MATH 323 Lecture 10

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Vector Space

Closure criteria:

1. ${\vec {x}}+{\vec {y}}\in V\quad \forall {\vec {x}},{\vec {y}}\in V$
2. $\alpha \,{\vec {x}}\in V\quad \forall {\vec {x}}\in V\quad \forall \alpha \in \mathbb {R}$

Subspace

A vector space $V$ automatically has two subspaces:

1. $S_{1}=\{0\}$
2. $S_{2}=V$

When $S\neq \{0\}$ and $S\neq V$, then $S$ is called a proper subspace.

Null Space of Matrix

Let $A$ be a $m\times n$ matrix, and let $N(A)$ be the set of all solutions to the system $A\,{\vec {x}}={\vec {0}}$:

$N(A)=\left\{{\vec {x}}\in \mathbb {R} ^{n}~\mid ~A\,{\vec {x}}={\vec {0}}\right\}$

$N(A)$ is clalled the null space (also called nullspace or kernel) of $A$.

$N(A)$ is a vector space since $A({\vec {x}}+{\vec {y}})=A\,{\vec {x}}+A\,{\vec {y}}={\vec {0}}$ and $A(\alpha \,{\vec {x}})=\alpha \,A\,{\vec {x}}=\alpha \,{\vec {0}}={\vec {0}}$

Example

Determine $N(A)$ if $A={\begin{bmatrix}1&1&1&0\\2&1&0&1\end{bmatrix}}$

Gauss-Jordan reduction of $A$ gives $\left[{\begin{array}{cccc|c}1&0&-1&1&0\\0&1&2&-1&0\end{array}}\right]$, so the solutions of $A\,{\vec {x}}={\vec {0}}$ are

${\begin{aligned}{\vec {x}}&=\left\langle \alpha -\beta ,-2\alpha +\beta ,\alpha ,\beta \right\rangle \quad \forall \alpha ,\beta \in \mathbb {R} \\&=\alpha {\begin{pmatrix}1\\-2\\1\\0\end{pmatrix}}+\beta {\begin{pmatrix}-1\\1\\0\\1\end{pmatrix}}\end{aligned}}$

Linear Combinations and Span

Let ${\vec {v}}=v_{1},\ldots ,v_{n}\in V$.

The sum $\alpha _{1}\,v_{1}+\alpha _{2}\,v_{2}+\dots +\alpha _{n}\,v_{n}$, where ${\vec {\alpha }}\in \mathbb {R} ^{n}$, is called a linear combination of ${\vec {v}}$.

The set of all such linear combinations is called the span of ${\vec {v}}$:

$\mathrm {Span} (v_{1},\ldots ,v_{n})=\{v\in V~\mid ~V=\alpha _{1}\,v_{1}+\dots +\alpha _{n}\,v_{n}\}$

Example

${\vec {b}}_{1}=\left\langle 1,0,0\right\rangle$ and ${\vec {b}}_{2}=\left\langle 0,1,0\right\rangle$ are in $\mathbb {R} ^{3}$

The span of ${\vec {b}}_{1}$ and ${\vec {b}}_{2}$ is the set of all vectors ${\vec {x}}=\alpha \,{\vec {b}}_{1}+\beta \,{\vec {b}}_{2}=\left\langle \alpha ,\beta ,0\right\rangle$.

These two vectors form the standard basis in 2D space.

Theorem

$\mathrm {Span} (v_{1},\ldots ,v_{n})$ is a subspace of $V$

${\begin{aligned}{\vec {x}}&=\alpha _{1}\,v_{1}+\dots +\alpha _{n}\,v_{n}\\{\vec {y}}&=\beta _{1}\,v_{1}+\dots +\beta _{n}\,v_{n}\\{\vec {x}}+{\vec {y}}&=(\alpha _{1}+\beta _{1})v_{1}+\dots +(\alpha _{n}+\beta _{n})v_{n}\in \mathrm {Span} (v_{1},\ldots ,v_{n})\\\lambda \,{\vec {x}}&=(\lambda \,\alpha _{1})v_{1}+\dots +(\lambda \,\alpha _{n})v_{n}\in \mathrm {Span} (v_{1},\ldots ,v_{n})\end{aligned}}$

Spanning Set

The set $S=\{v_{1},\ldots ,v_{n}\}$ is a spanning set for $V$ if $V=\mathrm {Span} (v_{1},\ldots ,v_{n})$

If a subset of $S$ is a spanning set of $V$, then $S$ itself is a spanning set of $V$ (set linear coefficient of other terms in set to 0).

Example

$\mathrm {Span} ({\hat {\imath }},{\hat {\jmath }},{\hat {k}})=\mathbb {R} ^{3}$

These are the standard basis vectors for 3D space.

Example

Is $\left\{{\begin{pmatrix}1\\1\\1\end{pmatrix}},{\begin{pmatrix}1\\1\\0\end{pmatrix}},{\begin{pmatrix}1\\0\\0\end{pmatrix}}\right\}$ a spanning set of $\mathbb {R} ^{3}$?

${\begin{aligned}\alpha _{1}\left\langle 1,1,1\right\rangle +\alpha _{2}\left\langle 1,1,0\right\rangle +\alpha _{3}\left\langle 1,0,0\right\rangle &=\left\langle x_{1},x_{2},x_{3}\right\rangle \\\alpha _{1}+\alpha _{2}+\alpha _{3}&=x_{1}\\\alpha _{1}+\alpha _{2}&=x_{2}\\\alpha _{1}&=x_{3}\\\end{aligned}}$

This has a solution for any ${\vec {x}}\in \mathbb {R} ^{3}$, so it is indeed a spanning set (the vectors in the original problem are noncoplanar, so that's easier to visualize)

Example

$P_{3}$ is the set of polynomials of degree < 3.

$P_{3}=\mathrm {Span} (1,x,x^{2})$

$c\,v_{1}+b\,v_{2}+a\,v_{3}=a\,x^{2}+b\,x+c\in P_{3}$

Linear Independence

Given ${\vec {x}}_{1}=\left\langle 1,-1,2\right\rangle$, ${\vec {x}}_{2}=\left\langle -2,3,1\right\rangle$, and ${\vec {x_{3}}}=\left\langle -1,3,8\right\rangle$,

$S=\mathrm {Span} ({\vec {x}}_{1},{\vec {x}}_{2},{\vec {x}}_{3})=\mathrm {Span} ({\vec {x}}_{1},{\vec {x}}_{2})$

This is true since ${\vec {x}}_{3}=3{\vec {x}}_{1}+2{\vec {x}}_{2}$

The set $\{{\vec {x}}_{1},{\vec {x}}_{2},{\vec {x}}_{3}\}$ is called linearly dependent

1. If $v_{1},\ldots ,v_{n}$ span $V$ and one of these vectors can be written as a linear combination of the $n-1$ others, then those $n-1$ vectors span $V$.
2. Given $n$ vectors $v_{1},\ldots ,v_{n}$, it is possible to write one of the vectors as a linear combination of the other $n-1$ vectors iff there exist scalars $c_{1},\ldots ,c_{n}$ (not all zero!) such that $c_{1}\,v_{1}+\dots +c_{n}\,v_{n}=0$.