How to prove span does not change with change of basis This is one of my tutorial questions and I am having trouble solving it.
Let $\mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_m$ be vectors in $\mathbb{R}^n$ such that span$\{\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_m\}=\mathbb{R}^n$.
$A$ be an invertible matrix of order $n$.
Show that span$\{A\mathbf{u}_1, A\mathbf{u}_2, \ldots, A\mathbf{u}_m\}=\mathbb{R}^n$.
To me it seems that the left multiplication of A changes the basis of the vector space. How can I from here show that, for every vector $\mathbf{v}\in\mathbb{R}^n$ , $\mathbf{v}$ is a linear combination of $A\mathbf{u}_1, A\mathbf{u}_2, \ldots, A\mathbf{u}_m$?
 A: Suppose that $\text{span} (u_i) = \mathbb{R}^n$. Let $A$ be an invertible matrix. We want to show that every $x \in \mathbb{R}^n$ can be written as a linear combination of $A(u_i)$.
First, note that since $A$ is invertible, it is onto. Hence, there exists $y$ such that $Ay =x$. 
Since $\text{span} (u_i) = \mathbb{R}^n$, $y$ can be written as a linear combination of $u_i$:
$$
y = \sum_{i=1}^m \alpha_i u_i
$$ 
Now, note that because $A$ is invertible, $A^{-1}A(u_i) = u_i$ for all $i$. This leads to:
$$
y = \sum_{i=1}^m \alpha_i A^{-1}(A u_i)
$$
Since $A^{-1}$ is a linear operator, we can absorb the sum inside:
$$
y = A^{-1} \bigg(\sum_{i=1}^m \alpha_i(A u_i)\bigg)
$$
Applying $A$ on both sides,
$$
x = Ay = \sum_{i=1}^m \alpha_i(A u_i)
$$
Hence, $x \in \text{span} (Au_i)$. Furthermore, we also have a way to write $x$ in terms of $u_i$ : find the inverse of $x$, write it in terms of $u_i$ and change $u_i$ to $Au_i$, keeping the coefficients the same.
Hence, $\mathbb{R}^n = \text{span} (Au_i)$.
You can go the other way as well, namely that if $A$ is an operator such that $\text{span}\{Au_i\} = \text{span}\{u_i\} = \mathbb{R}^n$, then $A$ is invertible. 
Therefore, if $A$ is singular, then it will not preserve this property.
And the proof of that, is as follows: Whenever $A$ is singular, it's rank is less than $n$ (or it is not full rank). Let $\text{im}(A) = \{Ax : x \in \mathbb{R}^n\}$. Then, for $\{ u_i\} \subset \mathbb{R}^n$, $\{ Au_i\} \subset \text{im}(A)$. Hence, $\text{span}(Au_i) \subset \text{im}A \subsetneq \mathbb{R}^n$. Hence, the span of a set of vectors under a singular map, cannot give the whole space, it can only give a subspace with dimension equal to the rank of the singular matrix (which we can only say is less than $n$, unless we inspect the matrix itself).
