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4.1 Vector Spaces and Subspaces

We all know what vectors in \(\mathbb{R}^n\) are.

Other things that behave according to a set of rules can also be called "vectors" - they are in vector space.

Vector Space

A nonempty set of objects called "vectors" on which operations called "addition" and "multiplication by scalars" subject to the following 10 axioms. \(\vec{u}, \vec{v}, \vec{w}\) are vectors and \(c, d\) are scalars.

\(\vec{u} + \vec{v}\) is in the set
\(\vec{u} + \vec{v} = \vec{v} + \vec{u}\)
\((\vec{u} + \vec{v}) + \vec{w} = \vec{u} + (\vec{v} + \vec{w})\)
\(\vec{0}\) defined such that \(\vec{0} + \vec{u} = \vec{u}\)
\(-\vec{u}\) defined such that \(\vec{u} + (-\vec{u}) = \vec{0}\)

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\(c\vec{u}\) is in the set
\(c(\vec{u} + \vec{v}) = c\vec{u} + c\vec{v}\)
\((c + d)\vec{u} = c\vec{u} + d\vec{u}\)
\(c(d\vec{u}) = (cd)\vec{u}\)
\(1\vec{u} = \vec{u}\)

Some examples:

The real number system \(\mathbb{R}\) (but the set of natural numbers is not)
\(M_{2 \times 2}\), all \(2 \times 2\) matrices are in a vector space
Set of all 2nd degree polynomials:
\[ \vec{p}(t) = a_0 + a_1 t + a_2 t^2 \]

If a vector space is a subset of another vector space, then it's called a subspace. Only 3 of the 10 axioms need to be checked.

a) existence of \(\vec{0}\) (#4)

b) \(\vec{u} + \vec{v}\) is in set (#1)

c) \(c\vec{u}\) is in set (#6)

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the rest are satisfied automatically

a subspace is a vector space

a vector space is a subspace

the subspace it belongs to

objects in subspace must be similar to the vector space it belongs to

for example, \(\mathbb{R}^2\) is a subset of \(\mathbb{R}^3\) but NOT a subspace of \(\mathbb{R}^3\)

\[ \mathbb{R}^2 : \begin{bmatrix} a \\ b \end{bmatrix} \]

\[ \mathbb{R}^3 : \begin{bmatrix} a \\ b \\ c \end{bmatrix} \]

vectors of the form \[ \begin{bmatrix} a \\ b \\ 0 \end{bmatrix} \] are in a subspace of \(\mathbb{R}^3\)

\[ \begin{bmatrix} a \\ b \end{bmatrix} \text{ does behave just like } \begin{bmatrix} a \\ b \\ 0 \end{bmatrix} \]

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example

Is the set of all polynomials of the form \[ \vec{p}(t) = at^4 \] a subspace of \(\mathbb{P}_4\) (all fourth-deg polynomials)?

a) Is \(\vec{0}\) defined?
yes, \(\vec{0} = 0t^4\)
b) closed under addition?
\[ \vec{p}(t) = at^4 \quad \vec{g}(t) = bt^4 \]\[ \vec{p}(t) + \vec{g}(t) = (a+b)t^4 = ct^4 \]
still in the form of \(at^4\)
c) closed under scalar multiplication?
\[ \vec{p}(t) = at^4 \]\[ c\vec{p}(t) = cat^4 = (ca)t^4 = dt^4 \]

\(\vec{p}(t) = at^4\) are in a subspace

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Subspaces and Spanning Sets

Example: Polynomials in \(\mathbb{P}_2\)

Are all polynomials in the form \(\vec{p}(t) = a + t^2\) a subspace of \(\mathbb{P}_2\)?

Is \(\vec{0}\) defined?
\[ \vec{0} = 0 + 0t^2 \]
\(\rightarrow\) is not in the form of \(a + t^2\) (lost the \(t^2\))

So these do not live in a subspace of \(\mathbb{P}_2\).

Theorem: Spanning Sets as Subspaces

If \(\vec{v}_1, \vec{v}_2, \dots, \vec{v}_p\) are in a vector space \(V\), then \(\text{span} \{ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_p \}\) is a subspace of \(V\).

\(\{ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_p \}\) is called the spanning set

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Proof Sketch: Why?

Suppose \(\vec{v}_1, \vec{v}_2\) are in \(V\). Let \(H = \text{span} \{ \vec{v}_1, \vec{v}_2 \}\).

\(\vec{0} = 0\vec{v}_1 + 0\vec{v}_2\) is in \(H\)
Let \(\vec{u} = a\vec{v}_1 + b\vec{v}_2\) and \(\vec{w} = c\vec{v}_1 + d\vec{v}_2\)
\(\vec{u} + \vec{w} = (a+c)\vec{v}_1 + (b+d)\vec{v}_2\) is just another linear combo of \(\vec{v}_1, \vec{v}_2\)
\(c\vec{u} = (ca)\vec{v}_1 + (cb)\vec{v}_2\)

\(H\) is a subspace

Example: Vector Form

\(W\) is the set of all vectors of the form:

\[ \begin{bmatrix} 4b - 2c \\ -b \\ 9c \end{bmatrix} = b \begin{bmatrix} 4 \\ -1 \\ 0 \end{bmatrix} + c \begin{bmatrix} -2 \\ 0 \\ 9 \end{bmatrix} \]

\(W\) is \(\text{span} \left\{ \begin{bmatrix} 4 \\ -1 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 9 \end{bmatrix} \right\} \) and using the result from above, \(W\) is a subspace (of \(\mathbb{R}^3\)).