# 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...

• You have to start with the second equation. Is there some way back to the first, so that you can apply lin.ind. of the $u_i$? – Lord_Farin Oct 18 '12 at 13:55
• I get what you mean, but if i could get the answer without using the suposition, isn tthat better? That is, so i have shown that no matter if a is invertible or not, the answer will always be linearly independent – Yellow Skies Oct 18 '12 at 14:01
• @SingaporeanDude.: But that result is clearly false (take $A=0$ for example), so your reasoning must be flawed. – Chris Eagle Oct 18 '12 at 14:02
• Im still unsure how is it flawed :( – Yellow Skies Oct 18 '12 at 14:04
• Just because the coefficients have to be $0$ for the $\mathbf{u_i}$ to sum to $\mathbf{0}$ doesn't mean the same holds for $A\mathbf{u_i}$. – EuYu Oct 18 '12 at 14:07

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

• Thanks that makes things extremely clear!! – Yellow Skies Oct 18 '12 at 14:17
• @SingaporeanDude. - I'm very glad it helps! :) – Belgi Oct 18 '12 at 14:21

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.

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.