# Invertible Matrix and Linearly Independent Vectors Proof

Trying to do this one:

Suppose $A$ is an invertible $n$ x $n$ matrix and the vectors $v_1$, $v_2$, ..., $v_n$ are linearly independent. Show that the vectors $Av_1$, $Av_2$, ..., $Av_n$ are linearly independent.

I know that A's column vectors are linearly independent since A is invertible. I also know there is no relation amongst the $v_i$ because they're linearly independent.

My idea is to write the product of A and a given $v_i$ in terms of the columns of A. Not sure if this is right, any guidance much appreciated!

Thanks, Mariogs

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Maybe you can do a proof by contradiction? If they are lin dependent, then the zero vector can be written as a lin combination (with nonzero scalars) of your $Av's$ That means that a particular $Av$ vector can be written as a lin combination of the remaining $Av's$. Now what does that mean? – imranfat Feb 27 '14 at 16:33

Let $w_i=Av_i$ for $i=1,\dots,n$. To show linear independence, we must show that $$c_1w_1+\cdots+c_nw_n=0 \Longrightarrow c_i=0,\quad i=1,\dots n.$$ Since $A$ is invertible, we can left-multiply $c_1w_1+\dots=0$ by $A^{-1}$ to get...

Can you take it from there? Remember that the vectors $v_1,\dots v_n$ are linearly independent.

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Ah this makes sense to me, thanks! – bclayman Feb 27 '14 at 16:56
To finish it: We multiply on the left by $A^-1$ and get $c_1 v_1$ + ... + $c1 v_n$ = 0. Since $v_1, ..., v_n$ are linearly independent, we must have $c_i$ = 0 for all i. Yes? – bclayman Feb 27 '14 at 17:00
That's right, yes. – user12477 Mar 3 '14 at 12:05

Notice that since $A$ is invertible then $$Ax=0\iff x=0$$ Now let $a_1,a_2,\ldots,a_n\in \Bbb R$ such that $$\sum_{k=1}^na_k Av_k=0\iff A\left(\sum_{k=1}^na_k v_k\right)=0\iff \sum_{k=1}^na_k v_k=0\Rightarrow a_i=0\;\forall i$$ since $v_1$, $v_2$, ..., $v_n$ are linearly independent. Conclude.

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Whoa! You've been busy with answers! – amWhy Feb 28 '14 at 13:34

Why not use the definition directly? Justify/explain the following:

$$\text{For scalars}\;\;c_1,...,c_n\;:\;\;0=\sum_{k=1}^nc_kAv_k=\sum_{k=1}^nA(c_kv_k)=A\left(\sum_{k=1}^nc_kv_k\right)\iff$$

$$\iff\sum_{k=1}^nc_kv_k=0\iff c_1=c_2=\ldots=c_n=0$$

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