Show that if A is invertible, then $Au_1,Au_2,\ldots, Au_k$ are linearly independent. There is an added condition of
Suppose $u_1,u_2,\ldots,u_k$ are linearly independent.
Show that if A is invertible, then $Au_1,Au_2,\ldots,Au_k$ are linearly independent.

my solution
since $u_1,u_2,\ldots,u_k$ are linearly independent, then $c_1 u_1+c_2 u_2+\cdots+c_k u_k=0$ implies $c_1,\ldots$ are all zero.
by multiplying A on both sides, we have $c_1 A  u_1+c_2 A u_2+\cdots =c_k A u_k=0$ since all those c can only be 0, therefore $Au_1,Au_2,\ldots,Au_k$ are linearly independent.

Notice how i didnt use the supposition if A is invertive and still get the answer. I am wondering if my answer is right or wrong...
 A: The problem is that you haven’t paid enough attention to just what it is that you’re trying to prove. You want to show that the vectors $Au_1,\dots,Au_k$ are linearly independent. Write down exactly what this means: you want to show that if some linear combination $c_1Au_1+\ldots+c_kAu_k=0$, then $c_1=\ldots=c_k=0$. The most straightforward approach to proving an if-then statement is to assume the hypothesis and work your way to the conclusion. It isn’t always the best approach, or even a feasible approach, but it’s the most straightforward and therefore worth a look.
Assume, then, that $c_1Au_1+\ldots+c_kAu_k=0$. How can you reasonably hope to show that $c_1=\ldots=c_k=0$? One way would be somehow to find linearly independent vectors $v_1,\dots,v_k$ such that $c_1v_1+\ldots+c_kv_k=0$; then you could use the linear independence of the $v_i$ to conclude that $c_1=\ldots=c_k=0$. But where might these $v_i$ come from? Well, we know that $u_1,\dots,u_k$ are linearly independent, and they’re the only vectors about which we’ve been given any information, so perhaps we should try to show that $c_1u_1+\ldots+c_ku_k=0$. Is there any way to deduce that from the hypothesis that $c_1Au_1+\ldots+c_kAu_k=0$? Here is where you’ll use the information that $A$ is invertible.
A: Suppose $c_1Au_1+\ldots+c_kAu_k=0$. Then $A(c_1u_1+\ldots+c_ku_k)=0$. Since $A$ is invertible, we have $c_1u_1+\ldots+c_ku_k=0$. Since $u_1,\ldots,u_k$ are linearly independent, we have $c_1=\ldots=c_k=0$. Hence $Au_1,\ldots,Au_k$ are linearly independent.
A: Hint: Assume by negation they are linear dependent and write
a non-trivial linear combination that sums to zero.
Now this is the step where if the $A$ weren't there then you had
a non-trivial linear combination of the $u_{i}$'s that sums to zero,
can you get rid of the $A$ ? remember that $A$ is invertible means
there exist $A^{-1}$  
ADDED: Your "proof" is incorrect. What you did is take something of the form $0+...+0=0$ and multiplied it by $A$ (each $c_i$ is zero so basicly you summed zeros).
You started out with all the coefficients as zero (before multiplying by $A$), this does not show how any linear combination is zero.
To see this more clearly say I give you all the coefficient as $1$, that is : 
$Au_{1}+...+Au_{k}$ and I claim that $Au_{1}+...+Au_{k}=0$ can you
trace from your proof that this can not happen ?
Clearly it is wrong for take $A=0$ , then it is clear that $\{Au_{i}\}_{i=1}^{i=k}$
is linear dependent since all the vectors are $0$. The assumption
about $A$ is important
