# Columns of a matrix linearly independent and spans

Let $\mathbf v_1, \mathbf v_2, ...,\mathbf v_n \in \Bbb R^n$ and let $P$ be the $n\times n$ matrix whose columns are $\mathbf v_1, \mathbf v_2, ...,\mathbf v_n$

I'm wondering why the followings are equivalent?
(1) $\{\mathbf v_1, \mathbf v_2, ...,\mathbf v_n \}$ is linearly independent.
(2) $\{\mathbf v_1, \mathbf v_2, ...,\mathbf v_n \}$ is a basis for $\Bbb R^n$.
(3) P is invertible.

In particular, why the columns are linearly independent implies that they span?

• Just the linear independence doesn't imply that they span. The linear independence along with the fact that the cardinality of that linearly independent set is $n=\textrm{dim}(\Bbb R^n)$ implies that the set spans $\Bbb R^n$. This actually follows as a corollary of the Replacement Theorem. Apr 30, 2016 at 14:08
• Thank you @Liad. Do you mean this form an expression for any vectors in this space by the n linearly independent vectors? Apr 30, 2016 at 14:23
• Recall the definition of Dim. this will give you 1,2. for 3 - define linear transformation from Rn to Rn defined by Tp(v) = Pv.
– user335501
Apr 30, 2016 at 14:26
• and remember that Im(T) is defined by activating T on basis's vectors
– user335501
Apr 30, 2016 at 14:28

(1)==>(2) Because a basis is a maximal linearly independent set = a minimal generating set, so if you have $\;n\;$ linearly ind. vectors in a space of dimension $\;n\;$ then they're automatically a basis there.
(2)==>(3) If $\;\{v_1,...,v_n\}\;$ are a basis of $\;\Bbb R^n\;$ then Im$\,P=\Bbb R^n\iff P\;$ is surjective $\;\iff P\;$ is injective $\;\iff P\;$ is invertible.
(3)==>(1) If $\;\{v_1,...,v_n\}\;$ weren't linearly independent then upon reducing to echelon form $\;P\;$ at least one row (or column) would be come all zeros, which of course would mean $\;P\;$ isn't invertible, contradiction.