# Revision of Linear Algebra prerequisites

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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