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Wize University Linear Algebra Textbook > Vector Spaces and Subspaces

Span

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Span

The span of a set of vectors SS is the set of all linear combinations of the vectors in SS.

Given the set S={u⃗1,u⃗2,…,u⃗n}S=\{\vec u_1,\vec u_2,\dots,\vec u_n\}, and allowing c1,c2,…,cnc_1,c_2,\dots,c_n to be any real numbers, the span of SS is:
span(S)=span({u⃗1,u⃗2,…,u⃗n})={c1u⃗1+c2u⃗2+⋯+cnu⃗n}\text{span}(S)=\text{span}(\{\vec u_1,\vec u_2,\dots,\vec u_n\})=\{c_1\vec u_1+c_2\vec u_2+\cdots+c_n\vec u_n\}

Wize Concept
SS is a finite set of vectors.
span(S)\text{span}(S) is an infinite set of vectors: it contains every possible linear combination of the vectors in SS.

Geometrically
  • The span of one non-zero vector u⃗\vec{u} is the line through the origin in the direction of u⃗\vec{u}:
span⁡({u⃗})={ tu⃗ ∣ t∈R}\operatorname{span}(\{\vec u\}) = \{\ t\vec u \ |\ t\in\reals \}

  • The span of two linearly independent vectors is a plane:
span⁡({u⃗,v⃗})={ su⃗+tv⃗ ∣ s,t∈R}\operatorname{span}(\{ \vec u, \vec v \}) = \{\ s\vec u + t \vec v \ |\ s,t\in\reals \}

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Determining if a Vector is in the Span

Given a finite set S={u⃗1,u⃗2,…,u⃗n}S=\{\vec u_1, \vec u_2,\dots, \vec u_n\} and a vector b⃗\vec b, determining if b⃗\vec b is in span(S)\text{span}(S) is equivalent to solving the linear system with augmented matrix:
[u⃗1u⃗2⋯u⃗nb⃗]\left[\begin{array}{cccc|c} \\ \vec u_1 & \vec u_2 & \cdots & \vec u_n & \vec b\\ \\ \end{array}\right]
If this system has:
  • No solution  ⟹  b⃗∉span(S)\boxed{\text{No solution} \implies \vec b \notin \text{span}(S)} since no linear combination of the vectors produces b⃗\vec b.
  • At least one solution  ⟹  b⃗∈span(S)\boxed{\text{At least one solution} \implies \vec b \in \text{span}(S)}
Wize Tip
This is very similar to determining whether a set of vectors is linearly independent!
Instead of augmenting with the zero vector, we check whether b⃗∈span(S)\vec b \in \text{span}(S) by augmenting with b⃗\vec b.

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Spanning Sets or Generating Sets

If VV is a vector space and span(S)=V\text{span(S)}=V, then SS is called a generating set (or spanning set) of VV.
Wize Tip
A set of nn linearly independent vectors always spans (is a generating set of) Rn\reals^n.

Example
The following are spanning sets of the vector space R2\mathbb{R}^2, since linear combinations of the vectors can produce any vector in R2\reals^2.
S1={(1,0),(0,1)}→linearly independent spanning setS2={(1,0),(1,1),(2,1)}→linearly dependent spanning set\begin{aligned} S_1&=\{(1,0),(0,1)\} && \rightarrow \text{linearly independent spanning set}\\[0.5em] S_2&=\{(1,0),(1,1),(2,1)\} && \rightarrow \text{linearly \textbf{dependent} spanning set} \end{aligned}
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Example: Span

