# Why isn't this set of vectors a basis of planar subspace in ${\bf R}^3$

This is a question in my study guide for linear algebra:

A set of vectors is a basis if it's linearly independent and spans the space. Why isn't the set $$\left\{ \left( \begin{array}{l} 1 \\ 0 \\ 0 \\ \end{array}\right) , \left( \begin{array}{l} 0 \\ 1 \\ 0 \\ \end{array}\right), \left( \begin{array}{l} 0 \\ 0 \\ 1 \\ \end{array}\right) \right\}$$ a basis of a planar subspace of ${\bf R}^3$?

(I will refer to the planar subspace as S)

I am currently pretty lost, I know the requirements for a basis are linear independence and spanning S, and the set given is clearly linearly independent but it also spans R3, and therefore because S is a subspace of R3 also spans S. S however is not stated to span R3 and because it is a planar subspace I suspect that it in fact does not span R3. If this is true then is this the reason the set cannot be a basis for S - because it is "overqualified" and spans more than just S?

Thank you for any help

• By "planar subspace" is probably meant "subspace of dimension $2$" (hence like a plane). The word "dimension" does not appear in your Question. Do you see the relevance? Can you decide what dimension the spanned subspace has? – hardmath Dec 14 '16 at 1:19

A basis of $S$ is a set of maximally linear independant vectors in $S$ which minimally generate $S$. You are right, the canonical basis of $\mathbb{R^3}$ is "overqualified".

• So it could be said that the span of a basis of a subspace is limited to the span of the subspace? – Winguh Dec 14 '16 at 1:23
• Yes, that is what minimally generating means – B.Swan Dec 14 '16 at 1:25
• Excellent, thank you for the help – Winguh Dec 14 '16 at 1:27
• @B.Schnebbler More importantly, as you noted, it must be a set of vectors in S. – Henry Swanson Dec 14 '16 at 1:33

From Wikipedia: "A basis $B$ of a vector space $V$ over a field $F$ is a linearly independent subset of $V$ that spans $V$."

From Wolfram: "A basis of a vector space $V$ is defined as a subset $v_1, \ldots , v_n$ of vectors in $V$ that are linearly independent and vector space span $V$."

You overlooked a small detail: a basis for $S$ must be a subset of $S$. At least one, and possibly all of your proposed basis vectors lie outside $S$.

and therefore because $S$ is a subspace of $\mathbb{R}^3$ also spans $S$.

This is incorrect. When we say "a finite set of vectors $\{v_i\}_1^m$ spans some vector space $V$", we not only mean that any element of $V$ can be written as a linear combination of the $v_i$'s, but also $V$ is the set of all the linear combinations of $v_i$'s. In notations, people usually write: $$V=span\{v_1,\cdots,v_m\}.$$