How to show an invertible matrix can span a space?

Question

(Vectors in this question are written as column vectors.)

Let $ U_1, U_2, ..., Um$ be vectors in $\mathbb{R}^n$ such that span$\{U_1, U_2,...,U_m\}$ = $\mathbb{R}^n$

a) Let $A$ be an invertible matrix of order $n$. Show that span$\{AU_1, AU_2,...,AU_m\}$ = $\mathbb{R}^n$

(Hint: For every vector $v$ $\in$ $\mathbb{R}^n$, you need to show that $v$ is a linear combination of $AU_1, AU_2,...,AU_m$.)

b) Suppose $A$ is singular. Can span$\{AU_1, AU_2,...,AU_m\}$ = $\mathbb{R}^n$?

For part a, what I have done is that since $A$ is an invertible matrix, it can be obtained by

$A$ = $E_1E_2,...,E_kB$

Since $U_1,U_2,...,Um$ are column vectors and $A$ has no zero rows, span$\{AU_1, AU_2,...,AU_m\}$ = $\mathbb{R}^n$

for part b, I understand that if $A$ is singular, it doesn't have an inverse. However, I don't really know how to link this to the concept in part a nor do I have any idea if I am even on the right track.

Any help is greatly appreciated!

Answer 1

You know that a matrix has an inverse if and only if it is injective and surjective. So if $A$ is singular, it is not injective (A is an endomorphism so it is injective if and only if it is surjective). So you can extract from ${U1,U2,...,Um}$ a base of $\mathbb R^n$, and now you know that ${AU1,AU2,...,AUn}$ is not a base of $\mathbb R^n$ because A is not injective.

Answer 2

Hint for a: if $V\in\mathbb{R}^n$, then $A^{-1}V$ is a linear combination of $U_1,U_2,\dots,U_m$.

Hint for b: the span of $\{AU_1,\dots,AU_m\}$ is contained in the column space of $A$.