Answer
Apply Th.8 and the fact that a product of invertible matrices is an invertible matrix (Th.6.b.).
When the columns of $A$ span $\mathbb{R}^{\mathrm{n}}$, $A$ is invertible.
Then, $\mathrm{A}^{2}$is invertible.
Then, the columns of $A^{2}$ span $\mathbb{R}^{n}$
Work Step by Step
Applying Th.8,
$\mathrm{e}.\quad$ The columns of $A$ form a linearly independent set.
and
$\mathrm{a}. \quad A$ is an invertible matrix.
are both true, so the product
$\mathrm{A}\cdot \mathrm{A}=\mathrm{A}^{2}$
is a product of invertible matrices, and is also invertible (by Th.6.b.)
Since
$\mathrm{a}. \quad A^{2}$ is an invertible matrix
is true, then so is
$\mathrm{h}.\ \quad$The columns of $A^{2}$ span $\mathbb{R}^{n}$,
by Theorem 8.
