Span – Serlo

In linear algebra, the span of a subset $M$ of a vector space $V$ over a field $K$ is the set of all linear combinations with vectors from $M$ and scalars from $K$. The span is often called the linear hull of $M$ or the span of $M$.

The span forms a subspace of the vector space $V$, namely the smallest subspace that contains $M$.

Derivation of the span

Generating vectors of the $xy$-plane

We consider the vector space $\mathbb {R} ^{3}$ and restrict to the $xy$-plane. I.e., the set of all vectors of the form $(a,b,0)^{T}$ with $a,b\in \mathbb {R}$:

Each vector of this plane can be written as a linear combination of the vectors $(1,0,0)^{T}$ and $(0,1,0)^{T}$:

${\begin{pmatrix}a\\b\\0\end{pmatrix}}=a\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}+b\cdot {\begin{pmatrix}0\\1\\0\end{pmatrix}}$

With the set of these linear combinations, every point of the $xy$-plane can be reached. In particular, the two vectors $(1,0,0)^{T}$ and $(0,1,0)^{T}$ lie in the $xy$-plane. Furthermore, all linear combinations of the two vectors $(1,0,0)^{T}$ and $(0,1,0)^{T}$ lie in the $xy$-plane. This is because the $z$ component of the two vectors under consideration is $0$ and thus the third component of the linear combination of the vectors must also be $0$.

In summary, we can state: Every vector of the $xy$-plane is a linear combination of $(1,0,0)^{T}$ and $(0,1,0)^{T}$. Every linear combination of these two vectors is also an element of this plane. So the vectors $(1,0,0)^{T}$ and $(0,1,0)^{T}$ generate the $xy$-plane. Or as a mathematician would say, they span the $xy$-plane (like two rods spanning a side of a tent).

The $xy$-plane is a subspace of the vector space $\mathbb {R} ^{3}$. We call this subspace $U$. Our two vectors span the subspace (=plane) $U$. So we write

$U=\operatorname {span} \left\{{\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}}\right\}.$

We say that "$U$ is substpace generated by the two vectors $(1,0,0)^{T}$ and $(0,1,0)^{T}$" or that "$U$ is the linear hull of the two vectors $(1,0,0)^{T}$ and $(0,1,0)^{T}$ or even better: $U$ is the span of the two vectors $(1,0,0)^{T}$ and $(0,1,0)^{T}$.

Are these generating vectors unique? The answer is no, because the plane $U$ can also be spanned by two vectors like $(1,0,0)^{T}$ and $(1,1,0)^{T}$:

${\begin{pmatrix}a\\b\\0\end{pmatrix}}=(a-b)\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}+b\cdot {\begin{pmatrix}1\\1\\0\end{pmatrix}}.$

There is hence also

$U=\operatorname {span} \left\{{\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}1\\1\\0\end{pmatrix}}\right\}.$

Thus, the two vectors spanning a plane are not necessarily unique.

Intuitively, we can think of the span of vectors as the set of all possible linear combinations that can be built from these vectors. In our example this means

$\operatorname {span} \left\{{\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}}\right\}=\left\{a\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}+b\cdot {\begin{pmatrix}0\\1\\0\end{pmatrix}}{\Bigg |}\,a,b\in \mathbb {R} \right\}.$

Another intuition is the following: The span of a set $M$ describes the vector space where all combinations of directions represented by elements from $M$ are merged.

The span of even monomials

We now examine a slightly more complicated example: Consider the vector space $V$ of polynomials over $\mathbb {R}$. Let $M=\{x^{n}|\,n\in \mathbb {N} _{0}{\text{ is even}}\}\subset V$. The elements from $M$ are the monomials $1$, $x^{2}$, $x^{4}$, $x^{6}$ ans so on. In other words, all monomials that have an even exponent. For odd exponents, however $x,x^{3},x^{5},...\notin M$. We consider $\operatorname {span} (M)$, the set of all linear combinations with vectors of $M$. For example $2x^{2}+5x^{4}+9x^{8}+7x^{12}$ is an element in $\operatorname {span} (M)$. In particualr, $\operatorname {span} (M)$ is a subspace of $V$ since it contains polynomials.

Further, the set $\operatorname {span} (M)$ is not empty, since it contains for instance $x^{2}\in M$.

