Application: Coordinate Vectors

Section 4.5 Application: Coordinate Vectors

In abstract vector spaces, we often work with vectors that are not written as column vectors with numerical entries. However, once we choose a basis for the vector space, we can represent any vector uniquely as a coordinate vector with respect to that basis. This provides a powerful connection between abstract vector spaces and the familiar \(\mathbb{R}^n\text{.}\)

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Subsection 4.5.1 Definition of Coordinate Vectors

Definition 4.5.1. Coordinate Vector.

Let \(V\) be a vector space and let \(\mathcal{B} = \{\mathbf{b}_1, \mathbf{b}_2, \ldots, \mathbf{b}_n\}\) be a basis for \(V\text{.}\) For any vector \(\mathbf{v} \in V\text{,}\) there exist unique scalars \(c_1, c_2, \ldots, c_n\) such that

\begin{equation*} \mathbf{v} = c_1\mathbf{b}_1 + c_2\mathbf{b}_2 + \cdots + c_n\mathbf{b}_n. \end{equation*}

The coordinate vector of \(\mathbf{v}\) relative to the basis \(\mathcal{B}\) is

\begin{equation*} [\mathbf{v}]_{\mathcal{B}} = \begin{bmatrix}c_1\\c_2\\\vdots\\c_n\end{bmatrix}. \end{equation*}

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The coordinate vector \([\mathbf{v}]_{\mathcal{B}}\) is an element of \(\mathbb{R}^n\) and tells us exactly how to express \(\mathbf{v}\) as a linear combination of the basis vectors.

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Subsection 4.5.2 Properties of Coordinate Vectors

Theorem 4.5.2. Properties of Coordinate Vectors.

Let \(V\) be a vector space with basis \(\mathcal{B} = \{\mathbf{b}_1, \mathbf{b}_2, \ldots, \mathbf{b}_n\}\text{.}\) For vectors \(\mathbf{u}, \mathbf{v} \in V\) and scalar \(k\text{:}\)

\([\mathbf{u} + \mathbf{v}]_{\mathcal{B}} = [\mathbf{u}]_{\mathcal{B}} + [\mathbf{v}]_{\mathcal{B}}\) (Additivity)

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\([k\mathbf{v}]_{\mathcal{B}} = k[\mathbf{v}]_{\mathcal{B}}\) (Homogeneity)

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Proof.

Property 1: Let \(\mathbf{u} = a_1\mathbf{b}_1 + a_2\mathbf{b}_2 + \cdots + a_n\mathbf{b}_n\) and \(\mathbf{v} = c_1\mathbf{b}_1 + c_2\mathbf{b}_2 + \cdots + c_n\mathbf{b}_n\text{.}\) Then

\begin{align*} \mathbf{u} + \mathbf{v} &= (a_1\mathbf{b}_1 + a_2\mathbf{b}_2 + \cdots + a_n\mathbf{b}_n) + (c_1\mathbf{b}_1 + c_2\mathbf{b}_2 + \cdots + c_n\mathbf{b}_n)\\ &= (a_1 + c_1)\mathbf{b}_1 + (a_2 + c_2)\mathbf{b}_2 + \cdots + (a_n + c_n)\mathbf{b}_n \end{align*}

Therefore, \([\mathbf{u} + \mathbf{v}]_{\mathcal{B}} = \begin{bmatrix}a_1 + c_1\\a_2 + c_2\\\vdots\\a_n + c_n\end{bmatrix} = \begin{bmatrix}a_1\\a_2\\\vdots\\a_n\end{bmatrix} + \begin{bmatrix}c_1\\c_2\\\vdots\\c_n\end{bmatrix} = [\mathbf{u}]_{\mathcal{B}} + [\mathbf{v}]_{\mathcal{B}}\text{.}\)

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Property 2: Let \(\mathbf{v} = c_1\mathbf{b}_1 + c_2\mathbf{b}_2 + \cdots + c_n\mathbf{b}_n\text{.}\) Then

\begin{align*} k\mathbf{v} &= k(c_1\mathbf{b}_1 + c_2\mathbf{b}_2 + \cdots + c_n\mathbf{b}_n)\\ &= (kc_1)\mathbf{b}_1 + (kc_2)\mathbf{b}_2 + \cdots + (kc_n)\mathbf{b}_n \end{align*}

Therefore, \([k\mathbf{v}]_{\mathcal{B}} = \begin{bmatrix}kc_1\\kc_2\\\vdots\\kc_n\end{bmatrix} = k\begin{bmatrix}c_1\\c_2\\\vdots\\c_n\end{bmatrix} = k[\mathbf{v}]_{\mathcal{B}}\text{.}\)

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These properties show that the coordinate mapping preserves the linear structure of the vector space. This is why coordinate vectors provide such a powerful tool for working with abstract vector spaces.

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Subsection 4.5.3 An example in \(\mathbb{R}^4\)

Example 4.5.3.

