# Revision of Linear Algebra prerequisites

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In this section, we review the concepts of linear combination, spanning set, linear independence, basis and dimension for a vector space. We will illustrate the concepts with examples from ${\mathbb R}^2$, ${\mathbb R}^3$ and polynomials.

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## Examples of vector spaces

Let ${\mathbb R}^2$ be the vector space of all $2 \times 1$ column vectors $\left ( \begin{array}{c}x\\y\end{array} \right )$ with entries in ${\mathbb R}$.

Let ${\mathbb R}^3$ be the vector space of all $3 \times 1$ column vectors $\left ( \begin{array}{c}x\\y\\z\end{array} \right )$ with entries in ${\mathbb R}$.

Let $F$ be a field and let $P_n$ be the vector space of all polynomials $a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$ of degree less than or equal to $n$, where the coefficients $a_0, a_1, \ldots , a_n$ are in the field $F$.

Recall that a vector is simply an element of a vector space. Thus $1+x+3x^4$ is a vector in $P_4$.

## Linear combinations and Spanning sets

Let $u$ and $v$ be vectors in a vector space $V$ over the field $F$. A linear combination of $u$ and $v$ is any vector of the form $r_1u+r_2v$ where $r_1$ and $r_2$ are in $F$.

Examples:

• The vector $\left ( \begin{array}{r}-16\\13\end{array} \right )$ is a linear combination of $\left ( \begin{array}{r}4\\-1\end{array} \right )$ and $\left ( \begin{array}{r}-2\\5\end{array} \right )$ since $\left ( \begin{array}{r}-16\\13\end{array} \right ) = -3 \left ( \begin{array}{r}4\\-1\end{array} \right )+ 2\left ( \begin{array}{r}-2\\5\end{array} \right )$.
• More generally, any element $\left ( \begin{array}{c}x\\y\end{array} \right )$ in ${\mathbb R}^2$ is a linear combination of $\left ( \begin{array}{r}4\\-1\end{array} \right )$ and $\left ( \begin{array}{r}-2\\5\end{array} \right )$ since we may write :$\left ( \begin{array}{c}x\\y\end{array} \right ) = \frac{5x+2y}{18} \left ( \begin{array}{r}4\\-1\end{array} \right )+ \frac{x+4y}{18}\left ( \begin{array}{r}-2\\5\end{array} \right ).$

Let $v_{1}, v_{2}, \ldots , v_{n}$ be vectors in a vector space $V$ over the field $F$. The set ${v_{1}, v_{2}, \ldots , v_{n}}$ is a spanning set for $V$ if every element in $V$ can be written as a linear combination of the vectors $v_{1}, v_{2}, ..., v_{n}$. That is, given $v$ in $V$, there exist scalars $a_{1}, a_{2}, \ldots , a_{n}$ in $F$ such that $v = a_{1}v_{1} + a_{2}v_{2} + \cdots + a_{n}v_{n}$. Note that some of the $a_{i}$ may be zero.

Examples:

• The set $\left\lbrace p_{1} = x^5 + x^4 + x^3 + x^2 + x + 1; p_{2} = x^4 - x^3 - x^2; p_{3} = x^3 - 1; p_{4} = x^2 - 3x; p_{5} = x + 1; p_{6} = x^2 – 1 \right\rbrace$ is a spanning set for $P_{5}$. This is since any vector in $P_{5}$, :$p = c_{0} + c_{1}x +c_{2}x^2 + c_{3} x^3 + c_{4} x_4 + c_{5} x^5$

$\qquad$ can be expressed as:

$p = c_{5} p_{1} + (c_{4} - c_{5}) p_{2} + (c_{3} + c_{4} - 2c_{5}) p_{3} + \frac{1}{4} (c_{0} - c_{1} + c_{2} + c_{3} + 2c_{4} - 4c_{5}) p_{4} +$
$\qquad \qquad \frac{1}{4} (3c_{0} + c_{1} + 3c_{2} + 3c_{3} + 6c_{4} - 16c_{5}) p_{5} + \frac{1}{4} (-c_{0} + c_{1} + 3c_{2} - c_{3} + 2c_{4} - 4c_{5}) p_{6}.$
• The set $\left\lbrace x^5, x^4, x^3, x^2, x, 1 \right\rbrace$ is also a spanning set for $P_{5}$.
• The set $\left\lbrace \left( \begin{array}{c}1\\0\\0\\\end{array}\right), \left( \begin{array}{c}0\\1\\0\\\end{array}\right), \left( \begin{array}{c}0\\0\\1\\\end{array}\right)\right\rbrace$ is a spanning set for ${\mathbb R}^3$, as any vector $\left( \begin{array}{c}a\\b\\c\\\end{array}\right)$ can be written as :$\left( \begin{array}{c}a\\b\\c\\\end{array}\right) = a \left( \begin{array}{c}1\\0\\0\\\end{array}\right) + b \left( \begin{array}{c}0\\1\\0\\\end{array}\right) + c \left( \begin{array}{c}0\\0\\1\\\end{array}\right).$

## Linear Independence

Let $v_{1}, v_{2}, \ldots , v_{n}$ be vectors in a vector space $V$ over a field $F$.

