How does this proof of the Nullity Plus Rank theorem work?

Question

In Apostol's Calculus Volume 2, he states the Nullity Plus Rank Theorem as:

If $V$ is finite-dimensional, then $T(V)$ is also finite-dimensional, and we have
$\dim N(T)+\dim T(V)= \dim V~~~~~$(2.1).
In other words, the nullity plus rank of a linear transformation is equal to the dimension of it domain.

His proof of this is as follows:

Let $n=\dim V$ and let $e_1,...,e_k$ be a basis for $N(T)$ where $k=\dim N(T) ≤ n$.
By Theorem 1.7, these elements are part of some basis for $V$, say the basis
$e_1,...,e_k,e_{k+1},...,e_{k+r},~~~~~$(2.2)
where $k+r=n$. We shall prove that the $r$ elements
$T(e_{k+1}),...,T(e_{k+r})~~~~~$(2.3)
form a basis for $T(V)$, thus proving that $\mathrm {dim}~T(V)=r$. Since $k+r=n$, this also proves (2.1).
First we show that the $r$ elements in (2.3) span $T(V)$. If $y\in T(V)$, we have $y=T(x)$ for some $x$ in $V$, and we can write $x=c_1e_1+...+c_{k+r}e_{k+r}$. Hence, we have
$$y=T(x)=\sum_{i=1}^{k+r}c_iT(e_i)=\sum_{i=1}^kc_iT(e_i)+\sum_{i=k+1}^{k+r}c_iT(e_i)=\sum_{i=k+1}^{k+r}c_iT(e_i) $$ since $T(e_1)=\ldots=T(e_k)=0$. This shows that the elements in (2.3) span $T(V)$.
Now we show that these elements are independent. Suppose that there are scalars $c_{k+1},...,c_{k+r}$ such that $$\sum_{i=k+1}^{k+r}c_iT(e_i)=0\text{.}$$ This implies that $$T\left( \sum_{i=k+1}^{k+r}c_i e_i\right)=0$$ so the element $x=c_{k+1}e_{k+1}+\ldots+c_{k+r}e_{k+r}$ is in the null space $N(T)$. This means that there are scalars $c_1,\ldots,c_k$ such that $x=c_1e_1+\ldots+c_ke_k$, so we have $$x-x=\sum_{i=1}^k c_i e_i- \sum_{i=k+1}^{k+r} c_i e_i= 0\text{.}$$ But since the elements in (2.2) are independent, this implies that all the scalars $c_i$ are zero. Therefore, the elements in (2.3) are independent.

Now I understand most of the proof, until the proof of independence of the elements of (2.3). Why do the $c_i$ have to be zero in the last equation. How does the fact that $c_i$ are $0$ there show that the $c_i$ in $\sum_{i=k+1}^{k+r}c_iT(e_i)=0$ are also $0$?

Answer 1

Recall the definition of linear independence:

$v_1, \ldots, v_n$ are linearly independent iff for every time we have scalars $c_1, \ldots,c_n$ such that $\sum_{i=1}^n c_i v_i = 0$ we can conclude that $\forall {1 \le i \le n} : c_i=0$.

So in order to start the proof of linear independence of $\{T(e_{k+1}), \ldots, T(e_{k+r})\}$ we have to start with an arbitrary linear combination of them that yields $0$, so $$\sum_{i=k+1}^{k+n} c_i T(e_i)=0$$ for some scalars $c_i, k+1\le i \le k+r$.

It is then shown that $$x= \sum_{i=k+1}^{k+r} c_i e_i \in N(T)$$ so $x$ is a linear combination of the first $k$ many $e_i$, so indeed we can also write $$x= \sum_{i=1}^{k} c_i e_i$$ for some extra scalars $c_i, 1 \le i \le k$.

But then (using $x-x = 0)$:

$$\sum_{i=1}^{k} c_i e_i + \sum_{i=k+1}^{k+r} (-c_i) e_i = 0\tag{1}$$

so we have in $(1)$ some new linear combination of independent vectors (we know $\{e_1, \ldots, e_{k+r}\}$ is a base of $V$, so independent for sure!) and the definition I quoted then tells us that necessarily:

$$c_1 = \ldots c_k = -c_{k+1} = \ldots -c_{k+n}=0$$

and in particular all $c_i$ are $0$ for $i \in \{k+1, \ldots, k+r\}$, as $-c=0$ iff $c=0$ (in any additive group) and this was the goal from the start.

Answer 2

The last equation says that a linear combination of the $e_i$, a basis for $V$, is zero. Since the $e_i$ are linear independent we get that all these coefficients are zero. The $c_i$ are still the same as in the equation $$\sum_{i=k+1}^{k+r}c_iT(e_i) = 0$$for all $k+1 \leq i \leq k+r$. Therefore that is exactly what we wanted to show.

Answer 3

Since the $e_i$ are chosen to be a basis of $V$, they must be independent (because that is part of the definition of a basis). Then, the only way for a linear combination of independent vectors to be zero is if all of the coefficients are zero, so the $c_i$ have to be zero.

The $c_{k+1},\dots,c_{k+r}$ are the same, since for a linear transformation $T$ we have that $\sum_{k+1}^{k+r} c_i T(e_i) = T(\sum_{k+1}^{k+r} c_ie_i)$