# Invertible Matrices and basis

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.

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

• 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? Jun 6, 2020 at 22:47
• It’s the other way around
– lhf
Jun 6, 2020 at 23:04

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.