Linear independence and basis within linear transformations

Question

I am new to linear algebra, and without a teacher. I just want to check my understanding based on the following question:

Let $\{v_1,…,v_k\}$ be a subset in the linear space V and $T:V \to V$ a linear transformation.

1. If $\{Tv_1,…,Tv_k\}$ is independent, then $\{v_1,…,v_k\}$ is independent
2. If $\{Tv_1,…,Tv_k\}$ spans V, then $\{v_1,…,v_k\}$ is a basis of $V$.

My answer was "yes" to no. 1 (by definition), and "no" to 2. (because we do not know if $\{v_1,…,v_k\}$ is linearly independent)...

Am I on the right path?

Thank you!

Answer 1

1. It is true, but not “by definition”. If $\alpha_1,\ldots,\alpha_k$ are such that $\alpha_1v_1+\cdots+\alpha_kv_k=0$, then\begin{align}0&=T(0)\\&=T(\alpha_1v_1+\cdots+\alpha_kv_k)\\&=\alpha_1T(v_1)+\cdots+\alpha_kT(v_k).\end{align}Since $\{T(v_1),\ldots,T(v_k\}$ is linearly independent, the $\alpha_i$'s are all equal to $0$.
2. It is false, but you should provide a counterexample.

Answer 2

Your answers are correct but your reasoning is invalid.

As a rule of thumb, if you want to justify "yes" answers, you (usually) need a conventional proof using theorems, definitions, and so forth. But if you want to justify a "no" answer you (usually) need to exhibit a specific counter-example.

So, in your case for #2, you need to give a vector space $V$, a set $\{v_1,\cdots,v_k\}$ which is not a basis for $V$, and a linear map $T:V\to V$ such that $\{Tv_1,\cdots,Tv_k\}$ spans $V$.

The easiest way to do this is to let $T$ be the identity map on $V$ and let $\{v_1,\cdots,v_k\}$ be any overcomplete set.

For #1, it is true but not by definition. If $\{v_1,\cdots v_k\}$ were not independent we would have a nontrivial linear combination $\sum_{i=1}^ka_iv_i=0$ to which we could apply $T$, violating independence of $\{Tv_1,\cdots,Tv_k\}$.