Let us now consider two polynomials $p,q\in \operatorname {span} (M)$. By construction of $\operatorname {span} (M)$, $p$ and $q$ consist exclusively of monomials with an even exponent. Thus, of addition of $p$ and $q$ also results in a polynomial with exclusively even exponents. The set $\operatorname {span} (M)$ is therefore closed with respect to addition.

The same argument gives us completeness with respect to scalar multiplication. Thus the set $\operatorname {span} (M)$ is a subspace of the vector space of all polynomials. As we will see later, it is even the smallest subspace that contains $M$.

Definition of the span

Above, we found out that the span of a set $M$ is the set of all linear combinations with vectors from $M$. Intuitively, the span is the subspace resulting from the union of all directions given by vectors from $M$. Now, we make this intuition mathematically precise.

Definition (Span of a set)

Let $V$ be a vector space over the field $K$. Let $M\subseteq V$ be a non-empty set. We define the span of $M$ as the set of all vectors from $V$ which can be represented as a finite linear combination of vectors from $M$ and denote it as $\operatorname {span} (M)$:

$\operatorname {span} (M)=\left\{\sum _{i=1}^{n}\lambda _{i}\cdot m_{i}{\Bigg |}\ n\in \mathbb {N} ,\,\lambda _{1},\ldots ,\lambda _{n}\in K,\,m_{1},\ldots ,m_{n}\in M\right\}$

For the empty set we define:

$\operatorname {span} (\emptyset )=\{0\}$

Alternatively, one can call the span of a set the generated subspace or linear hull.

Hint

The sum always has only finitely many summands, even if M is infinite.

Hint

Occasionally, the notation $\langle M\rangle _{K}$ is also used for the span. The advantage of this notation is that it is clear which field defines the vector space. It makes a difference which field we use as a basis. For the example for $M:=\{1\}$ we have that $\pi \cdot 1\in \langle M\rangle _{\mathbb {R} }$, but $\pi \cdot 1\notin \langle M\rangle _{\mathbb {Q} }$. It can be shown that $\langle M\rangle _{\mathbb {Q} }=\mathbb {Q}$ and $\langle M\rangle _{\mathbb {R} }=\mathbb {R}$.

Example

Example (Line through the origin is a certain span)

Let $x=(1,2)^{T}\in \mathbb {R} ^{2}$. We consider the set $\lbrace x\rbrace$ as a subset of the vector space $\mathbb {R} ^{2}$. The span $\operatorname {span} (\lbrace x\rbrace )$ is the straight line through the origin pointing in the direction of the vector $(1,2)^{T}$.

$\operatorname {span} (\lbrace x\rbrace )=\operatorname {span} \left(\left\lbrace {\begin{pmatrix}1\\2\end{pmatrix}}\right\rbrace \right)=\left\lbrace \rho \cdot {\begin{pmatrix}1\\2\end{pmatrix}}\,{\bigg |}\,\rho \in \mathbb {R} \right\rbrace$

Example (plane through the origin as a span)

Let $(5,0,0)^{T}$ and $(0,3,0)^{T}$ be two vectors from $\mathbb {R} ^{3}$. The span of these two vectors is the $xy$-plane. The following transformation shows

${\begin{aligned}\operatorname {span} \left(\left\lbrace {\begin{pmatrix}5\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\3\\0\end{pmatrix}}\right\rbrace \right)&=\left\lbrace \lambda \cdot {\begin{pmatrix}5\\0\\0\end{pmatrix}}+\mu \cdot {\begin{pmatrix}0\\3\\0\end{pmatrix}}{\Bigg |}\ \lambda ,\mu \in \mathbb {R} \right\rbrace \\[0.3em]&=\,\left\lbrace 5\cdot \lambda \cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}+3\cdot \mu \cdot {\begin{pmatrix}0\\1\\0\end{pmatrix}}{\Bigg |}\ \lambda ,\mu \in \mathbb {R} \right\rbrace \\[0.3em]&\ {\color {OliveGreen}\left\downarrow {\text{set }}{{\tilde {\lambda }}=5\cdot \lambda ,\ {\tilde {\mu }}=3\cdot \mu }\ {\text{ bzw. }}\ \lambda ={\frac {1}{5}}\cdot {\tilde {\lambda }},\ \mu ={\frac {1}{3}}\cdot {\tilde {\mu }}\right.}\\[0.3em]&=\,\left\lbrace {\tilde {\lambda }}\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}+\cdot {\tilde {\mu }}\cdot {\begin{pmatrix}0\\1\\0\end{pmatrix}}{\Bigg |}\ {\tilde {\lambda }},{\tilde {\mu }}\in \mathbb {R} \right\rbrace \\[0.3em]&=\,\left\{{\begin{pmatrix}{\tilde {\lambda }}\\{\tilde {\mu }}\\0\end{pmatrix}}{\Bigg |}\ {\tilde {\lambda }},{\tilde {\mu }}\in \mathbb {R} \right\}\end{aligned}}$

