MATH347 L10: Vector spaces and subspaces

• New concepts:

  • Vector space

  • Vector subspace

  • Span of a set of vectors

  • Linear dependence and independence

Vector space

• Formalize linear combinations

  Addition rules for $\forall 𝒂,𝒃,𝒄\in V$ $𝒂+𝒃\in V$ Closure $𝒂+(𝒃+𝒄)=(𝒂+𝒃)+𝒄$ Associativity $𝒂+𝒃=𝒃+𝒂$ Commutativity $𝟎+𝒂=𝒂$ Zero vector $𝒂+(-{𝒂}_{})=𝟎$ Additive inverse Scaling rules for $\forall 𝒂,𝒃\in V$, $\forall x,y\in S$ $x𝒂\in V$ Closure $x(𝒂+𝒃)=x𝒂+x𝒃$ Distributivity $(x+y)𝒂=x𝒂+y𝒂$ Distributivity $x(y𝒂)=(xy)𝒂$ Composition $1𝒂=𝒂$ Scalar identity

  Table 1. Vector space properties

• Example: $V={ℝ}^m,S=ℝ$

Vector subspace

• Consider projection in ${ℝ}^2$ onto the $x_1$ axis

  ${\displaystyle 𝑷={𝒆}_1{𝒆}_1^T=\left[\begin{array}{l}1\\ 0\end{array}\right]\left[\begin{array}{ll}1& 0\end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 0\end{array}\right],𝑷𝒙=\left[\begin{array}{ll}1& 0\\ 0& 0\end{array}\right]\left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{l}{x}_1\\ 0\end{array}\right]}$

• Let $V={ℝ}^2$ and consider $U$ be the set of all vectors in ${ℝ}^2$ with zero second component. Notice that $U\subset V$ and $U$ is also a vector space, i.e., any linear combination of vectors in $U$ stays within $U$

• In general if vectors in $V$ form a vector space, $U\subset V$ and if for any $\alpha ,\beta \in ℝ$, $𝒖,𝒗\in U$, $\alpha 𝒖+\beta 𝒗\in U$, then $U$ is a vector subspace of $V$

• Example

  ${\displaystyle V=\{\left[\begin{array}{l}{x}_1\\ x_2\end{array}\right],x_1,x_2\in ℝ\},U=\{\left[\begin{array}{l}{x}_1\\ 0\end{array}\right],x_1\in ℝ\}}$

  Note that vectors in $U$ still have two components, just like those in $V$

• Choose $\alpha =-\beta$, $𝒖=𝒗$ to set that $\alpha 𝒖+\beta 𝒗=\alpha 𝒖-\alpha 𝒖=𝟎$ must be within the subspace, i.e., the zero element is always a member of a subspace

Vector span

Definition. The span of vectors ${𝒂}_1,{𝒂}_2,\dots ,{𝒂}_n\in 𝒱,$is the set of vectors reachable by linear combination

The notation used for set on the right hand side is read: “those vectors $𝒃$ in $𝒱$ with the property that there exist $n$ scalars $x_1,\dots ,x_n$ to obtain $𝒃$ by linear combination of ${𝒂}_1,{𝒂}_2,\dots ,{𝒂}_n$.

• A linear combination is conveniently expressed as a matrix-vector product leading to a different formulation of the same concept

Definition. The column space (or range) of matrix $𝑨\in {ℝ}^{m\times n}$ is the set of vectors reachable by linear combination of the matrix column vectors

Linearly dependent vectors

In the example (?)

since ${𝒂}_3={𝒂}_1+{𝒂}_2\iff {𝒂}_1+{𝒂}_2-{𝒂}_3=𝟎$. Introduce a concept to capture the idea that a vector can be expressed in terms of other vectors.

Definition. The vectors ${𝒂}_1,{𝒂}_2,\dots ,{𝒂}_n\in 𝒱,$are linearly dependent if there exist $n$ scalars, $x_1,\dots ,x_n\in 𝒮$, at least one of which is different from zero such that

Note that $\left\{𝟎\right\}$, with $𝟎\in 𝒱$ is a linearly dependent set of vectors since $1\cdot 𝟎=𝟎$.

Linearly independent vectors

The converse of linear dependence is linear independence, a member of the set cannot be expressed as a non-trivial linear combination of the other vectors

Definition. The vectors ${𝒂}_1,{𝒂}_2,\dots ,{𝒂}_n\in 𝒱,$are linearly independent if the only $n$ scalars, $x_1,\dots ,x_n\in 𝒮$, that satisfy

are $x_1=0$, $x_2=0$,…,$x_n=0$.

The choice $𝒙={\left(\begin{array}{lll}{x}_1& \dots & x_n\end{array}\right)}^T=𝟎$ that always satisfies (1) is called a trivial solution. We can restate linear independence as (1) being satisfied only by the trivial solution.