The set $\left\lbrace v_{1}, v_{2}, \ldots , v_{n}\right\rbrace$ is linearly independent if, whenever $a_{1}v_{1} + a_{2}v_{2} + \cdots + a_{n}v_{n} = 0$, then all of the scalars $a_{1}, a_{2}, \ldots , a_{n}$ are zero.

Examples:

• The set $\left\lbrace \left( \begin{array}{r}1\\3\\-2\\\end{array}\right), \left( \begin{array}{c}2\\0\\3\\\end{array}\right)\right\rbrace$ is linearly independent in ${\mathbb R}^3$. For if $a_{1} \left( \begin{array}{r}1\\3\\-2\\\end{array}\right) +a_{2}\left( \begin{array}{c}2\\0\\3\\\end{array}\right) = \left(\begin{array}{c}0\\0\\0\\\end{array}\right),$

then

$\left( \begin{array}{c}a_{1}+2a_{2}\\3a_{1}\\-2a_{1}+3a_{2}\\\end{array} \right) = \left( \begin{array}{c}0\\0\\0\\\end{array} \right)$ and so $a_{1}+2a_{2} = 0$, $3a_{1} = 0$ and $-2a_{1}+3a_{2} = 0$. Then $a_{1} = 0$ and $a_2 = 0$.
• The set $\left\lbrace \left( \begin{array}{c}1\\0\\0\\\end{array}\right), \left( \begin{array}{c}0\\1\\0\\\end{array}\right), \left ( \begin{array}{c}0\\0\\1\\\end{array}\right)\right\rbrace$ is a linearly independent set in ${\mathbb R}^3$. For suppose $a_{1} \left( \begin{array}{c}1\\0\\0\\\end{array}\right) + a_{2}\left( \begin{array}{c}0\\1\\0\\\end{array}\right) + a_{3}\left( \begin{array}{c}0\\0\\1\\\end{array}\right) = \left(\begin{array}{c}0\\0\\0\\\end{array}\right).$

Then

$\left( \begin{array}{c}a_{1}\\a_{2}\\a_{3}\\\end{array} \right) = \left( \begin{array}{c}0\\0\\0\\\end{array} \right)$ and hence $a_{1} = 0, a_{2} = 0$ and $a_3 = 0$.
• The set $\left\lbrace p_{1} = x^5 + x^4 + x^3 + x^2 + x + 1; p_2 = x^4 - x^3 - x^2; p_3 = x^3 - 1; p_4 = x^2 - 3x; p_5 = x + 1; p_6 = x^2 – 1 \right\rbrace$ is a linearly independent set in $P^5.$ For suppose :$a_{1} (x^5 + x^4 + x^3 + x^2 + x + 1) + a_{2} (x^4 - x^3 - x^2) + a_{3} (x^3 - 1) + a_{4} (x^2 - 3x) + a_{5} (x + 1) + a_{6} (x^2 - 1) = 0.$

Then

$a_{1} x^5 + (a_{1} + a_{2}) x^4 + (a_{1} - a_{2} +a_{3}) x^3 + (a_{1} - a_{2} + a_{4} + a_{6}) x^2 + (a_{1} - 3a_{4} +a_{5}) x + (a_{1} +a_{5} - a_{6}) = 0.$ Equating coefficients of $x^5$, $x^4$ and $x^3$ gives $a_{1} = 0$; $a_{1} + a_{2}=0$ and $a_{1} - a_{2} +a_{3}=0$. Therefore $a_{1} = 0$, $a_{2} = 0$, $a_{3} =0$. Using this and equating coefficients of $x^2$, $x$ and constants gives $a_{4} + a_{6} = 0$; $- 3a_{4} +a_{5}$ and $a_{5} - a_6 = 0$. Therefore $a_{4} = -a_{6}$, $a_{5} = a_{6}$ and $3a_{4} = a_{5}$. Hence $-3a_{6} = a_{6}$ so $a_{6} = 0$ and thus $a_{4} =0$, $a_{5} = 0$. Therefore it is a linearly independent set.

## Basis and Dimension

A basis for a vector space $V$ is a linearly independent spanning set for $V$.

Examples:

• The set $\left\lbrace \left( \begin{array}{c}1\\0\\0\\\end{array}\right), \left( \begin{array}{c}0\\1\\0\\\end{array}\right), \left( \begin{array}{c}0\\0\\1\\\end{array}\right)\right\rbrace$ is a linearly independent spanning set for ${\mathbb R}^3$, so is a basis of ${\mathbb R}^3$.
• The set $\left\lbrace p_{1} = x^5 + x^4 + x^3 + x^2 + x + 1; p_2 = x^4 - x^3 - x^2; p_3 = x^3 - 1; p_4 = x^2 - 3x; p_5 = x + 1; p_6 = x^2 – 1 \right\rbrace$ is a linearly independent spanning set in $P^5$, so is a basis for $P^5$.
• The set $\left\lbrace x^5, x^4, x^3, x^2, x,1 \right\rbrace$ is also a basis for $P^5$.

Theorem: All bases in a given vector space have the same number of elements.

The dimension of $V$ is the number of elements in any basis.

Examples:

• The dimension of ${\mathbb R}^3$ is 3 (and that of ${\mathbb R}^n$ is $n$).
• The dimension of $P^5$ is 6 (and that of $P^n$ is $n+1$).

Useful facts:

• a minimal spanning set is a basis;
• a maximal linearly independent set is a basis.

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