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 [math]{\mathbb R}^2[/math], [math]{\mathbb R}^3[/math] and polynomials.

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References

Examples of vector spaces

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

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

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

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

Linear combinations and Spanning sets

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

Examples:

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

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

Examples:

  • The set [math]\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 [/math] is a spanning set for [math]P_{5}[/math]. This is since any vector in [math]P_{5}[/math], :[math]p = c_{0} + c_{1}x +c_{2}x^2 + c_{3} x^3 + c_{4} x_4 + c_{5} x^5[/math]

$\qquad$ can be expressed as:

[math]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} + [/math]
[math]\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}.[/math]
  • The set [math]\left\lbrace x^5, x^4, x^3, x^2, x, 1 \right\rbrace[/math] is also a spanning set for [math]P_{5}[/math].
  • The set [math]\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[/math] is a spanning set for [math]{\mathbb R}^3[/math], as any vector [math]\left( \begin{array}{c}a\\b\\c\\\end{array}\right)[/math] can be written as :[math]\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).[/math]

Linear Independence

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

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

Examples:

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

then

[math]\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)[/math] and so [math]a_{1}+2a_{2} = 0[/math], [math]3a_{1} = 0[/math] and [math]-2a_{1}+3a_{2} = 0[/math]. Then [math]a_{1} = 0[/math] and [math]a_2 = 0[/math].
  • The set [math]\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[/math] is a linearly independent set in [math]{\mathbb R}^3[/math]. For suppose [math]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).[/math]

Then

[math]\left( \begin{array}{c}a_{1}\\a_{2}\\a_{3}\\\end{array} \right) = \left( \begin{array}{c}0\\0\\0\\\end{array} \right)[/math] and hence [math]a_{1} = 0, a_{2} = 0[/math] and [math]a_3 = 0[/math].
  • The set [math]\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[/math] is a linearly independent set in [math]P^5.[/math] For suppose :[math]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.[/math]

Then

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

Basis and Dimension

A basis for a vector space [math]V[/math] is a linearly independent spanning set for [math]V[/math].

Examples:

  • The set [math]\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[/math] is a linearly independent spanning set for [math]{\mathbb R}^3[/math], so is a basis of [math]{\mathbb R}^3[/math].
  • The set [math]\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[/math] is a linearly independent spanning set in [math]P^5[/math], so is a basis for [math]P^5[/math].
  • The set [math]\left\lbrace x^5, x^4, x^3, x^2, x,1 \right\rbrace [/math] is also a basis for [math]P^5[/math].

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

The dimension of [math]V[/math] is the number of elements in any basis.

Examples:

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

Useful facts:

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




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