Overview: Properties of the span

Let $V$ be a $K$-vector space, $M$, $N\subseteq V$ subsets of $V$ and $W\subseteq V$ a subspace of $V$. Then, we have

• For a vector $v\in V$ we have $\operatorname {span} (\{v\})=\{\lambda \cdot v|\lambda \in K\}$
• If $N\subseteq M$, then $\operatorname {span} (N)\subseteq \operatorname {span} (M)$
• From $\operatorname {span} (N)=\operatorname {span} (M)$ one can usually not conclude $N=M$
• $M\subseteq \operatorname {span} (M)$
• $\operatorname {span} (M)$ is a subspace of $V$
• For a subspace $W$ we have $\operatorname {span} (W)=W$
• $\operatorname {span} (M)$ is the smallest subspace of $V$ including $M$
• $N\subseteq \operatorname {span} (M)\iff \operatorname {span} (M)=\operatorname {span} (M\cup N)$
• $\operatorname {span} (\operatorname {span} (M))=\operatorname {span} (M)$

Properties of the span

The span of a vector $v$ in $V$

For a vector $v\in V$ we have that $\operatorname {span} (\{v\})=\{\lambda \cdot v\,|\ \lambda \in K\}$. For the zero vector $v=0$ the span again consists only of the zero vector, so $\operatorname {span} (\{0\})=\{0\}$. If $v\neq 0$ holds, then $\operatorname {span} (\{v\})$ is exactly the set of elements that lie on the line through the origin im direction of the vector $v$.

Span preserves subsets

Theorem (Span preserves subsets)

Let $V$ be a $K$-vector space and let $M,N\subseteq V$. If $N\subseteq M$, then also $\operatorname {span} (N)\subseteq \operatorname {span} (M)$.

Proof (Span preserves subsets)

Since $\operatorname {span} (\emptyset )=\{0\}$ and $0$ is an element in the span of every set, we have $\operatorname {span} (\emptyset )=\{0\}\subseteq M$.

Thus we can assume without loss of generality that $\emptyset \neq N\subseteq M$. We consider any element $v\in \operatorname {span} (N)$. By the definition of the span, vectors $v_{1},...,v_{n}\in N$ and $\lambda _{1},...,\lambda _{n}\in K$ exist such that $v=\sum _{i=1}^{n}\lambda _{i}v_{i}$. Because of $N\subseteq M$ we have for all $v_{i}$ with $1\leq i\leq n$ that $v_{i}\in M$. Hence also $v\in \operatorname {span} (M)$. Consequently, $\operatorname {span} (N)\subseteq \operatorname {span} (M)$.

Hint

The converse of the above theorem does not hold true in general! By this we mean: From $\operatorname {span} (N)\subseteq \operatorname {span} (M)$ we cannot conclude $N\subseteq M$.

A possible counterexample is:

$N=\left\{{\begin{pmatrix}1\\0\end{pmatrix}},{\begin{pmatrix}2\\0\end{pmatrix}}\right\}{\text{ and }}M=\left\{{\begin{pmatrix}1\\0\end{pmatrix}}\right\}$

Here,

${\begin{aligned}\operatorname {span} (N)&=\left\{\lambda _{1}\cdot {\begin{pmatrix}1\\0\end{pmatrix}}+\lambda _{2}\cdot {\begin{pmatrix}2\\0\end{pmatrix}}{\Bigg |}\ \lambda _{1},\lambda _{2}\in K\right\}=\left\{(\lambda _{1}+2\lambda _{2})\cdot {\begin{pmatrix}1\\0\end{pmatrix}}{\Bigg |}\ \lambda _{1},\lambda _{2}\in K\right\}\\[0.3em]\operatorname {span} (M)&=\left\{\mu \cdot {\begin{pmatrix}1\\0\end{pmatrix}}{\Bigg |}\ \mu \in K\right\}\end{aligned}}$

