Serlo: EN: Span

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

The xy-plane in three-dimensional space

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

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 ${ℝ}^3$. We call this subspace $U$. Our two vectors span the subspace (=plane) $U$. So we write

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 $(1,1,0{)}^T$:

There is hence also

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

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 $ℝ$. 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 $\mathrm{span}(M)$, the set of all linear combinations with vectors of $M$. For example $2x^2+5x^4+9x^8+7x^{12}$ is an element in $\mathrm{span}(M)$. In particualr, $\mathrm{span}(M)$ is a subspace of $V$ since it contains polynomials.

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

Let us now consider two polynomials $p,q\in \mathrm{span}(M)$. By construction of $\mathrm{span}(M)$, $p$ 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 $\mathrm{span}(M)$ is therefore closed with respect to addition.

The same argument gives us completeness with respect to scalar multiplication. Thus the set $\mathrm{span}(M)$ 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.

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

Properties of the span

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

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

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.

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