T/F Question of linear independence

[Mr Davis 97]

Homework Statement

T/F: Let $T: V \rightarrow W$. If $\{v_1,v_2,...,v_k \}$ is a linearly independent set, then $\{T(v_1), T(v_2),..., T(v_k) \}$ is linearly independent.

Homework Equations

The Attempt at a Solution

This seems to be true, because we know that $a_1v_1 + a_2v_2 + \cdots + a_k v_k = 0$ has only the trivial solution, since they are linearly independent. Then if we apply $T$ to both sides, we get that $a_1T(v_1) + a_2T(v_2) + \cdots + a_kT(v_k) = 0$, which should mean that this only has the trivial solution as well, right? Meaning that they are linearly independent?

 

[fresh_42]

What if $T=0$?

 

[Mr Davis 97]

What if $T=0$?

So does a problem arise only when $\{v_1,v_2,...,v_k \} \in \text{Ker}(T)$?

 

[Mark44]

So does a problem arise only when $\{v_1,v_2,...,v_k \} \in \text{Ker}(T)$?

No. Think about this in terms of simple examples, such as T(x, y) = <x, 0>; i.e. T is the projection onto the x-axis. The vectors <1, 0> and <0, 1> form a basis for R². Do the vectors T(<1, 0>) and T(<0, 1>) also form a basis for the same space?

 

[fresh_42]

Then if we apply $T$ to both sides, we get that $a_1T(v_1) + a_2T(v_2) + \cdots + a_kT(v_k) = 0$, which should mean that this only has the trivial solution as well, right?

The point is: $a_1=\ldots =a_k=0$ is always a solution, but to be linearly independent, it has to be - as you correctly pointed out - the only possible solution. So every summand $a_i T(v_i) = 0$ allows two possibilities to become zero. And even worse, what if they are all unequal to zero, but add up to zero? Do you have an idea which property of $T$ would be needed, to conclude, that all $a_i=0$?

Your mistake was to start with $a_1v_1+\ldots+a_kv_k=0$. But you want to find out, what $x_1T(v_1)+\ldots+x_kT(v_k)=0$ means for the $x_i$, so better proceed the other way around.

 

[Mr Davis 97]

The point is: $a_1=\ldots =a_k=0$ is always a solution, but to be linearly independent, it has to be - as you correctly pointed out - the only possible solution. So every summand $a_i T(v_i) = 0$ allows two possibilities to become zero. And even worse, what if they are all unequal to zero, but add up to zero? Do you have an idea which property of $T$ would be needed, to conclude, that all $a_i=0$?

Your mistake was to start with $a_1v_1+\ldots+a_kv_k=0$. But you want to find out, what $x_1T(v_1)+\ldots+x_kT(v_k)=0$ means for the $x_i$, so better proceed the other way around.

So, taking your advice, we start with $a_1T(v_1)+\ldots+a_kT(v_k)=0$, which means that $T(a_1v_1 + \cdots + a_nv_n) = 0$. We can only then conclude $a_1v_1 + \cdots + a_nv_n = 0$ from this if $T$ is injective, right?

So is it true that you need $T$ to be injective to preserve linear independence?

 

[fresh_42]

So, taking your advice, we start with $a_1T(v_1)+\ldots+a_kT(v_k)=0$, which means that $T(a_1v_1 + \cdots + a_nv_n) = 0$. We can only then conclude $a_1v_1 + \cdots + a_nv_n =$ from this if $T$ is injective, right?

So is it true that you need $T$ to be injective to preserve linear independence?

That's right. Beside the missing $0$. And if you have $a_1v_1 + \cdots + a_nv_n = 0$, then you can apply the given fact, that the $v_i$ are linearly independent.

Edit: One more important remark. If $T$ is not injective, the $\{T(v_1),\ldots,T(v_k)\}$ could still be linear independent, e.g. if $k=1$ and $T(v_1) \neq 0$. But in general, we just don't know, if $T$ isn't injective. The statement, however, could still be true - or wrong. It all depends on what $\left. T\right|_{span\{v_i\}}$ does.