# Are all $n\times n$ invertible matrices change-of-coordinates matrices in $\mathbb{R}^n$?

More precisely, I'm trying to prove the following problem:

Assume that $\text{Span}\{\vec{v}_{1},\dots,\vec{v}_{k}\}=\mathbb{R}^{n}$ and that A is an invertible matrix. Prove that $\text{Span}\{A\vec{v}_{1},\dots,A\vec{v}_{k}\}=\mathbb{R}^{n}$.

but I think I need the lemma in the title. I find it difficult to prove without large amounts of handwaving. Is the proposition in the title true, and if so, how to prove it?

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## 4 Answers

Actually, the opposite is true: the problem in your question is half of the proof of the proposition in the title, namely, that $\{Av_1,\dotsc,Av_k\}$ is a spanning set whenever $\{v_1,\dotsc,v_k\}$ is. So, let $x \in \mathbb{R}^n$. Then, of course, $A^{-1}x \in \mathbb{R}^n$, so $A^{-1}x = \sum_{i=1}^k \alpha_i v_i$ for some $\alpha_i \in \mathbb{R}$ -- note that I'm only using the fact that $\{v_1,\dotsc,v_k\}$ is a spanning set, and am assuming absolutely nothing concerning linear independence. Since $x = A(A^{-1}x)$, what follows?

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Yes, but what you're trying to prove is essentially a lemma towards that fact.

Towards the question you've been asked, note that if $\mathrm{Span}\{\vec v_1,\ldots,\vec v_k\}=\mathbb R^n$ then some subset of $\{\vec v_1,\ldots,\vec v_k\}$ is a basis for $\mathbb R^n$, say $\{\vec v_1,\ldots,\vec v_n\}$. Suppose $\text{Span}\{A\vec{v}_{1},\dots,A\vec{v}_{k}\}\neq\mathbb{R}^{n}$. Then $\text{Span}\{A\vec{v}_{1},\dots,A\vec{v}_{n}\}\neq\mathbb{R}^{n}$, so $\{A\vec v_1,\ldots,A\vec v_n\}$ must not be linearly independent. Thus we have some $c_1,\ldots,c_n\in \mathbb R$ not all $0$ such that $c_1A\vec v_1+\cdots+c_nA\vec v_n=0$, thus $A(c_1\vec v_1+\cdots+c_n\vec v_n)=0$. But since $A$ is invertible this gives us $c_1\vec v_1+\cdots+c_n\vec v_n=0$, contradicting the fact that $\{\vec v_1,\ldots,\vec v_n\}$ is a basis. Thus $\text{Span}\{A\vec{v}_{1},\dots,A\vec{v}_{k}\}=\mathbb{R}^{n}$ after all.

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Actually, this is a very roundabout proof which relies fundamentally on the properties of finite-dimensional spaces so doesn't generalize. Branimir's is better. I should probably get some sleep. – Alex Becker Mar 19 '13 at 15:01

Let $v$ be a general vector and set $w = A^{-1} v$. Write $w = \displaystyle\sum_i c_i v_i$ since $\{v_i\}$ is a spanning set. Then $v = A w = A \displaystyle\sum_i c_i v_i = \displaystyle\sum_i c_i (A v_i)$ and so we can write any vector $v$ as a combination of the vectors $\{A v_i\}$

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To span $\mathbb{R}^n$ there need to be at least $n$ linearly independent vectors, so take $k \ge n$.

Assume $k=n$ and remember the definition for linear independence. Define each $\vec{u}_i \equiv A\vec{v}_i$ and consider two vectors, for some $i\ne j$ such that $\vec{u}_i=k\vec{u}_j$ (i.e. assume that there is a linear dependence between two of them) $$A\vec{v}_i = Ak \vec{v}_j \Leftrightarrow A^{-1}A \vec{v}_i=A^{-1}A k\vec{v}_j$$ $\Leftrightarrow\vec{v}_i=k\vec{v}_j$, which is a contradiction.

It follows that no vector exists in $\{\vec{u}_1,\cdots ,\vec{u}_n\}\equiv\{A\vec{v}_1,\cdots ,A\vec{v}_n\}$ such that $\vec{u}_i=k \vec{u}_j$ given the set of linearly independent $\{\vec{v}_1,\cdots,\vec{v}_n\}$ provided that $A$ is an invertible matrix.

In the case where $k>n$ we have

$$\{\vec{v}_1,\cdots ,\vec{v}_n\}\subset\{\vec{v}_1,\cdots ,\vec{v}_k\} \to\{\vec{u}_1,\cdots ,\vec{u}_k\} \supset \{\vec{u}_1,\cdots ,\vec{u}_n\}$$

If $\text{Span}\{A\vec{v}_1,\cdots ,A\vec{v}_n\}=\mathbb{R}^n$ then $\text{Span}\{A\vec{v}_1,\cdots ,A\vec{v}_k\}=\mathbb{R}^n$ since the latter contains, additionally, only linearly dependent vectors.

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