Could a set of $3$ vectors in $\mathbb{R}^4$ span all of $\mathbb{R}^4$?

Question

Could a set of $3$ vectors in $\mathbb{R}^4$ span all of $\mathbb{R}^4$; is this the same as asking if a 4 x 3 matrix could span $\mathbb{R}^4$ or if a 3 x 4 matrix could span $\mathbb{R}^4$?

Answer 1

For $m \lt n$, no set of $m$ vectors will span all of $\mathbb R^n$

In your case, it is not possible for $3$ linearly independent vectors to span all of $\mathbb R^4$: see the Wikipedia Entry on the dimension of a vector space.

Answer 2

Suppose that the three vectors $\vec{v}_1,\vec{v}_2, \vec{v}_3 \in \mathbb{R}^4$ form a linearly independent set $\{\vec{v}_1,\vec{v}_2, \vec{v}_3\}$, then this set can be a basis for some vector space. Because $\{\vec{v}_1,\vec{v}_2, \vec{v}_3\} \subset \mathbb{R}^4$, its span is a subspace of $\mathbb{R}^4$. Now the dimension of a subspace is the number of elements of any of its basis, so the dimension of the subspace spaning by set $\{\vec{v}_1,\vec{v}_2, \vec{v}_3\}$ has dimension $3$. Therefore, because $\mathbb{R}^4$ has dimension $4$, this set can not span it. Note that indeed this set can span subspaces with dimension less than or equal $3$, less than $3$ when there is any linear relationship among them.

Answer 3

If $\vec{v}_1,\vec{v}_2,\vec{v}_3 \in \mathbb{R}^4$, we can find a vector in $\mathbb{R}^4$ outside of the span of the three vectors.

Let $\mathbf{u} \in \mathbb{R}^4$ be such that $\vec{v}_1 \cdot \mathbf{u}=0$, $\vec{v}_2 \cdot \mathbf{u}=0$ and $\vec{v}_3 \cdot \mathbf{u}=0$. The Rank-Nullity Theorem implies non-zero $\mathbf{u}$ exists, since the $3 \times 4$ matrix $$ \left(\begin{matrix} v_1[1] & v_1[2] & v_1[3] & v_1[4] \\ v_2[1] & v_2[2] & v_2[3] & v_2[4] \\ v_3[1] & v_3[2] & v_3[3] & v_3[4] \\ \end{matrix}\right) $$ where $v_i[j]$ is the $j$-th component of vector $\vec{v}_i$, has nullity $\geq 1$. I.e., the above matrix has a non-trivial null space.

Geometrically, $\mathbf{u}$ is orthogonal to $\vec{v}_1$, $\vec{v}_2$ and $\vec{v}_3$.

Answer 4

Compare this question with "n+1 vectors in $\mathbb{R}^n$ cannot be linearly independent" which asks a more or less dual question. Once you know that dimension is well defined, namely that all possible bases of a give vector space have equally many elements, it is quite easy to see that this is so, and that what you ask (having a spanning set of$~\Bbb R^n$ consisting of strictly less than $n$ vectors) is not possible either. Namely for your question, if some number $m<n$ of vectors would span $\Bbb R^n$, which space also has its standard basis of $n~$vectors, then after possibly throwing out some of these$~m$ vectors (in case they should be linearly dependent), one would have a basis of $\Bbb R^n$ with strictly less that $n$ vectors, contradicting the cited fact.

However, proving that dimension is well defined (all bases have the same number of elements), inf fact really comes down to answering both your question (generalised to arbitrary$~n$) and the question cited about linear dependence, so proving those facts using a dimension argument is somewhat circular and hides the need to truly prove something directly from the definitions. One way you can do this is to show that, given any finite basis$~B$ of a vector space$~V$ and also a separate linearly independent set $S$ of $k$ vectors, one can construct a basis of$~V$ consisting of $S$ together with a subset of$~B$ obtained by throwing out exactly$~k$ vectors (this is essentially the Steinitz exchange lemma). For the negative answer to your question you would apply this with for $B$ a basis of strictly less than $n$ vectors extracted from your hypothetic spanning set, and for $S$ the standard basis of$~\Bbb R^n$ that contains $k=n$ vectors; the state result then says that $n$ vectors can be thrown out from $S$, but there are not enough vectors in$~S$ to do that, so this gives a contradiction.