Spanning Sets
A Spanning Set is the smallest set of vectors where any vector in a certain space can be represented as a Linear Combination of the spanning set vectors.
The Spanning Set can span a line (1-dimension), a plane (2-dimension) or a space (3-dimensions).
Look how the Standard Basis unit vectors can be used to create Spanning Set:
| Spanning Set | Line | Spanning Set | Plane | Spanning Set | Space |
|---|---|---|---|---|---|
| \(\hat{\imath}\) | \(x\)-dir | \(\hat{\imath}\), \(\hat{\jmath}\) | \(xy\)-plane | \(\hat{\imath}\), \(\hat{\jmath}\), \(\hat{k}\) | 3D space |
| \(\hat{\jmath}\) | \(y\)-dir | \(\hat{\imath}\), \(\hat{k}\) | \(xz\)-plane | ||
| \(\hat{k}\) | \(z\)-dir | \(\hat{\jmath}\), \(\hat{k}\) | \(yz\)-plane |
Given a spanning set of just \(\hat{\imath}\), the only vectors we can make using linear combnations are vectors along the x-axis. Therefore, it spans a line.
\(2\hat{\imath}\),\(-1.5\hat{\imath}\), etc.
Given a spanning set of \(\hat{\jmath}\), \(\hat{k}\), the only vectors we can make using linear combinations are vectors in the yz-plane. Therefore, they span a plane.
\(2\hat{\jmath} + \hat{k}\), \(-2\hat{\jmath} + \hat{k}\), etc.
Given a spanning set of \(\hat{\imath}\), \(\hat{\jmath}\), \(\hat{k}\), we can make any vector in 3D space using linear combinations. Therefore, they space a space.
\(\hat{\imath} + \hat{\jmath} - 0.5\hat{k} \), etc.
To span a plane, we need at least two vectors.
We can remove vectors from the spanning set if they don't help achieve our goal of spanning a plane. Here, \( \left<10, 4 \right> \) is a linear combination of \( \left<1, 0 \right> \) and \( \left<0, 1 \right> \). Therefore, it is Coplanar or on the plane spanned by \( \left<1, 0 \right> \) and \( \left<0, 1 \right> \).
The only vectors we need are \(\left<1, 0 \right> \) and \( \left<0, 1 \right> \). Alternatively, we could use any two of the three vectors.
Let's see a few more examples:
 |
 |
 |
| The lines are Coplanar. They both lie on a plane. | The lines are Coplanar. They all lie on a plane. | The lines are non-Coplanar. They all cannot lie on the same plane. |
 |
 |
 |
| Spans a plane. | Span a plane. | Span a space. |
We need to check if the vectors are collinear or coplanar. If they are all collinear, they will span a line, if they are coplanar, they will span a plane. Otherwise, they span a space:
\(\vec{u} = k_1\vec{v} \)
\(\left<3,-1,4 \right> = k_1\left<6,-4,-8 \right> \)
\(3=6k_1 => k_1 = 3/6 \)
\( -1=-4k_1 => k_1 = -1/-4 \)
Therefore, \(\vec{u} \) and \(\vec{v} \) are not collinear. Using the same steps, you can see that none of the vectors are collinear.
Now let's see if they are coplanar:
\(\vec{u} = a\vec{v} + b\vec{w} \)
\(\left<3,-1,4 \right> = a\left<6,-4,-8\right> + b\left<7,-3,4 \right> \)
\(3 = 6a + 7b \tag{1}\)
\(-1 = -4a-3b \tag{2}\)
\(4=-8a+4b \tag{3}\)
We can pick two questions to solve two unknowns, then see if the third equation works outs.
\(1=-2a+1b \tag{3}\)
\(1+2a=b \tag{3}\)
\(-1 = -4a-3b \tag{2}\)
\(-1 = -4a-3(1+2a) \tag{2}\)
\(-1 = -4a-3-6a \tag{2}\)
\(-1 = -10a-3 \tag{2}\)
\(-1+3 = -10a \tag{2}\)
\(2 = -10a \tag{2}\)
\(a = \cfrac{-1}{5} \tag{2}\)
\(1+2\left(\cfrac{-1}{5}\right)=b \tag{3}\)
\(\cfrac{3}{5}=b \tag{3}\)
\(3 = 6\left(\cfrac{-1}{5}\right) + 7\left(\cfrac{3}{5}\right) \tag{1}\)
\(3 = \cfrac{-6}{5} + \cfrac{21}{5} \tag{1}\)
\(3 = \cfrac{15}{5} \tag{1}\)
\(3 = 3 \tag{1}\)
There exists a solution to this system of three equations. Therefore, the vectors are coplanar and span a plane.