Thus $\operatorname {span} (N)=\operatorname {span} (M)$, since in both cases we get exactly the multiples of the vector $(1,0)^{T}$. Since the two subsets are equal, we have in particular $\operatorname {span} (N)\subseteq \operatorname {span} (M)$, but $N\nsubseteq M$. Therefore, the converse of the theorem cannot hold true in general.

The set $M$ is contained in its span

Theorem ($M$ is contained in its span)

Let $V$ ein $K$-vector space and $M\subseteq V$. Then, there is $M\subseteq \operatorname {span} (M)$.

Proof ($M$ is contained in its span)

If $M=\emptyset$, then $\operatorname {span} (M)=\{0\}$ , and the assertion is true.

Otherwise, let $m\in M$ be arbitrary. Then $m$ can be represented by $m=1\cdot m$. In particular, $m=1\cdot m$ is a linear combination with a summand of $M$. Thus there is $m\in \operatorname {span} (M)$, since $\operatorname {span} (M)$ contains all linear combinations of elements from $M$.

This establishes the assertion $M\subseteq \operatorname {span} (M)$.

The span of $M$ is a subspace of $V$

Theorem (The span of $M$ is a subspace of $V$)

$\operatorname {span} (M)$ is a subspace of $V$

Proof (The span of $M$ is a subspace of $V$)

If $M$ is the empty set, then by definition $\operatorname {span} (M)=\{0\}$, and that is a subspace of $V$. From now on we may therefore assume that $M$ is not empty.

First, it is clear that $\operatorname {span} (M)\subseteq V$. But this is obvious according to the definition of vector space and span.

We still have to show $\operatorname {span} (M)$ is subspace of $V$. In other words,we have to show that

• $\operatorname {span} (M)\neq \emptyset$
• for two elements $u,v\in \operatorname {span} (M)$ we have also $u+v\in \operatorname {span} (M)$ (completeness of addition)
• $\rho \cdot u\in \operatorname {span} (M)$ (completeness of scalar multiplication)

Proof step: $\operatorname {span} (M)\neq \emptyset$

Since $\operatorname {span} M$ is not empty, there exists at least one $u\in M$. Then $u$ can be written as $u=1\cdot u$, and therefore $u$ itself is in $\operatorname {span} M$. So this condition is fulfilled.

Proof step: Completeness of addition

We show the completeness concerning the vector addition. Let $u,v\in \operatorname {span} (M)$. Then there are vectors $m_{1},\ldots ,m_{n}\in M$ and $n_{1},\ldots ,n_{k}\in M$, so that $u=\sum _{i=1}^{n}\lambda _{i}\cdot m_{i}$ and $v=\sum _{i=1}^{k}\mu _{i}\cdot n_{i}$. So

$u+v=\sum _{i=1}^{n}\lambda _{i}\cdot m_{i}+\sum _{i=1}^{k}\mu _{i}\cdot n_{i}\in \operatorname {span} (M)$

Hence, $\operatorname {span} (M)$ is complete with respect to addition.

Proof step: Completeness of scalar multiplication

Establish completeness of scalar multiplication is done easily:

$\rho \cdot u=\rho \cdot \sum _{i=1}^{n}\lambda _{i}\cdot m_{i}=\sum _{i=1}^{n}(\rho \cdot \lambda _{i})\cdot m_{i}\in \operatorname {span} (M)$

Thus we have proved that $\operatorname {span} (M)$ is a subspace of the vector space $V$.

The span of a subspace $W$ is again $W$

Theorem (The span of $M$ is the smallest subspace von $V$)

The span of a subspace $W$ is again $W$

Proof (The span of $M$ is the smallest subspace von $V$)

Since $W$ is a subspace, for some vectors $m_{1},\ldots ,m_{n}\in W$ also all linear combinations of the $m_{1},\ldots ,m_{n}$ are contained in $W$. Therefore $\operatorname {span} (W)\subseteq W$. Together with $W\subseteq \operatorname {span} (W)$ our assertion follows.

