The following problem is in Chapter 4 Section 4.3 entitled "Least Squares Approximations" from Gilbert Strang's Introduction to Linear Algebra.
- Usually there will be exactly one hyperplane in $R^n$ that contains the $n$ given points $x=0, a_1, a_{n-1}$. (Example for $n=3$: There will be one plane containing $0,a_1,a_2$ unless $a_1$ and $a_2$ are dependent). That is the test to have exactly one plane in $R^n$?
The solution manual says
Only one plane contains $0, a_1, a_2$ unless $a_1, a_2$ are dependent. Same test for $a_1, ..., a_{n-1}$. If they are dependent, there is a vector $v$ perpendicular to all $a$'s. Then they all lie on the plane $v^Tx=0$ going through the origin.
I'd like to understand why it is that if these $n-1$ vectors are dependent then there is a vector perpendicular to all of them.
Intuitively, it seems that the dimension of the subspace spanned by the vectors is $<n-1$. Therefore, since the vectors are in $\mathbb{R}^n$ then there is another subspace that is orthogonal (and has dimension $>1$; this subspace is the left nullspace of the matrix $A$ that has the initial $n-1$ vectors as columns). Any vector in this subspace is perpendicular to every vector in the subspace spanned by the initial $n-1$ vectors.
Is this the reason?
Even if the $n-1$ vectors were independent, however, the dimension of the orthogonal subspace would still be one. Ie, we could still affirm that there is a vector $v$ that is perpendicular to all the initial vectors.
My original answer was a bit wordy, but here it is
Consider $\mathbb{R}^3$ and three points $0,a_1,a_2$.
If these points are on a line, then there are infinite planes passing through all three points.
One way to see this is that to span a plane two independent vectors are required, and to span a line one independent vector (ie, not the zero vector) is required. In constructing a plane, we can choose as one basis vector a vector on the line, and the other basis vector can be any vector not one the line. The latter means we have infinite possibilities for different planes, one for each basis vector not on the line.
In $\mathbb{R}^4$ with four points, the points cannot all be on the same plane or line.
Intuitively, we have a similar story as before: two independent vectors span a plane, and three independent vectors span a three dimensional vector space. If we select two independent vectors from a specific plane (the one that our points are on), then we can choose any third independent vector to form a three-dimensional hyperplane. This means there are infinite 3d hyperplanes passing through the plane the points are on.
Note that this means that not even two of the three non-zero points can be on the same line passing through zero. If they were, then whatever the third non-zero point, all points would be on a plane passing through zero.
The four points need to span a hyperplane in $\mathbb{R}^4$. Such a hyperplane is a subspace with three dimensions. Thus, when we view the non-zero points as vectors, they need to be independent. Any hyperplane containing these three non-zero points will thus contain this basis, and will thus be the same subspace.
In $\mathbb{R}^n$ with $n$ points, the vectors represented by the $n-1$ nonzero points need to be independent.
Thus, if we form an $n$ by $n-1$ matrix with such $n-1$ vectors as the columns, then the rank must be $n-1$, ie full column rank.