Determine if x⃗=[2−45]\vec x = \begin{bmatrix} 2\\ -4\\ 5\\ \end{bmatrix} is in span({u⃗,v⃗,w⃗})\text{span}(\{\vec u, \vec v, \vec w\}) where u⃗=[111],v⃗=[110],w⃗=[101]\vec u = \begin{bmatrix} 1\\ 1\\ 1\\ \end{bmatrix} ,\quad \vec v = \begin{bmatrix} 1\\ 1\\ 0\\ \end{bmatrix} ,\quad \vec w = \begin{bmatrix} 1\\ 0\\ 1\\ \end{bmatrix}.
We are looking to find values of c1,c2,c_1,c_2, and c3c_3 that are solutions to: c1u⃗+c2v⃗+c3w⃗=x⃗(∗)c_1\vec u + c_2 \vec v + c_3\vec w = \vec x \quad \colorTwo{(*)}
This is equivalent to solving the linear system with augmented matrix: [1112110−41015]\left[\begin{array}{rrr|r}1&1&1&2\\1&1&0&-4\\1&0&1&5\end{array}\right]
The reduced row echelon form of this matrix is: [100−1010−30016]\left[\begin{array}{rrr|r}1&0&0&-1\\0&1&0&-3\\0&0&1&6\end{array}\right]
Hence the solution is c1=−1c_1=-1, c2=−3c_2=-3, c3=6c_3=6.
This means we can write x⃗\vec x as a linear combination of the vectors: x⃗=−1u⃗−3v⃗+6w⃗\vec x = -1\vec u - 3\vec v + 6\vec w.
Therefore x⃗\vec x is indeed in span({u⃗,v⃗,w⃗})\text{span}(\{\vec u, \vec v, \vec w\}).
Alternatively
We can use the determinant of the matrix whose columns are the vectors of interest:
Let A=[u⃗v⃗w⃗]=[111110101]A =\left[\begin{array}{rrr}\vec u & \vec v & \vec w\end{array}\right] =\left[\begin{array}{rrr}1&1&1\\1&1&0\\1&0&1\end{array}\right]
Then det(A)=−1≠0\text{det}(A)=-1\textcolor{red}{\ne}0, so AA is invertible, which means the linear system (∗)\colorTwo{(*)} has exactly one solution.
This method does not tell us the solution, but since we know one exists, we conclude that x⃗∈span({u⃗,v⃗,w⃗})\vec x \in \text{span}(\{\vec u, \vec v, \vec w\}).

Practice: Span

Let u⃗=⟨1,0,1⟩, v⃗=⟨1,1,1⟩, w⃗=⟨1,−1,1⟩\vec u = \lang 1,0,1\rang,\ \vec v = \lang 1,1,1\rang,\ \vec w = \lang 1,-1,1 \rang be vectors in R3\reals^3, and suppose b⃗=⟨3,y,z⟩∈span({u⃗,v⃗,w⃗})\vec b = \lang 3,y,z \rang \in \text{span}(\{\vec u, \vec v, \vec w\}).
Find the value of zz.
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Practice: Spans are Subspaces

