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Linear Algebra For Undergraduate Students

Yilan(Peter) Tan

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
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\begin{equation*} \mathbf{v} = c_1\mathbf{b}_1 + c_2\mathbf{b}_2 + \cdots + c_n\mathbf{b}_n. \end{equation*}
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The coordinate vector of \(\mathbf{v}\) relative to the basis \(\mathcal{B}\) is
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\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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Example 4.5.2. Finding Coordinate Vectors in \(\mathbb{R}^4\).

Let
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\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*}
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  1. Show that \(v_1, v_2, v_3\) and \(v_4\) is a basis of \(\mathbb{R}^4\text{.}\)
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  2. Let \(\mathbf{B}=\{v_1,v_2,v_3,v_4\}\text{.}\) Compute \([v_5]_{\mathbf{B}}\text{.}\)
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  3. Suppose that \([v]_{\mathbf{B}}=\begin{bmatrix}1\\-2\\1\\2\end{bmatrix}\text{.}\) Find \(v\text{.}\)
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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
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\begin{equation*} v_5 = c_1v_1 + c_2v_2 + c_3v_3 + c_4v_4. \end{equation*}
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Part 3: see Part 2, but in reverse. We are given the coordinate vector and need to find the original vector \(v\) by computing
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\begin{equation*} v = 1\cdot v_1 + (-2)\cdot v_2 + 1\cdot v_3 + 2\cdot v_4. \end{equation*}
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Summary:
  1. 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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  2. 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 4.5.2 Properties of Coordinate Vectors

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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Proof.

(\(\Rightarrow\)) Forward Direction: Assume \(S = \{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\}\) is linearly independent. We want to show that \(\{[\mathbf{v}_1]_\alpha, [\mathbf{v}_2]_\alpha, \ldots, [\mathbf{v}_k]_\alpha\}\) is linearly independent in \(\mathbb{R}^n\text{.}\)
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Suppose \(c_1[\mathbf{v}_1]_\alpha + c_2[\mathbf{v}_2]_\alpha + \cdots + c_k[\mathbf{v}_k]_\alpha = \mathbf{0}\text{.}\) By the properties of coordinate vectors,
\begin{align*} c_1[\mathbf{v}_1]_\alpha + c_2[\mathbf{v}_2]_\alpha + \cdots + c_k[\mathbf{v}_k]_\alpha &= [c_1\mathbf{v}_1]_\alpha + [c_2\mathbf{v}_2]_\alpha + \cdots + [c_k\mathbf{v}_k]_\alpha\\ &= [c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k]_\alpha\\ &= \mathbf{0}. \end{align*}
Since the coordinate map is one-to-one (each vector has a unique representation in the basis \(\alpha\)), we have
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\begin{equation*} c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{0}_V. \end{equation*}
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Since \(S\) is linearly independent, this implies \(c_1 = c_2 = \cdots = c_k = 0\text{.}\) Therefore, \(\{[\mathbf{v}_1]_\alpha, [\mathbf{v}_2]_\alpha, \ldots, [\mathbf{v}_k]_\alpha\}\) is linearly independent.
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(\(\Leftarrow\)) Backward Direction: Assume \(\{[\mathbf{v}_1]_\alpha, [\mathbf{v}_2]_\alpha, \ldots, [\mathbf{v}_k]_\alpha\}\) is linearly independent. We want to show that \(S\) is linearly independent.
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Suppose \(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{0}_V\text{.}\) Taking coordinate vectors with respect to \(\alpha\text{:}\)
\begin{align*} [c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k]_\alpha &= [\mathbf{0}_V]_\alpha\\ c_1[\mathbf{v}_1]_\alpha + c_2[\mathbf{v}_2]_\alpha + \cdots + c_k[\mathbf{v}_k]_\alpha &= \mathbf{0}. \end{align*}
Since \(\{[\mathbf{v}_1]_\alpha, [\mathbf{v}_2]_\alpha, \ldots, [\mathbf{v}_k]_\alpha\}\) is linearly independent, we must have \(c_1 = c_2 = \cdots = c_k = 0\text{.}\) Therefore, \(S\) is linearly independent.
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Insight 4.5.5. Key Insight: Coordinate Vectors Connect Abstract and Concrete.

Theorem 4.5.4 is remarkably powerful: it tells us that linear independence is preserved under coordinate transformations. This means we can test whether vectors in any vector space are linearly independent by simply checking whether their coordinate vectors in \(\mathbb{R}^n\) are linearly independent.
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For example:
  • To check if polynomials \(p_1(x), p_2(x), \ldots, p_k(x)\) are linearly independent in \(\mathcal{P}_n(x)\text{,}\) we can convert them to coordinate vectors and check linear independence in \(\mathbb{R}^{n+1}\) using techniques we already know (row reduction, determinants, etc.).
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  • To check if matrices \(M_1, M_2, \ldots, M_k\) are linearly independent in \(\mathcal{M}_{m \times n}\text{,}\) we can "flatten" them to coordinate vectors in \(\mathbb{R}^{mn}\) and apply standard tests.
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  • To check if functions \(f_1, f_2, \ldots, f_k\) are linearly independent in a function space, we can use their coordinate representations with respect to a basis.
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This theorem is the bridge that allows us to use computational linear algebra (row reduction, matrix operations) to solve problems in abstract vector spaces (polynomials, matrices, functions). It transforms abstract questions into concrete calculations.
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Example 4.5.6. Application to Polynomial Space \(\mathcal{P}_3(x)\).

