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Section 3.1 Span

In Subsection 2.1.4 we defined the notion of a linear combination of vectors. At that time we saw that determining which vectors can be expressed as linear combinations of a given set of vectors amounts to solving a certain system of linear equations. Take a moment to review Subsection 2.1.4. We are now going to introduce some important terminology related to these same ideas.

Definition 3.1.1.

Let \(\vec{v_1}, \ldots, \vec{v_k}\) be vectors in \(\mathbb{R}^n\text{.}\) The span of \(\vec{v_1}, \ldots, \vec{v_k}\) is the collection of all linear combinations of \(\vec{v_1}, \ldots, \vec{v_k}\text{.}\) Symbolically,

\begin{equation*} \SpanS(\vec{v_1},\ldots,\vec{v_k}) = \{a_1\vec{v_1} + \cdots + a_k\vec{v_k} : a_1, \ldots, a_k \text{ are scalars}\}\text{.} \end{equation*}

Note 3.1.2.

It is important to note that \(\SpanS(\vec{v_1}, \ldots, \vec{v_k})\) is a collection of vectors, and is not a vector. An expression like \(\SpanS(\vec{v_1}, \ldots, \vec{v_k}) = a_1\vec{v_1}+\cdots+a_k\vec{v_k}\) is meaningless, because the left side is a collection of vectors and the right side is a single vector.

Determine whether or not \(\begin{bmatrix}1\\-1\\2\end{bmatrix}\) is in \(\SpanS\left(\begin{bmatrix}3\\-1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\\-1\end{bmatrix}\right)\text{.}\)

Solution.

The question is really asking whether \(\begin{bmatrix}1\\-1\\2\end{bmatrix}\) can be written as a linear combination of \(\begin{bmatrix}3\\-1\\1\end{bmatrix}\) and \(\begin{bmatrix}1\\-1\\-1\end{bmatrix}\text{.}\) That is, we want to know whether or not there are scalars \(a, b\) such that

\begin{equation*} \begin{bmatrix}1\\-1\\2\end{bmatrix} = a\begin{bmatrix}3\\-1\\1\end{bmatrix} + b\begin{bmatrix}1\\-1\\-1\end{bmatrix}\text{.} \end{equation*}

Unpacking this vector equation we arrive at a system of linear equations:

\begin{gather*} 3a+b = 1 \\ -a-b=-1 \\ a-b = 2 \end{gather*}

We can then solve the system by row reducing.

\begin{equation*} \matr{cc|c}{3 \amp 1 \amp 1 \\ -1 \amp -1 \amp -1 \\ 1 \amp -1 \amp 2} \to \matr{cc|c}{1 \amp 0 \amp 3/2 \\ 0 \amp 1 \amp -1/2 \\ 0 \amp 0 \amp -3/4} \text{.} \end{equation*}

From here we see that the system of equations has no solutions, and therefore we conclude that \(\begin{bmatrix}1\\-1\\2\end{bmatrix}\) is not in \(\SpanS\left(\begin{bmatrix}3\\-1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\\-1\end{bmatrix}\right)\text{.}\)

Notice that in the above example the matrix we arrived at had columns \(\begin{bmatrix}3\\-1\\1\end{bmatrix}\) and \(\begin{bmatrix}1\\-1\\-1\end{bmatrix}\) (the vectors whose span we were considering) for its coefficient matrix and the augmented column was the vector \(\begin{bmatrix}1\\-1\\2\end{bmatrix}\) (the vector we were testing for being in the span of the others). There was nothing particularly special about this example: Whenever we want to know if \(\vec{w}\) is in \(\SpanS(\vec{v_1}, \ldots, \vec{v_k})\) we will end up looking at the augmented matrix \([\vec{v_1} \cdots \vec{v_k} | \vec{w}]\text{.}\) The following theorem makes this connection precise.

