# 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.

## Contents

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## References

## 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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