Let $T: V \to V$ be a linear operator.
- Prove that if T is injective and $\{v_i\}$ is linearly independent, then $\{Tv_i\}$ is also linearly independent.
- Prove that if T is surjective and $\{v_i\}$ spans V, then $\{Tv_i\}$ also spans V.
Proof of 1: Assume that T is injective and $\{v_i\}$ is linearly independent. By definitions,
- A linear operation T is called injective if $x=0$ whenever $Tx=0$. (i.e. If $Tx=0$ then $x=0$)
- $\{v_i\}_{i=1}^k$ is linearly independent if the only set of scalars $\{ c_1, c_2, \dots, c_k\}$ which gives $\sum_{i=1}^k c_iv_i=0$ is the zero set $c_1=c_2=\dots=c_k=0$.
\end{itemize}
We need to show that $\{Tv_i\}_{i=1}^k$ is linearly independent if the only set of scalars $\{ a_1, a_2, \dots, a_k\}$ which gives $\sum_{i=1}^k a_iTv_i=0$ is the zero set $a_1=a_2=\dots=a_k=0$.
How do I apply the definitions?
Proof of 2: What do it mean $\{ v_i \}$ spans V?
Best Answer
Ok, so let's check which scalars make $\sum \limits_{i = 1}^{k} \alpha_{i} T(v_{i})$ equal to $0$.
Note that $T$ is linear, so we know $\sum \limits_{i = 1}^{k} \alpha_{i} T(v_{i}) = T(\sum \limits_{i = 1}^{k} \alpha_{i}v_{i})$, and we want to know which $\alpha_{i}$ make this equal to $0$.
So we want to know for which $\alpha_{i}$ is $T(\sum \limits_{i = 1}^{k} \alpha_{i}v_{i}) = 0$. But $T$ is injective, and by your definition, that means we must have:
$T(\sum \limits_{i = 1}^{k} \alpha_{i}v_{i}) = 0 \implies \sum \limits_{i = 1}^{k} \alpha_{i}v_{i} = 0$.
Finally, since $v_{i}$'s are linearly independent, we know that the only scalars that make the $v_{i}$'s all combine to equal $0$ are the scalars that are all equal to $0$. So this means we must have $\alpha_{i} = 0$ for all $i$.
But this means the only scalar solution to $\sum \limits_{i = 1}^{k} \alpha_{i} T(v_{i}) = 0$ is when all of the $\alpha_{i}$'s are $0$, which is precisely what it means for the set $\{ T(v_{i}) \}$ to be linearly independent.