\(\vec{w}\) is in \(\SpanS(\vec{v_1}, \ldots, \vec{v_k})\) if and only if there are scalars \(a_1, \ldots, a_k\) such that \(\vec{w} = a_1\vec{v_1} + \cdots + a_k\vec{v_k}\text{,}\) and this vector equation is equivalent to the linear system represented by \([\vec{v_1} \cdots \vec{v_k} | \vec{w}]\text{,}\) so such scalars exist if and only if this system is consistent.

So far we have a method for answering the question "is \(\vec{w}\) in \(\SpanS(\vec{v_1}, \ldots, \vec{v_k})\)", where \(\vec{w}, \vec{v_1}, \ldots, \vec{v_k}\) are specific vectors. Very often, however, what we actually want to do is answer questions of the form "describe all of the vectors in \(\SpanS(\vec{v_1}, \ldots, \vec{v_k})\)". Our next examples illustrate how we can approach questions like this.

Describe all of the vectors in \(\SpanS\left(\begin{bmatrix}3\\-1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\\-1\end{bmatrix}\right)\text{.}\)

Solution.

Let \(\vec{w} = \begin{bmatrix}x\\y\\z\end{bmatrix}\text{.}\) By Theorem 3.1.4 we have that \(\vec{w}\) is in \(\SpanS\left(\begin{bmatrix}3\\-1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\\-1\end{bmatrix}\right)\) if and only if the system of linear equations represented by \(\matr{cc|c}{3 \amp 1 \amp x \\ -1 \amp -1 \amp y \\ 1 \amp -1 \amp z}\) is consistent. We therefore row-reduce:

\begin{equation*} \matr{cc|c}{3 \amp 1 \amp x \\ -1 \amp -1 \amp y \\ 1 \amp -1 \amp z} \to \matr{cc|c}{1 \amp 0 \amp \frac{-y+z}{2} \\ 0 \amp 1 \amp \frac{-y-z}{2} \\ 0 \amp 0 \amp x+2y-z}\text{.} \end{equation*}

From here we can see that the system is consistent if and only if \(x+2y-z=0\text{.}\) Thus what we've shown is that the vectors in \(\SpanS\left(\begin{bmatrix}3\\-1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\\-1\end{bmatrix}\right)\) are exactly those vectors \(\begin{bmatrix}x\\y\\z\end{bmatrix}\) where \(x+2y-z=0\text{.}\) As you'll recognize from Section 2.3, these vectors form a plane in \(\mathbb{R}^3\text{.}\)

Describe all of the vectors in \(\SpanS\left(\begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\end{bmatrix}, \begin{bmatrix}3\\2\end{bmatrix}\right)\text{.}\)

Solution.

By Theorem 3.1.4 the vectors in \(\SpanS\left(\begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\end{bmatrix}, \begin{bmatrix}3\\2\end{bmatrix}\right)\) are exactly those \(\begin{bmatrix}x\\y\end{bmatrix}\) where the system represented by \(\matr{ccc|c}{1 \amp 1 \amp 3 \amp x \\ 1 \amp -1 \amp 2 \amp y}\) is consistent. So we row-reduce:

\begin{equation*} \matr{ccc|c}{1 \amp 1 \amp 3 \amp x \\ 1 \amp -1 \amp 2 \amp y} \to \matr{ccc|c}{1 \amp 0 \amp 1/2 \amp \frac{x+y}{y} \\ 0 \amp 1 \amp 5/2 \amp \frac{x-y}{2}}\text{.} \end{equation*}

We now see that this system is always consistent, no matter what \(x\) and \(y\) are, so every vector in \(\mathbb{R}^2\) is in \(\SpanS\left(\begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\end{bmatrix}, \begin{bmatrix}3\\2\end{bmatrix}\right)\text{.}\) That is,

\begin{equation*} \SpanS\left(\begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\end{bmatrix}, \begin{bmatrix}3\\2\end{bmatrix}\right) = \mathbb{R}^2\text{.} \end{equation*}

In the next section it will be useful to know when we can remove a vector from a set without changing the span of that set.