Let

\begin{equation*} S=\{v_1,v_2,v_3,v_4,v_5\}=\left\{\begin{bmatrix}1\\0\\-1\\0\end{bmatrix},\begin{bmatrix}1\\1\\0\\2\end{bmatrix},\begin{bmatrix}0\\3\\1\\-2\end{bmatrix},\begin{bmatrix}0\\1\\-1\\2\end{bmatrix}, \begin{bmatrix}1\\1\\1\\1\end{bmatrix}\right\}. \end{equation*}

Show that \(v_1, v_2, v_3\) and \(v_4\) is a basis of \(\mathbb{R}^4\text{.}\)
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Let \(\mathbf{B}=\{v_1,v_2,v_3,v_4\}\text{.}\) Compute \([v_5]_{\mathbf{B}}\text{.}\)
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Solution.

Part 1: To show that \(\{v_1, v_2, v_3, v_4\}\) is a basis of \(\mathbb{R}^4\text{,}\) we need to prove that these four vectors are linearly independent and span \(\mathbb{R}^4\text{.}\) Since we have exactly 4 vectors in a 4-dimensional space, it suffices to show they are linearly independent.

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Part 2: Once we establish that \(\mathbf{B} = \{v_1, v_2, v_3, v_4\}\) is a basis, we can find the coordinate vector \([v_5]_{\mathbf{B}}\) by solving for scalars \(c_1, c_2, c_3, c_4\) such that

\begin{equation*} v_5 = c_1v_1 + c_2v_2 + c_3v_3 + c_4v_4. \end{equation*}

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Summary:

The set \(\{v_1, v_2, v_3, v_4\}\) forms a basis for \(\mathbb{R}^4\) because the \(4 \times 4\) matrix \(A = [v_1 | v_2 | v_3 | v_4]\) has a non-zero determinant, which means the vectors are linearly independent. Since we have 4 linearly independent vectors in a 4-dimensional space, they automatically span the entire space.

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The coordinate vector \([v_5]_{\mathbf{B}}\) is found by solving the matrix equation \(A\mathbf{c} = v_5\text{,}\) where \(\mathbf{c}\) contains the coefficients that express \(v_5\) as a linear combination of the basis vectors.

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Subsection 4.5.4 Application to Polynomial Space \(\mathcal{P}_3(x)\)

Example 4.5.4.

Consider the polynomial space \(\mathcal{P}_3(x)\) and the following set of polynomials:

\begin{align*} p_1(x) &= 1 - x^2\\ p_2(x) &= 1 + x + 2x^3\\ p_3(x) &= 3x + x^2 - 2x^3\\ p_4(x) &= x - x^2 + 2x^3\\ p_5(x) &= 1 + x + x^2 + x^3 \end{align*}

Show that \(\{p_1(x), p_2(x), p_3(x), p_4(x)\}\) is a basis of \(\mathcal{P}_3(x)\text{.}\)
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Let \(\mathbf{B}=\{p_1(x), p_2(x), p_3(x), p_4(x)\}\text{.}\) Compute \([p_5(x)]_{\mathbf{B}}\text{.}\)
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Solution.

Key Insight: We can solve this problem by converting the polynomials to their coordinate vectors relative to the standard basis \(\{1, x, x^2, x^3\}\text{,}\) which transforms this into the previous \(\mathbb{R}^4\) problem.

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Part 1: To show that \(\{p_1(x), p_2(x), p_3(x), p_4(x)\}\) is a basis of \(\mathcal{P}_3(x)\text{,}\) we convert each polynomial to its coordinate vector and check linear independence.

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Part 2: Once we establish the basis, we find \([p_5(x)]_{\mathbf{B}}\) by expressing \(p_5(x)\) as a linear combination of the basis polynomials.

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Connection to Previous Example: Notice that this polynomial problem has exactly the same solution as the \(\mathbb{R}^4\) problem! This demonstrates the power of coordinate vectors:

The polynomials \(p_1, p_2, p_3, p_4\) have coordinate vectors \(v_1, v_2, v_3, v_4\) from the previous example

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The polynomial \(p_5(x) = 1 + x + x^2 + x^3\) has coordinate vector \([1,1,1,1]^T\)

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The linear relationships are preserved: \([p_5(x)]_{\mathbf{B}} = [v_5]_{\mathbf{B}}\)

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Same coefficients, same linear algebra, different vector spaces!

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Summary:

The set \(\{p_1(x), p_2(x), p_3(x), p_4(x)\}\) forms a basis for \(\mathcal{P}_3(x)\) because their coordinate vectors relative to the standard basis form a linearly independent set in \(\mathbb{R}^4\text{.}\)

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The coordinate vector \([p_5(x)]_{\mathbf{B}}\) is found by solving the same matrix equation as in the \(\mathbb{R}^4\) case, showing how coordinate representations unify different vector spaces.

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Subsection 4.5.5 Conclusion

Coordinate vectors provide a bridge between abstract vector spaces and concrete numerical computations. They allow us to:

Represent abstract vectors as familiar column vectors in \(\mathbb{R}^n\)

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Perform computations using matrix algebra

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Transfer results between different vector spaces

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Solve problems in the most convenient setting

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This connection is fundamental to many applications of linear algebra in engineering, computer science, and mathematics.

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