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*}
  1. 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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  2. 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:
  1. 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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  2. 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 4.5.3 Change of Basis and Transition Matrices

So far, we have seen how to represent vectors using coordinate vectors relative to a fixed basis. But what happens when we want to change from one basis to another? This is a common situation in applications — we might start with the standard basis for convenience, but then switch to a different basis that reveals important structure in our problem.
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The key question is: if we know the coordinate vector of \(\mathbf{v}\) with respect to one basis \(\alpha\text{,}\) how do we find its coordinate vector with respect to a different basis \(\beta\text{?}\)
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Definition 4.5.7. Transition Matrix (Change of Basis Matrix).

Let \(V\) be a finite-dimensional vector space with two bases:
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\begin{equation*} \alpha = \{\mathbf{\alpha}_1, \mathbf{\alpha}_2, \ldots, \mathbf{\alpha}_n\} \quad \text{and} \quad \beta = \{\mathbf{\beta}_1, \mathbf{\beta}_2, \ldots, \mathbf{\beta}_n\}. \end{equation*}
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The transition matrix from \(\alpha\) to \(\beta\) (also called the change of basis matrix) is defined as
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\begin{equation*} P_{\alpha \to \beta} = \begin{bmatrix} [\mathbf{\alpha}_1]_\beta & [\mathbf{\alpha}_2]_\beta & \cdots & [\mathbf{\alpha}_n]_\beta \end{bmatrix}. \end{equation*}
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In other words, the columns of \(P_{\alpha \to \beta}\) are the coordinate vectors of the \(\alpha\)-basis vectors expressed in the \(\beta\)-basis.
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Proof.

Let \([\mathbf{v}]_\alpha = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix}\text{.}\) This means
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\begin{equation*} \mathbf{v} = c_1 \mathbf{\alpha}_1 + c_2 \mathbf{\alpha}_2 + \cdots + c_n \mathbf{\alpha}_n. \end{equation*}
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Taking coordinate vectors with respect to \(\beta\) on both sides:
\begin{align*} [\mathbf{v}]_\beta &= [c_1 \mathbf{\alpha}_1 + c_2 \mathbf{\alpha}_2 + \cdots + c_n \mathbf{\alpha}_n]_\beta\\ &= c_1 [\mathbf{\alpha}_1]_\beta + c_2 [\mathbf{\alpha}_2]_\beta + \cdots + c_n [\mathbf{\alpha}_n]_\beta\\ &= \begin{bmatrix} [\mathbf{\alpha}_1]_\beta & [\mathbf{\alpha}_2]_\beta & \cdots & [\mathbf{\alpha}_n]_\beta \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix}\\ &= P_{\alpha \to \beta} [\mathbf{v}]_\alpha. \end{align*}
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Important Properties:
  1. \(P_{\alpha \to \beta}\) is always invertible, and \((P_{\alpha \to \beta})^{-1} = P_{\beta \to \alpha}\text{.}\)
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  2. If \(\alpha\) is the standard basis, then \(P_{\alpha \to \beta} = [[\mathbf{\beta}_1]_\alpha \,|\, [\mathbf{\beta}_2]_\alpha \,|\, \cdots \,|\, [\mathbf{\beta}_n]_\alpha]^{-1}\text{.}\)
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  3. Transition matrices satisfy \(P_{\alpha \to \gamma} = P_{\beta \to \gamma} P_{\alpha \to \beta}\) (matrix multiplication corresponds to composition of coordinate changes).
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Example 4.5.9. Change of Basis in \(\mathbb{R}^3\).

Let \(\alpha\) be the standard basis of \(\mathbb{R}^3\text{,}\) and let
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\begin{equation*} \beta = \left\{\begin{bmatrix}1\\1\\0\end{bmatrix}, \begin{bmatrix}1\\0\\1\end{bmatrix}, \begin{bmatrix}0\\1\\1\end{bmatrix}\right\}. \end{equation*}
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  1. Verify that \(\beta\) is a basis of \(\mathbb{R}^3\text{.}\)
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  2. Find the transition matrix \(P_{\beta \to \alpha}\) (from \(\beta\) to the standard basis).
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  3. Find the transition matrix \(P_{\alpha \to \beta}\) (from standard basis to \(\beta\)).
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  4. If \([\mathbf{v}]_\beta = \begin{bmatrix}2\\-1\\3\end{bmatrix}\text{,}\) find \([\mathbf{v}]_\alpha\text{.}\)
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Solution.
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Key Takeaway: The transition matrix \(P_{\alpha \to \beta}\) acts as a "translator" between coordinate systems. It allows us to work with the same vector \(\mathbf{v}\) using different coordinate representations. This is essential in many applications where certain bases reveal structure or simplify computations.
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Subsection 4.5.4 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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