One direction is easy: If \(\SpanS(\vec{v_1}, \ldots, \vec{v_k}) = \SpanS(\vec{v_1}, \ldots, \vec{v_{i-1}}, \vec{v_{i+1}}, \ldots, \vec{v_k})\) then, since \(\vec{v_i}\) is certainly in \(\SpanS(\vec{v_1}, \ldots, \vec{v_k})\text{,}\) we also have that \(\vec{v_i}\) is in \(\SpanS(\vec{v_1}, \ldots, \vec{v_{i-1}}, \vec{v_{i+1}}, \ldots, \vec{v_k})\text{.}\)

For the other direction, suppose that \(\vec{v_i}\) is in \(\SpanS(\vec{v_1}, \ldots, \vec{v_{i-1}}, \vec{v_{i+1}}, \ldots, \vec{v_k})\text{.}\) Since any linear combination of \(\vec{v_1}, \ldots, \vec{v_{i-1}}, \vec{v_{i+1}}, \ldots, \vec{v_k}\) is also a linear combination of \(\vec{v_1}, \ldots, \vec{v_k}\) (by making the coefficient of \(\vec{v_i}\) be \(0\)) to prove that \(\SpanS(\vec{v_1}, \ldots, \vec{v_k}) = \SpanS(\vec{v_1}, \ldots, \vec{v_{i-1}}, \vec{v_{i+1}}, \ldots, \vec{v_k})\) all we need to do is show that every linear combination of \(\vec{v_1}, \ldots, \vec{v_k}\) can be written as a linear combination where \(\vec{v_i}\) does not appear.

By our assumption that \(\vec{v_i}\) is in \(\SpanS(\vec{v_1}, \ldots, \vec{v_{i-1}}, \vec{v_{i+1}}, \ldots, \vec{v_k})\text{,}\) there are scalars \(b_1, \ldots, b_{i-1}, b_{i+1}, \ldots, b_k\) such that \(\vec{v_i} = b_1\vec{v_1} + \cdots + b_{i-1}\vec{v_{i-1}} + b_{i+1}\vec{v_{i+1}}\cdots + b_k\vec{v_k}\text{.}\) Now given any element of \(\SpanS(\vec{v_1}, \ldots, \vec{v_k})\text{,}\) say \(a_1\vec{v_1} + \cdots + a_k\vec{v_k}\text{,}\) we have:

\begin{align*} a_1\amp \vec{v_1} + \cdots + a_k\vec{v_k} \\ \amp = a_1\vec{v_1} + \cdots + a_{i-1}\vec{v_{i-1}} + a_i\vec{v_i} + a_{i+1}\vec{v_{i+1}} + \cdots + a_k\vec{v_k}\\ \amp = a_1\vec{v_1} + \cdots + a_{i-1}\vec{v_{i-1}} + a_i(b_1\vec{v_1} + \cdots + b_{i-1}\vec{v_{i-1}} + b_{i+1}\vec{v_{i+1}}\cdots + b_k\vec{v_k}) + a_{i+1}\vec{v_{i+1}} + \cdots + a_k\vec{v_k}\\ \amp = (a_1+a_ib_1)\vec{v_1} + \cdots + (a_{i-1}+a_ib_{i-1})\vec{v_{i-1}} + (a_{i+1}+a_ib_{i+1})\vec{v_{i+1}} + \cdots + (a_k+a_ib_k)\vec{v_k} \end{align*}

Since the last expression is a linear combination that does not use \(\vec{v_i}\text{,}\) the proof is complete.

Exercises Exercises

1.

In each case, determine if \(\vec{x}\) is in \(\SpanS(\vec{y}, \vec{z})\text{.}\) If it is, write \(\vec{x}\) as a linear combination of \(\vec{y}\) and \(\vec{z}\text{.}\) If not, explain why not.

