Let's quickly review the key ideas of chapter 4. We examined the algorithm of Gauss–Jordan elimination for solving a system of linear equations. When the system involved three unknowns this was akin to finding the set of intersection of planes where each equation corresponded with a plane.
Next, we defined how to multiply an \(m \times n\) matrix by an \(n \times 1\) column vector. This allowed us to equate linear systems with matrix equations, for example:
Our key observation was that solving a matrix equation such as \(A\vec{x} = \vec{b}\) for the unknown vector \(\vec{x}\) was tantamount to finding weights whereby to scale or weight the column vectors of \(A\) to generate vector \(\vec{b}\text{.}\)
Definition5.1.1
A vector space, \(V\text{,}\) is a set of objects called vectors equipped with and closed under the two operations of
-
vector addition, and
-
vector scaling.
In addition to the above requirements, a vector space must also satisfy the following for all \(\vec{u}, \vec{v},
\vec{w}\in V\) and for all scalars, \(r,s \in \R\text{,}\)
- \(\vec{u} + \vec{v} = \vec{v} + \vec{u}\)
- \(\vec{u}+(\vec{v}+\vec{w}) =
(\vec{u}+\vec{v})+\vec{w}\)
- \(\vec{u} + \vec{0} = \vec{0} + \vec{u} = \vec{u}\)
- \(\vec{u} + (-\vec{u}) = (-\vec{u}) + \vec{u} = \vec{0}\)
- \(r(\vec{u} + \vec{v}) = r\vec{u} + r\vec{v}\)
- \((r+s)\vec{u} = r\vec{u} + s\vec{u}\)
- \(r(s\vec{u}) = (rs)\vec{u}\)
- \(1\vec{u} = \vec{u}\)
A vector space is a set with structure. The objects in the set are called vectors, the structure comes from the operations allowed on the set, and the fact that the set is closed under these operations. The operations are:
vector addition, and
scalar multiplication.
The phrase “closed under these operations” means the following. If \(\vec{u}, \vec{v} \in V\text{,}\) and \(c \in
\R\) is any scalar, then
Informally, you can add any two vectors in the set and the result will be a vector in the same set, and you can scale any vector by any real number and this will result in some vector still in the set.
You should associate the word space with “a set with structure”. And you should understand the phrases closed under addition and closed under scalar multiplication as meaning that the operations always result in a vector that is in the set. In other words, you can't get “outside” the set via these operations.
You are certainly familiar with the notion of a subset. As you might suspect, since vector spaces are fundamentally just sets of vectors, sometimes we will be interested in subsets of vector spaces.
Definition5.1.2
A vector subspace is a subset of a vector space that is itself a vector space.
If \(V\) is a vector space, and \(W\) is a subset of \(V\) such that for all \(\vec{u}, \vec{v} \in W\) and for all \(c \in \R\) the following are true:
\(\vec{u} + \vec{v} \in W\text{,}\) and
\(c\vec{u} \in W\text{.}\)
then \(W\) is a subspace of \(V\text{.}\)
Example5.1.3The Plane \(z=4\)
Consider the set \(W = \set{(x,y,4)}{x,y \in \R}\text{,}\) is this a subspace of \(\R^3\text{?}\)
SolutionThe elements of this set are triples which have 4 in the third, or \(z\) component. This is not a subspace because it is not closed under vector addition. If we take any two elements from \(W\text{,}\) say \(\vec{u} =
(0,0,4)\) and \(\vec{v} = (1,0,4)\text{,}\) then when we add them we get eight in the third component.
\begin{equation*}
\vec{u} + \vec{v} = (0,0,4) + (1,0,4) = (1,0,8) \not\in
W.
\end{equation*}
Thus this subset is not closed under vector addition, and therefore not a vector subspace.
This set is not closed under scalar multiplication either. For example, let \(c=0\text{,}\) then \(c\vec{u} =
c(0,0,4) = (0,0,0) \not\in W\text{.}\) Since the set is not closed under scalar multiplication it is not a subspace of \(\R^3\text{.}\)
Example5.1.4The Upper Half Plane
Is the upper half–plane, \(H = \left\{ (x,y)
\: \middle\vert \: y \ge 0 \right\}\)a subspace of \(\R^2\text{?}\)
When confronted with determining whether a set is a vector subspace, it is often wise to try to find a counterexample, i.e. a pair of vectors in the set whose sum is not in the set, or a vector which when scaled (often by a negative value) is not in the set. Upon failing to find such counterexamples it may be because you were not lucky in picking representative vectors or because it actually is a subspace.
SolutionClearly \((0,1)\) and \((1,0)\) are both in \(H\text{.}\) Their sum, \((1,1) \in H\) as well because the \(y\)-component is \(1 \ge 0\text{,}\) thus these two vectors do not provide a counterexample.
1
Show that the given set is a subspace of \(\R^3\text{.}\)
\begin{equation*}
V = \set{(x,y,z)}{x=0}.
\end{equation*}
2
Show that the given set is a subspace of \(\R^3\text{.}\)
\begin{equation*}
V = \set{(x,y,z)}{x+y+z=0}.
\end{equation*}
3
Show that the given set is not a subspace of \(\R^3\text{.}\)
\begin{equation*}
V = \set{(x,y,z)}{y=1}.
\end{equation*}
4
Show that the given set is not a subspace of \(\R^2\text{.}\) \(V = \set{(x,y)}{ \abs{x} = \abs{y} }.\)
5
let \(U\) and \(V\) be subspaces of the vector space \(W\text{.}\) Their intersection \(U \cap V\) is the set of all vectors that are both in \(U\) and \(V\text{.}\) Show that \(U \cap V\) is a subspace of \(W\text{.}\)
Notice how this parallels what we first did when solving systems of linear equations. For example, when we had a system of equations in \(\R^3\text{,}\) then each equation corresponded with a plane. For a plane to be a subspace of \(\R^3\) it must pass through the origin or in other words, its equation must be homogeneous. The intersection of two nonparallel planes which pass through the origin will be a line which passes through the origin, a one dimensional subspace.
HintTo prove this statement you should use \(\S 4.2\) Theorem 1 from the textbook. Let \(\vec{u} \in U\cap
V\) and \(\vec{v} \in U \cap V\) be two generic vectors in the intersection. What do you know about \(\vec{w} = \vec{u} + \vec{v}\text{?}\) Is \(\vec{w} \in
U\text{?}\) Is \(\vec{w} \in V\text{?}\)