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Consider the $n\times n$ matrix $A$ and the basis $\{\vec{v_1}\ldots \vec{v_n}\}$ for $\mathbb{R}^n$. Prove if $\{A\vec{v_1} \ldots A\vec{v_n}\}$ is a basis for $\mathbb{R}^n$, then A is invertible.

If we let $B=\{A\vec{v_1} \ldots A\vec{v_n}\}$, does this mean the column vectors form a basis and thus $B$ is invertible? How do we prove $A$ is invertible from there?

I think I have to start with $c_1(A\vec{v_1})+\ldots+c_n(A\vec{v_n})=\vec{0}$ where $c_1=\ldots=c_n=0$ but I am not sure where to go after that.

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2 Answers 2

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Hint: Write $v_1,\dots,v_n$ in the basis $Av_1,\dots,Av_n$.

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  • $\begingroup$ If I have $[A\vec{v_1}\ldots A\vec{v_n}]=A[\vec{v_1}\ldots \vec{v_n}]$, how do I manipulate it to show A is invertible? $\endgroup$ Jun 6, 2020 at 22:47
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    $\begingroup$ It’s the other way around $\endgroup$
    – lhf
    Jun 6, 2020 at 23:04
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Since $B = \{Av_1,\dots,Av_n\}$ is a basis, any vector $v\in\mathbb{R}^n$ may be written as $v = \sum_{i = 1}^{n}c_{i}Av_{i} = A\left(\sum_{i=1}^n c_{i}v_{i}\right)$ for some $c_i\in \mathbb{R}$, so $L_A(x):= Ax$ is surjective. Suppose there exists a $v\neq 0$ such that $Av = 0$. Then since $\{v_{1},\dots,v_{n}\}$ is a basis, $\sum_{i = 1}^{n}d_{i}Av_{i} = 0$ for some $d_i\in\mathbb{R}$. Since $B$ is linearly independent, $d_{i} = 0$ for $1\leq i\leq n$, so in fact $v = 0$ and therefore, $L_A$ is injective. So $A$ is invertible.

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