  1. \(\vec{x} = \begin{bmatrix}2\\-1\\0\\1\end{bmatrix}, \vec{y} = \begin{bmatrix}1\\0\\0\\1\end{bmatrix}, \vec{z} = \begin{bmatrix}0\\1\\0\\1\end{bmatrix}\text{.}\)

  2. \(\vec{x} = \begin{bmatrix}1 \\ 2 \\ 15 \\ 11\end{bmatrix}, \vec{y} = \begin{bmatrix}2\\-1\\0\\2\end{bmatrix}, \vec{z} = \begin{bmatrix}1\\-1\\-3\\1\end{bmatrix}\text{.}\)

  3. \(\vec{x} = \begin{bmatrix}8\\3\\-13\\20\end{bmatrix}, \vec{y} = \begin{bmatrix}2\\1\\-3\\5\end{bmatrix}, \vec{z} = \begin{bmatrix}-1\\0\\2\\-3\end{bmatrix}\text{.}\)

  4. \(\vec{x} = \begin{bmatrix}2\\5\\8\\3\end{bmatrix}, \vec{y} = \begin{bmatrix}2\\-1\\0\\5\end{bmatrix}, \vec{z} = \begin{bmatrix}-1\\2\\2\\-3\end{bmatrix}\text{.}\)

2.

In each case determine if the given vectors span all of \(\mathbb{R}^4\text{.}\) Support your answer.

  1. \(\begin{bmatrix}1\\1\\1\\1\end{bmatrix}, \begin{bmatrix}0\\1\\1\\1\end{bmatrix}, \begin{bmatrix}0\\0\\1\\1\end{bmatrix}, \begin{bmatrix}0\\0\\0\\1\end{bmatrix}\text{.}\)

  2. \(\begin{bmatrix}1\\3\\-5\\0\end{bmatrix}, \begin{bmatrix}-2\\1\\0\\0\end{bmatrix}, \begin{bmatrix}0\\2\\1\\-1\end{bmatrix}, \begin{bmatrix}1\\-4\\5\\0\end{bmatrix}\text{.}\)

Answer.

(a) Yes (b) No.

3.

Suppose that \(\vec{v}, \vec{w}, \vec{z}\) are vectors in \(\mathbb{R}^n\text{,}\) and \(2\vec{v}\) is in \(\SpanS(\vec{w}, \vec{z})\text{.}\) Show that \(\SpanS(\vec{v}, \vec{w}, \vec{z}) = \SpanS(\vec{w}, \vec{z})\text{.}\)
Hint.
Theorem 3.1.7 is very helpful, but you need to do a little bit of work before you can apply it.
Solution.
By definition of \(2\vec{v}\) being in \(\SpanS(\vec{w}, \vec{z})\text{,}\) there are scalars \(a, b\) such that
\begin{equation*} 2\vec{v} = a\vec{w} + b\vec{z}\text{.} \end{equation*}
Therefore
\begin{equation*} \vec{v} = \frac{a}{2}\vec{w} + \frac{b}{2}\vec{z}\text{,} \end{equation*}
so \(\vec{v}\) is in \(\SpanS(\vec{w}, \vec{z})\text{.}\) Therefore \(\SpanS(\vec{v}, \vec{w}, \vec{z}) = \SpanS(\vec{w}, \vec{z})\) by Theorem 3.1.7.

4.

Describe \(\SpanS\left(\begin{bmatrix}1\\2\\-2\end{bmatrix}, \begin{bmatrix}1\\3\\-2\end{bmatrix}, \begin{bmatrix}1\\-2\\-2\end{bmatrix}, \begin{bmatrix}-1\\0\\2\end{bmatrix}, \begin{bmatrix}1\\3\\-1\end{bmatrix}\right)\) using as few vectors as possible.
Hint.
Theorem 3.1.7 will let you determine when you can remove one vector without changing the span, and then you can repeat that process until you can't remove anymore.

5.

What kind of geometric object is \(\SpanS\left(\begin{bmatrix}1\\0\\-3\end{bmatrix}, \begin{bmatrix}2\\1\\1\end{bmatrix}, \begin{bmatrix}0\\-1\\-7\end{bmatrix}, \begin{bmatrix}5\\3\\6\end{bmatrix}\right)\text{?}\) Justify your answer.
Hint.
Use the definition of span to find out what conditions on \(x, y, z\) are necessary for \(\begin{bmatrix}x\\y\\z\end{bmatrix}\) to be in the span of the given vectors.
Answer.
A plane.