$T:V \rightarrow W$ be a linear map and $v_1...v_n$ be basis for $V$. Prove $\{T(v_1)...(T(v_n\}$ is basis for range(T). Is this proof correct?

Question

So the theorem is

Let $T : V \rightarrow W$ be a linear transformation. Prove that if $B = \{v_1,\cdots,v_n\}$ is a basis for the domain $V$ then $S = \{T(v_1), T(v_2), \cdots , T(v_n\}$ is a basis for the range $T$.

The proof I came across: (Note that in the proof we assume that we already know $T(v_1),\cdots, T(v_n)$ span $range(T)$

When proving linear independence of $S$, he only considers the case

$$T(O) = O$$

What about the case where $b \in V$ and $b ≠ O$, but

$$T(b) = O$$?

Clearly, we can rewrite $b$ as the linear combination of vectors in $B$

$$b = \sum_{i=1}^{n}c_iv_i$$

And after substituting and rearranging, we have following

$$\sum_{i=1}^{n}c_1T(v_1) = O$$

where not all scalars $c$ are zero.

For the proposition to be correct, this case must be impossible. But why is so?

So in summary:

Question 1

Is the proof above correct? If it is so, why he didn't consider the case where $b \in V$, $b ≠ O$ but $T(b) = O$?

Question 2

Is there an alternative way to prove the theorem? (Preferably I would like see the proof that would start like : "Suppose $v_1,\cdots,v_n$ form a basis. Consider linear combiation $\sum_{i=1}^{n}k_iT(v_i) = O$. Then we have......")

Answer 1

This "theorem" is false, and the "proof" is nonsense for exactly the reason you pointed out.

However, there is an actual (true) theorem which is reminiscent.

Theorem. Let $T:V\to W$ be a linear map between vector spaces $V$ and $W$. Then the following are equivalent.

(i) For any $\mathcal{B}\subseteq V$, we have that $\mathcal{B}$ is a basis for $V$ if and only if $T(\mathcal{B})$ is a basis for $T(V)$.

(ii) $T$ is injective.

(iii) $\mathcal{N}(T)=\{0\}$.

Answer 2

That "proof" is blatantly false. Your worries are 100% justified. In fact, we can show that the conclusion of that theorem is true if and only if there is no $0\neq b\in V$ with $T(b)=0$, which happens if and only if $T$ is injective.

The reason why the "proof" doesn't work is because the author does not consider an arbitrary linear combination (to show linear independence, one can take a linear combination that sums to $0$ and has to show it is trivial), but only considers the linear combination where all the coefficients are already $0$. The "proof" of linear independence is merely an incoherent way of restating that $T(0)=0$.

To immediately see that the conclusion of the "theorem" is false, consider the linear map $T\colon V\rightarrow W,\,v\mapsto0$; if $\{v_1,...,v_n\}$ is a basis for $V$, then $\{T(v_1),...,T(v_n)\}=\{0\}$ is not a basis for $\mathrm{im}(T)=\{0\}$ (the zero vector is not linearly independent). Whatever the text you are reading is, I recommend being very wary of it in the forthcoming considering it has a mistake of this caliber.