Ch.6 Dimension

Dimension

vector space is finite-dimensional if it has a basis with only finitely many vectors

space $\mathbb{R}^n$ has a basis with $n$ elements so is finite-dimensional
space of all polynomials has infinitely many dimensions
will assume finite-dimensional spaces from here

Example of a infinite-dimensional space

The space of all polynomials has infinite dimensions because you can have a polynomial with as high of a degree as you want, with no limit

For any finite-dimensional space, all of its bases have the same number of elements

Proof

differs from textbook proof
take a vector space, choose one $B=\langle \vec{\beta}_1,...,\vec{\beta}_n\rangle$ that has minimal size
a different basis $D=\langle \vec{\delta}_1,...,\vec{\delta}_n\rangle$ has to have at least $n$ elements
We must show it cannot have more than $n$ elements
Suppose $D$ has an $n+1$ th element and is linearly independent
Then the only solution to $a_1\vec{\delta}_1+\cdots+a_{n+1}\vec{\delta}_{n+1}=\vec{0}$ is all $a_i=0$
Now, represent the vectors of $D$ in terms of $B$
$\text{Rep}_B(\delta_1)=\begin{pmatrix}c_{1,1}\\\vdots\\c_{n,1}\end{pmatrix}\text{ }\cdots\text{ }\text{Rep}_B(\delta_{n+1})=\begin{pmatrix}c_{1,n+1}\\\vdots\\c_{n,n+1}\end{pmatrix}$
The linear relationship $a_1\vec{\delta}_1+\cdots+a_{n+1}\vec{\delta}_{n+1}=\vec{0}$ should still hold even after substituting the representation. This gives the linear system
$c_{1,1}a_1+\cdots+c_{1,n+1}a_{n+1}=0\\\vdots\\c_{n,1}a_1+\cdots+c_{n,n+1}a_{n+1}=0$
which has $n+1$ unknowns and $n$ equations. This must have infinitely many solutions for $(a_1,...,a_{n+1})$ , but that contradicts the earlier statement that the solution that all $a_i=0$ , therefore, there must not be an $n+1$ th element.

The dimension of a vector space is the number of vectors in any of its bases

In $n$ -dimensional space, a set of $n$ vectors is linearly independent iff they span the space
should be intuitive

Space and Rank

$\begin{pmatrix}\textcolor{#F79256}{\text{row}}\\\textcolor{#F79256}{\text{column}}\end{pmatrix}$ space is the span of the set of its $\begin{pmatrix}\text{row}\\\text{column}\end{pmatrix}$
$\begin{pmatrix}\textcolor{#F79256}{\text{row}}\\\textcolor{#F79256}{\text{column}}\end{pmatrix}$ rank is the dimension of the space, i.e. the number of linearly independent $\begin{pmatrix}\text{row}\\\text{column}\end{pmatrix}$

If matrix $A\sim B$ by row operation, row spaces are equal
The rows in $B$ are linear combinations of rows in $A$ , so rows in $B\in[A]$
non-zero rows in a echelon form matrix are linearly independent $\rightarrow$ forms a basis
Gauss's Method produces a basis for the row space of a matrix

solution exists iff constant column vector is in the column space of the coefficient matrix

Example 6.1

Consider this reduced echelon form matrix
$\begin{pmatrix}1&0&2&1&0\\0&1&1&3&2\\0&0&0&0&0\end{pmatrix}$
The row space is the span of the first two rows:
$\left\{a\begin{pmatrix}1\\0\\2\\1\\0\end{pmatrix}+ b\begin{pmatrix}0\\1\\1\\3\\2\end{pmatrix}\middle|a,b\in\mathbb{R}\right\}$
Those two column vectors also form the basis

Example 6.2

Consider the linear system
$2x+y=c_1\\-x+3y=c_2$

This can be represented as

x\cdot\begin{pmatrix}2\\-1\end{pmatrix}+y\cdot\begin{pmatrix}1\\3\end{pmatrix}=\begin{pmatrix}c_1\\c_2\end{pmatrix}

This has a solution if

\begin{pmatrix}c_1\\c_2\end{pmatrix}

is a linear combination of the two vectors, i.e. the constant column vector is in the column space of

\begin{pmatrix}2&1\\-1&3\end{pmatrix}

transpose of a matrix changes its rows and columns $A_{ij}=A^T_{ji}$
The basis of the column space can be found by row reducing the transpose

Example 6.3

The basis for the colum space of the matrix
$\begin{pmatrix}2&1\\-1&3\end{pmatrix}$
can be found by taking the transpose
$\begin{pmatrix}2&1\\-1&3\end{pmatrix}^T=\begin{pmatrix}2&-1\\1&3\end{pmatrix}$
then row reducing
$\begin{pmatrix}2&-1\\1&3\end{pmatrix}\sim\begin{pmatrix}1&3\\0&-7\end{pmatrix}$
Then transpose back
$\begin{pmatrix}1&3\\0&-7\end{pmatrix}^T=\begin{pmatrix}1&0\\3&-7\end{pmatrix}$
The basis is
$B=\left\langle\begin{pmatrix}1\\3\end{pmatrix},\begin{pmatrix}0\\-7\end{pmatrix}\right\rangle$
So the span is $\mathbb{R}^2$

For any matrix, row rank equals column rank
in reduced echelon form, row rank equals number of leading entries, which equals column rank
since theyre the same, its called the rank

For linear systems with $n$ unknowns and coefficeint matrix $A$ , the statements

the rank of $A$ is $r$
the vector space of solutions of the associated homogeneous system has dimension $n-r$

are equivalent

Proof

(1) iff $r$ leading variables, which is true iff there are $n-r$ free variables $\rightarrow$ (2)

In an $n\times n$ matrix $A$ , the statements

the rank of $A$ is $n$
$A$ is nonsingular
rows form a linearly independent set
columns form a linearly independent set
any linear system whose matrix of coefficients is $A$ has exactly one solution

are equivalent

Review:

defined leading variable, leading entry (last column may contain leading variable in augmented matrix) $\rightarrow$ $\to$ pivot
- a system with $0=1$ has a third leading entry but no second leading variable
- pivots in linear system are same as pivots in augmented matrix that are not in the last column
defined echelon form as pivot goes right as you go down the rows
defined reduced row echelon form if it is in echelon form and each pivot is 1 and is the only non-zero entry in the column (must cancel up)
Three possibilities:

last column contains a pivot: inconsistent, no solutions
every column except the last contains a pivot: unique solution
the last column and some other column(s) does not contain a pivot: infinitely many solutions

defined homogeneous linear system as having augmented matrix $(A|\vec{0})$ $(A ∣ 0)$
- all echelon forms of $(A|\vec{0})$ are of type $(B|\vec{0})$
- always contains at least one solution (namely, $(0,...,0)$ )
- column with no pivot $\rightarrow$ infinitely many solutions
defined linear independence in a vector space $V$ $V$ as set of vectors which cannot make $\vec{0}$ $0$ with a linear combination, except if all constants were $0$ $0$
- linear independence can be checked when $V=\mathbb{R}^n$ $V = R^{n}$ by making a matrix with the columns as the vectors
  - corresponds to linear relation, must have unique solution ( $0$ s everywhere), so must have pivot in every column
- vectors form a basis of $\mathbb{R}^n$ if the reduced echelon form of $\text{Mat}(\vec{a}_1,...,\vec{a}_n)$ is the $n\times n$ identity matrix $\text{Id}_n$