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.
What is the minimum number of vectors to span a plane?
Show Answer
To span a plane, we need at least two vectors.
What is the minimum number of vectors to span a plane from \( \left<10, 4 \right> \),
\( \left<1, 0 \right> \) and \( \left<0, 1 \right> \)?
Show Answer
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. |
What do the following vectors span? \( \vec{u} = \left<3,-1,4 \right>,
\vec{v} = \left<6,-4,-8\right>, \vec{w} = \left<7,-3,4 \right>\)
Show Answer
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. Try
\( \vec{u} = k_1\vec{v} \)
\( \left<3,-1,4 \right> = k_1\left<6,-4,-8 \right> \)
\( 3=6k_1 => k_1 = 3/6 \)
\( -1=-4k_1 => k_1 = -1/-4 \)
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.
\(1=-2a+1b \tag{3}\)
\(1+2a=b \tag{3}\)
\(-1 = -4a-3b \tag{2}\)
\(-1 = -4a-3(1+2a) \tag{2}\)
\(-1 = -4a-3-6a \tag{2}\)
\(-1 = -10a-3 \tag{2}\)
\(-1+3 = -10a \tag{2}\)
\(2 = -10a \tag{2}\)
\(a = \frac{-1}{5} \tag{2}\)
\(1+2\frac{-1}{5}=b \tag{3}\)
\(\frac{3}{5}=b \tag{3}\)
\(3 = 6(\frac{-1}{5}) + 7(\frac{3}{5}) \tag{1}\)
\(3 = \frac{-6}{5} + \frac{21}{5} \tag{1}\)
\(3 = \frac{15}{5} \tag{1}\)
\(3 = 3 \tag{1}\)
There exists a solution to this system of three equations. Therefore, the vectors are
coplanar and span a plane.