The span of $M$ is the smallest subspace of $V$, containing $M$

Theorem (The span of $M$ is the smallest subspace of $V$)

Let $V$ be a $K$-vector space and let $M\subseteq V$.

Then, $\operatorname {span} (M)$ is the smallest subspace of $V$ including $M$.

Proof (The span of $M$ is the smallest subspace of $V$)

We already know that $\operatorname {span} (M)$ is a subspace. Now we show that $\operatorname {span} (M)$ is the smallest subspace containing $M$.

If $M=\emptyset$, the assertion is obviously true, since then $\operatorname {span} (M)=\{0\}$.

Let $W$ be a subspace of $V$ containing $M$. Our goal is to show that $\operatorname {span} (M)\subseteq isW$. Since this would imply that the subspace $\displaystyle \operatorname {span} (M)$ is smaller or equal to every other subspace $W$ containing $M$.

Now if $u\in \operatorname {span} (M)$, then there are some $m_{1},...,m_{n}\in M$ and $\lambda _{1},\ldots ,\lambda _{n}\in K$ such that $\displaystyle u=\sum _{i=1}^{n}\lambda _{i}m_{i}$ (by definition of the span).

Since $W$ is a subspace and $m_{1},\ldots ,m_{n}\in W$, all linear combinations of $m_{1},\ldots ,m_{n}$ are also contained in $W$. This implies our assertion $\operatorname {span} (M)\subseteq W$.

Idempotency of the span

Theorem (Idempotency of the span)

Let $V$ be a $K$-vector space and $M\subseteq V$. Then we have $\operatorname {span} (\operatorname {span} (M))=\operatorname {span} (M)$. This property of the span is called idempotency.

Proof (Idempotency of the span)

For $M=\emptyset$ we have $\operatorname {span} (\emptyset )=\{0\}$ and $\operatorname {span} (\operatorname {span} (\emptyset ))=\operatorname {span} (\{0\})=\{0\}$.

Therefore, we can now assume that $M$ is not empty.

We already know that $\operatorname {span} (M)\subseteq \operatorname {span} (\operatorname {span} (M))$. So it only remains to show that $\operatorname {span} (\operatorname {span} (M))\subseteq \operatorname {span} (M)$.

Let $v\in \operatorname {span} (\operatorname {span} (M))$. Then $v$ can be written as

$v=\lambda _{1}\cdot u_{1}+...+\lambda _{n}\cdot u_{n}$

with $\lambda _{1},...,\lambda _{n}\in K$ and $u_{1},...,u_{n}\in \operatorname {span} (M)$. Since $u_{i}\in \operatorname {span} (M)$ for all $1\leq i\leq n$, every $u_{i}$ can be written as a linear combination of elements in $M$:

$u_{i}=\mu _{1}\cdot w_{i,1}+...+\mu _{m_{i}}\cdot w_{i,m_{i}}$

where $\mu _{1},...,\mu _{m_{i}}\in K$ and $w_{i,1},...,w_{i,m_{i}}\in M$. We now write the $u_{i}$ within $v$as a linear combination of the $w_{i,j}$:

${\begin{aligned}v&=\sum _{i=1}^{n}\lambda _{i}u_{i}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ u_{i}=\mu _{1}\cdot w_{i,1}+...+\mu _{m_{i}}\cdot w_{i,m_{i}}\right.}\\[0.3em]&=\sum _{i=1}^{n}\lambda _{i}\left(\sum _{j=1}^{m_{i}}\mu _{j}w_{i,j}\right)\\[0.3em]&=\sum _{i=1}^{n}\sum _{j=1}^{m_{i}}(\lambda _{i}\mu _{j})w_{i,j}\end{aligned}}$

For all $1\leq i\leq n$, the sum $\sum _{j=1}^{m_{i}}\lambda _{i}\mu _{j}$ is an element of the field $K$. So we obtain $v\in \operatorname {span} (M)$, which was to be shown.

Alternative proof (Idempotency of the span)

We know that $\operatorname {span} (M)$ is a subspace of $V$, and that the span $\operatorname {span} (W)$ of a subspace $W$ is $W$ again.

Therefore, $\operatorname {span} (\operatorname {span} (M))$ is again $\operatorname {span} (M)$.