Suppose that v⃗1,…,v⃗k∈Rn\vec v_1, \dots, \vec v_k\in\reals^n. Prove that U=span⁡({v⃗1,...,v⃗k})U=\operatorname{span}(\{\vec v_1, ...,\vec v_k\}) is a subspace of Rn\mathbb{R}^n.
Extra Practice
Practice: Span
Show that for all w⃗,x⃗,y⃗,z⃗∈Rn\vec w,\vec x,\vec y, \vec z\in\mathbb{R}^n , if w⃗∈Span{x⃗,y⃗,z⃗}\vec w\in Span\{\vec x, \vec y, \vec z\} then w⃗∈Span{x⃗−y⃗,y⃗−z⃗,z⃗}\vec w\in Span\{\vec x-\vec y, \vec y-\vec z,\vec z\} .
When is a Vector in the Span of Others
Determine if x⃗=(16,−15,8)∈Span{u⃗,v⃗,w⃗}\vec x=(16,-15, 8) \in \text{Span}\{\vec u, \vec v, \vec w\} if:
u⃗=(1,1,1),v⃗=(1,1,0),w⃗=(1,0,−1)\vec u = (1,1,1), \vec v = (1,1,0), \vec w =(1,0,-1)
Span
Practice: Span
Let u⃗=⟨1,0,1⟩, v⃗=⟨1,1,1⟩, w⃗=⟨1,−1,1⟩\vec u = \lang 1,0,1\rang,\ \vec v = \lang 1,1,1\rang,\ \vec w = \lang 1,-1,1 \rang be vectors in R3\reals^3, and suppose b⃗=⟨3,y,z⟩∈span({u⃗,v⃗,w⃗})\vec b = \lang 3,y,z \rang \in \text{span}(\{\vec u, \vec v, \vec w\}).
Find the value of zz.
Practice: Span
Let u⃗=(2,2,1),v⃗=(2,1,2),w⃗=(2,5,−2),x⃗=(x,y,z)\vec u = (2,2,1), \vec v = (2,1,2), \vec w = (2,5,-2), \vec x = (x,y,z) be vectors in R3\mathbb{R}^3
Find the conditions that the values x,y,zx,y,z must satisfy in order for x⃗\vec x to be in Span({u⃗,v⃗,w⃗})\text{Span}(\{\vec u, \vec v, \vec w\})
Span
Practice: Span
Let u⃗=⟨1,0,1⟩, v⃗=⟨1,1,1⟩, w⃗=⟨1,−1,1⟩\vec u = \lang 1,0,1\rang,\ \vec v = \lang 1,1,1\rang,\ \vec w = \lang 1,-1,1 \rang be vectors in R3\reals^3, and suppose b⃗=⟨3,y,z⟩∈span({u⃗,v⃗,w⃗})\vec b = \lang 3,y,z \rang \in \text{span}(\{\vec u, \vec v, \vec w\}).
Find the value of zz.
Span
Select all of the vectors that are in Span(u⃗,v⃗)\text{Span}(\vec u, \vec v) where u⃗=(−1,1,2),v⃗=(2,1,−3)\vec u=(-1,1,2), \vec v=(2,1,-3)
Linear Independence and Span concepts
Which of the following statements are true about the set of vectors {(1,1,−1,0),(−1,3,−1,1),(2,0,−3,4)}\{(1,1,-1,0),(-1,3,-1,1),(2,0,-3,4)\}?
133 - FML 3 - 18.1W - e.g. 31
Knowing that
[06330−7151−113]→RREF[100010001000]\bcb{\begin{bmatrix}%[c] 0 & 6 & 3 \\ 3 & 0 & -7 \\ 1 & 5 & 1 \\ -1 & 1 & 3 \end{bmatrix} % \xrightarrow{\text{{\tiny RREF}}} % \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}},
determine whether the vector v⃗1=<0,3,1,−1>\bcb{\vec{v}_1 = \big< 0,3,1,-1\big>} is in the span of the vectors v⃗2=<6,0,5,1>\bcb{\vec{v}_2 = \big< 6,0,5,1\big>} and v⃗3=<3,−7,1,3>\bcb{\vec{v}_3 = \big< 3,-7,1,3\big>}.
Consider the space of polynomials of degree 3 or lower, P3\mathcal{P}_3 . Give an example of a linearly independent set of vectors whose span is the entirety of P3\mathcal{P}_3 .
19.4F_Final_Builder_Ch_4.6_Span_Line_From_FAQ_$\tkco{eg}$_$\key{Final}$_Builder_$\tkcth{4.6.}\tkcf{6}$_$\key{|Strc>5.2}$
Describe the set of all linear combinations (sums of constant multiples of) the vectors:
v⃗=[111]\vv = \colvecth{1}{1}{1} and w⃗=[−121]\vw = \colvecth{-1}{2}{1}.
133 - FML 3 - 18.1W - e.g. 30
Given the standard basis vectors for R2\bcb{\mathbb{R}^{2}}, viz. ı^\bcb{\ihat} and ȷ^\bcb{\jhat} and the vector u⃗=[02]\bcb{\vec{u} = \begin{bmatrix} 0 \\ 2 \end{bmatrix}}, determine whether ı^\bcb{\ihat} is in the span ⁣{u⃗,ȷ^}\bcb{ \vspan{\vec{u}, \jhat} }.
Briefly justify your answer.
Let (V,+,⋅)(V, +, \cdot) be a real vector space. If v1,…,vnv_1, \dots, v_n is a set of vectors that spans the entirety of VV , prove that the following set of vectors also spans VV .
v1−v2,v2−v3,…,vn−1−vn,vnv_1 - v_2, v_2 - v_3, \dots, v_{n - 1} - v_n, v_n
19.4F_Final_Builder_Ch_4.6_Span_Line_From_FAQ_$\tkco{eg}$_$\key{Final}$_Builder_$\tkcth{4.6.}\tkcf{4}$_$\key{|Strc>5.2}$
Describe the set of all constant multiples of the vector:
v⃗=[−121]\vv = \colvecth{-1}{2}{1}.
19.4F_Final_Builder_Ch_4.6_Span_Line_From_FAQ_$\tkco{eg}$_$\key{Final}$_Builder_$\tkcth{4.6.}\tkcf{2}$_$\key{|Strc>5.2}$
Describe the set of all constant multiples of the vector v⃗=[11]\vv = \colvec{1}{1}.
Practice Question: Arbitrary Vector as Linear Combination
Show that every vector in R2\mathbb{R}^2 is a linear combination of (1,−1)(1,-1) and (2,1)(2,1) .
Arbitrary Vector as Linear Combination
Show every vector in R3\mathbb{R}^3 is a combination of (8,0,0),(0,1,1),(0,2,3)(8,0,0),(0,1,1),(0,2,3).
Suppose that S=span([1−12],[−33−6],[121],[547])S=span\left(\begin{bmatrix}1\\-1\\2\end{bmatrix},\begin{bmatrix}-3\\3\\-6\end{bmatrix},\begin{bmatrix}1\\2\\1\end{bmatrix},\begin{bmatrix}5\\4\\7\end{bmatrix}\right) , which of the following statements is true?