# Linearly independent vectors spanning lower dimensions

The following statement is false:

Suppose the set $\{ v_1, v_2, v_3 \}$ is linearly independent, where $v_1$, $v_2$, and $v_3$ are vectors in $\mathbb{R}^5$. Then $\text{Span}\{ v_1, v_2, v_3 \} = \mathbb{R}^3$.

I completely understand why this would not span $\mathbb{R}^5$. Here's what the reduced-row echelon matrix would look like:

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

While this still contains vectors in $\mathbb{R}^5$, it appears to me that this linear combination would be able to generate all vectors in $\mathbb{R}^3$.

Is it ever possible for vectors of $\mathbb{R}^n$ to ever span a dimension lower than $n$?

Thanks so much for your help!

There is an abuse of notation there, because $\Bbb R^3 \not\subseteq \Bbb R^5$. You can only say that if $\{v_1,v_2,v_3\}$ is a linearly independent set in $\Bbb R^5$, then ${\rm span}\{v_1,v_2,v_3\}$ is a $3-$dimensional subspace of $\Bbb R^5$, which will be isomorphic to $\Bbb R^3$.