Adding elements of the span doesn't change the span

Theorem (Adding elements of the span doesn't change the span)

Let $V$ be a $K$-vector space and $M$, $N\subseteq V$. Then, we have

$N\subseteq \operatorname {span} (M)\Longleftrightarrow \operatorname {span} (M)=\operatorname {span} (M\cup N)$

Proof (Adding elements of the span doesn't change the span)

We will establish the two implications $\implies$ and $\Longleftarrow$:

Proof step: $N\subseteq \operatorname {span} (M)\implies \operatorname {span} (M)=\operatorname {span} (M\cup N)$

The statement $\operatorname {span} (M)\subseteq \operatorname {span} (M\cup N)$ does always hold, since $M\subseteq M\cup N$. So all that remains is to show that $\operatorname {span} (M\cup N)\subseteq \operatorname {span} (M)$ holds. In order to do this, we consider an element $u\in \operatorname {span} (M\cup N)$. We can write it as

$u=\sum _{i=0}^{n}\lambda _{i}v_{i}+\sum _{i=0}^{m}\mu _{i}w_{i},$

with $v_{1},...,v_{n}\in M$, $w_{1},...,w_{m}\in N$, $\lambda _{1},...,\lambda _{n}\in K$ and $\mu _{1},...,\mu _{m}\in K$. Since $N\subseteq \operatorname {span} (M)$, one can write $w_{i}$ for all $1\leq i\leq m$ as a linear combination of elements from $M$:

$w_{i}=\rho _{1}\cdot {\hat {v}}_{1}+...+\rho _{m_{i}}\cdot {\hat {v}}_{m_{i}}$

where $\rho _{1},...,\rho _{m_{i}}\in K$ and ${\hat {v}}_{1},...,{\hat {v}}_{m_{i}}\in M$. Now, we plug this expression for $w_{i}$ into the formula above:

${\begin{aligned}u&=\sum _{i=0}^{n}\lambda _{i}v_{i}+\sum _{i=0}^{m}\mu _{i}\left(\sum _{j=1}^{m_{i}}\rho _{j}{\hat {v}}_{j}\right)\\[0.3em]&=\sum _{i=0}^{n}\lambda _{i}v_{i}+\sum _{i=1}^{m}\sum _{j=1}^{m_{i}}(\mu _{i}\rho _{j}){\hat {v}}_{j}\end{aligned}}$

We have thus represented $u$ as a linear combination of vectors from $M$ and hence $\operatorname {span} (M)=\operatorname {span} (M\cup N)$.

Proof step: $N\subseteq \operatorname {span} (M)\Longleftarrow \operatorname {span} (M)=\operatorname {span} (M\cup N)$

We show this statement using a proof by contradiction. We assume that there is some $w\in N$ but $w\notin \operatorname {span} (M)$. We now define an element $u:=\sum _{i=0}^{n}\lambda _{i}v_{i}+w$, with $v_{1},...,v_{n}\in M$ and $\lambda _{1},...,\lambda _{n}\in K$.

Now $u$ is a linear combination of vectors from $M\cup N$. Thus $u\in \operatorname {span} (M\cup N)$, since $w\in N$. However, we also have $u\notin \operatorname {span} (M)$, since $w\notin \operatorname {span} (M)$. But this contradicts the assumption $\operatorname {span} (M)=\operatorname {span} (M\cup N)$.

Hence our assumption is false and $N\subseteq \operatorname {span} (M)$ must hold.

Alternative proof (First proof step)

One can also argue as follows: we have $M,N\subseteq \operatorname {span} (M)$. So also $M\cup N\subseteq \operatorname {span} (M)$.

We have already proved that then $\operatorname {span} (M\cup N)\subseteq \operatorname {span} (\operatorname {span} (M))$. This set is the same as $\operatorname {span} (M)$ because of the idempotency of the span, so $\operatorname {span} (M\cup N)=\operatorname {span} (M)$.

Check whether vectors are inside the span

After we have learned some properties of the span, we will show in this section how we can check whether a vector of $V$ lies within the span of $M\subseteq V$ or not. We will see that in order to answer this question we have to solve a linear system of equations.

