My textbook states:

If a subspace $V$ of $\mathbb{R^3}$ is a plane, our geometric intuition tells us that we can find at most two linearly independent vectors in $V$, and we need at least two vectors to span $V$.

What geometric intuition is required for the linearly independent part? I was thinking it had something to do with the dot product (i.e. the vectors have to be orthogonal), but I know of other vectors in $\mathbb{R^3}$ that are not orthogonal but are linearly independent. Is it the fact that two non-orthogonal vectors in $\mathbb{R^3}$ can be linearly independent, but unless two vectors in $\mathbb{R^3}$ are orthogonal, they don't define a plane?


Just to be sure we're working with the same set of definitions:

  • If you have a set of vectors, their span is the smallest vector space that contains all of them.

  • A set of vectors is dependent if one of them is already contained in the span of the others. A set of vectors is independent otherwise.

When only two vectors are concerned, deciding whether or not they're independent is a simple matter. If the two vectors lie on the same line, they are dependent, because either one of them spans the whole line. But if the two vectors do not lie on the same line, then they are independent, and together they span a plane.

When you have three vectors, it gets a little bit more complicated. First of all, notice that in order for all three to be independent, then we surely need the first two vectors to also be independent. So by the above comments, the first two vectors span a plane. Now, in order for all three vectors to be independent, then by definition the third vector must not be in the plane spanned by the first two vectors. So it is outside this plane and the three vectors span a three-dimensional space.

Thus, it is impossible to have three linearly independent vectors in a two-dimensional vector space $V$ -- they need at least three dimensions in order to have the appropriate breathing room. That explains the at most part of the statement.

EDIT: The at least part of the statement follows because if you have only one vector, it only spans a line. You would need an additional vector in order to span the plane $V$.

By the way, linear independence doesn't require orthogonality.

  • $\begingroup$ Thank you! Excellent explanation. Is it possible to explain the "at least two vectors to span $V$" part too? $\endgroup$ – Joel B Sep 28 '14 at 5:37
  • $\begingroup$ I added a quick explanation of that $\endgroup$ – user64480 Sep 28 '14 at 5:40
  • $\begingroup$ Thank you! I appreciate your help. $\endgroup$ – Joel B Sep 28 '14 at 5:55

Vectors in $\mathbb R^3$ are independent iff they are not scalar multiples of each other. So geometrically, you are looking for two vectors that are not parallel or antiparallel (exactly opposite direction).

Now, suppose you had 3 independent vectors. Then pairwise, they each determine a plane...and more importantly, the planes are NOT the same plane. So, with the appropriate constants, you could describe any point in space with those 3 vectors.

  • $\begingroup$ So why is it that, in our case, "we can find at most two linearly vectors in $V$" (emphasis on "at most")? $\endgroup$ – Joel B Sep 28 '14 at 5:15
  • $\begingroup$ Because any two linearly independant vectors form a plane: If you have a third linearly inedependant vector, it means its no longer on that plane, and you need a third dimension. (I'll edit the answer for more details) $\endgroup$ – Alan Sep 28 '14 at 5:18
  • $\begingroup$ Sorry, what do you mean by "need a third dimension?" Aren't we already working in $\mathbb{R^3}$? If we have three linearly independent vectors in $\mathbb{R^3}$, each pair of them forms a different plane (3 planes in total), but all three together form what? $\endgroup$ – Joel B Sep 28 '14 at 5:23

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