ be scalars such that $c_{m+1}Tv_{m+1}+\dots+c_nTv_n=0$. Then we have that $T(c_{m+1}v_{m+1}+\dots+c_nv_n)=0$. This implies that $c_{m+1}v_{m+1}+\dots+c_nv_n\in $ null $T$. Write this as a linear combination of the basis of null $T$. Then $c_{m+1}v_{m+1}+\dots+c_nv_n=b_1v_1+\dots,b_mv_m$. Subtracting from both sides we get $0=b_1v_1+\dots+b_mv_m-c_{m+1}v_{m+1}+\dots+c_nv_n$. Because the list $v_1,\dots, v_m,\dots,v_n$ is a basis of $V$, the list is linearly independent. This implies that $b_1=\dots=c_n=0$. Thus our original scalars $c_{m+1}= \dots =c_n=0$. Which shows that $Tv_{m+1},\dots,Tv_n$ is linearly independent and is a basis of range $T$. Then you get the formula $\dim V=\dim$ range $T$ + dim null $T$.
A: Suppose $v_1 \ldots v_k$ are a basis for the kernel of a linear operator $T:V \to W$. Suppose we extend this basis with vectors $v_{k+1}, \ldots, v_n$ to a basis for $V$. Then why do the vectors $T(v_{k+1}) \ldots T(v_n)$ span the image of $T$? Well, it is certainly true that the vectors $T(v_{1}), \ldots, T(v_n)$ span the image, since the vectors $v_1, \ldots, v_n$ span $V$. But the first $k$ vectors are in the kernel of $T$, so $$T(v_1) = \ldots = T(v_k)=0.$$ Therefore the span of $T(v_{1}), \ldots, T(v_n)$ is the same as the span of $T(v_{k+1}), \ldots, T(v_n)$.