Example (Plane and line through the origin)

Let's start with a simple example from the $\mathbb {R} ^{2}$. We consider the line through the origin $\operatorname {span} (M)$ with the one-element subset $M=\{(4,3)^{T}\}$ of the plane $\mathbb {R} ^{2}$. The question now is whether the vector $(12,9)^{T}$ lies in the span of $M$. One can immediately see that

${\begin{pmatrix}12\\9\end{pmatrix}}=3\cdot {\begin{pmatrix}4\\3\end{pmatrix}}$

holds. In other words

${\begin{pmatrix}12\\9\end{pmatrix}}\in \operatorname {span} (M)$

Mathematically, we have to solve a system of equations. In our simple example, the exercise is to find a $\lambda \in K$ such that

${\begin{pmatrix}12\\9\end{pmatrix}}=\lambda \cdot {\begin{pmatrix}4\\3\end{pmatrix}}$

From this equation we obtain the linear system of equations

${\begin{aligned}12&=\lambda \cdot 4\\9&=\lambda \cdot 3\end{aligned}}$

with the obvious solution $\lambda =3$.

Example (Polynomials)

Let us now examine an example whose solution is not obvious at first sight. For this we consider the subset of the monomials $N=\{x^{3},x^{1},x^{0}\}$ and the polynomial $p(x)=(x-2)^{3}+3(x-2)^{2}$. We want to show that the polynomial is not in the span of $N$. To do this, it suffices to prove that $p(x)$ cannot be represented as a linear combination of the monomials of $N$. We can see this by transforming the polynomial:

$(x-2)^{3}+3(x-2)^{2}=(x^{3}+12x-6x^{2}-8)+(3x^{2}-12x+12)=x^{3}-3x^{2}+4$

We see that a summand contains the monomial $x^{2}$, but this monomial is not contained in $N$. Thus the polynomial is not in the span of the set $N$.

Example (Vectors from $\mathbb {R} ^{4}$)

We consider the subset $M=\{(1,-2,3,2)^{T},(3,0,2,1)^{T},(0,-2,1,-3)^{T},(1,1,-2,2)^{T}\}$ of $\mathbb {R} ^{4}$ and want to prove that the vector $(2,-9,2,-3)^{T}\in \operatorname {span} (M)$. For this we have to show that there are coefficients $\lambda _{1},\lambda _{2},\lambda _{3},\lambda _{4}\in \mathbb {R}$ such that

${\begin{pmatrix}2\\-9\\2\\-3\end{pmatrix}}=\lambda _{1}\cdot {\begin{pmatrix}1\\-2\\3\\2\end{pmatrix}}+\lambda _{2}\cdot {\begin{pmatrix}3\\0\\2\\1\end{pmatrix}}+\lambda _{3}\cdot {\begin{pmatrix}0\\-2\\1\\-3\end{pmatrix}}+\lambda _{4}\cdot {\begin{pmatrix}1\\1\\-2\\2\end{pmatrix}}$

From this representation we get the linear system of equations

${\begin{aligned}I:&&2&=&1\cdot \lambda _{1}+3\cdot \lambda _{2}+0\cdot \lambda _{3}+1\cdot \lambda _{4}\\[0.3em]II:&&-9&=&-2\cdot \lambda _{1}+0\cdot \lambda _{2}-2\cdot \lambda _{3}+1\cdot \lambda _{4}\\[0.3em]III:&&2&=&3\cdot \lambda _{1}+2\cdot \lambda _{2}+1\cdot \lambda _{3}-2\cdot \lambda _{4}\\[0.3em]IV:&&-3&=&2\cdot \lambda _{1}+1\cdot \lambda _{2}-3\cdot \lambda _{3}+2\cdot \lambda _{4}\end{aligned}}$

with solution $\lambda _{1}=2$, $\lambda _{2}=-1$, $\lambda _{3}=4$, $\lambda _{4}=3$. Hence, we have that

${\begin{pmatrix}2\\-9\\2\\-3\end{pmatrix}}=2\cdot {\begin{pmatrix}1\\-2\\3\\2\end{pmatrix}}-1\cdot {\begin{pmatrix}3\\0\\2\\1\end{pmatrix}}+4\cdot {\begin{pmatrix}0\\-2\\1\\-3\end{pmatrix}}+3\cdot {\begin{pmatrix}1\\1\\-2\\2\end{pmatrix}}$

and therefore $(2,-9,2,-3)^{T}\in \operatorname {span} (M)$.