Linear Algebra

4.6 Change of Basis

Imron Rosyadi

Change of Basis

Adapting Coordinate Systems for Problem Solving

Welcome to this session on Change of Basis. In linear algebra, just like in many engineering problems, the choice of coordinate system can drastically simplify or complicate our work. While we’ve learned about various bases, sometimes a basis that’s perfect for one aspect of a problem might be inconvenient for another. This section explores how to effectively switch between different bases, which is a fundamental skill in fields like signal processing, control systems, and computer graphics.

The Need for Changing Bases

A suitable basis for one problem may not be suitable for another. Changing bases is analogous to changing coordinate axes in \(R^{2}\) and \(R^{3}\).

Coordinate Maps

If \(S = \{\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots , \mathbf{v}_{n}\}\) is a basis for \(V\), and \((\mathbf{v})_{S} = (c_{1}, c_{2}, \ldots , c_{n})\) is the coordinate vector of \(\mathbf{v}\) relative to \(S\), then the mapping: \[ \mathbf{v} \to (\mathbf{v})_{S} \tag{1} \] creates a one-to-one correspondence between vectors in \(V\) and vectors in \(R^{n}\). We call this the coordinate map relative to \(S\) from \(V\) to \(R^{n}\).

For convenience, coordinate vectors are often expressed in matrix form: \[ [\mathbf{v}]_{S} = \left[ \begin{array}{c}c_{1} \\ c_{2} \\ \vdots \\ c_{n} \end{array} \right] \tag{2} \] This matrix notation emphasizes its role in linear transformations.

Figure 4.6.1: Coordinate map from \(V\) to \(R^n\).

Think about how you might rotate your coordinate system to align with an object’s principal axes in physics or engineering mechanics. That’s a change of basis. In abstract vector spaces, we use “coordinate maps” to translate between a vector \(\mathbf{v}\) in a general space \(V\) and its numerical representation as a coordinate vector in \(R^n\). For calculations, we typically write these coordinate vectors as column matrices, denoted with square brackets like \([\mathbf{v}]_S\).

The Change of Basis Problem

A common problem arises when we have a vector \(\mathbf{v}\) and want to express its coordinates in a different basis.

The Change of Basis Problem

If \(\mathbf{v}\) is a vector in a finite-dimensional vector space \(V\), and if we change the basis for \(V\) from a basis \(B\) to a basis \(B'\), how are the coordinate vectors \([\mathbf{v}]_{B}\) and \([\mathbf{v}]_{B'}\) related?

Remark: We will refer to \(B\) as the “old basis” and \(B'\) as the “new basis.” Our objective is to find a relationship between the old and new coordinates of a fixed vector \(\mathbf{v}\) in \(V\).

The “Change of Basis Problem” is precisely about finding this relationship. If we have a vector \(\mathbf{v}\) expressed in terms of an “old basis” \(B\), and we want to know its coordinates in a “new basis” \(B'\), how do we convert \([\mathbf{v}]_B\) to \([\mathbf{v}]_{B'}\) or vice-versa? This is crucial for consistency when different parts of a system or different algorithms use different reference frames.

Deriving the Relationship (2D Case)

Let’s derive the relationship for a 2-dimensional space first.

Let \(B = \{\mathbf{u}_{1}, \mathbf{u}_{2}\}\) be the old basis and \(B' = \{\mathbf{u}_{1}', \mathbf{u}_{2}'\}\) be the new basis.

1. Express new basis vectors in terms of the old basis: Suppose the coordinate vectors of the new basis vectors relative to the old basis are: \[ [\mathbf{u}_{1}']_{B} = \begin{bmatrix} a \\ b \end{bmatrix} \quad \text{and} \quad [\mathbf{u}_{2}']_{B} = \begin{bmatrix} c \\ d \end{bmatrix} \tag{3} \] This means: \[ \begin{array}{r}\mathbf{u}_{1}' = a\mathbf{u}_{1} + b\mathbf{u}_{2} \\ \mathbf{u}_{2}' = c\mathbf{u}_{1} + d\mathbf{u}_{2} \end{array} \tag{4} \]

To understand the relationship, let’s consider a 2D space. We have an old basis \(B\) and a new basis \(B'\). The key step is to express each vector of the new basis as a linear combination of the vectors in the old basis. The coefficients of these linear combinations form the coordinate vectors of the new basis vectors relative to the old basis.

Deriving the Relationship (2D Case Continued)

2. Express \(\mathbf{v}\) in terms of the new basis: Let \([\mathbf{v}]_{B'} = \begin{bmatrix} k_{1} \\ k_{2} \end{bmatrix}\) be the new coordinate vector, so: \[ \mathbf{v} = k_{1}\mathbf{u}_{1}^{\prime} + k_{2}\mathbf{u}_{2}^{\prime} \tag{6} \]

3. Substitute and find old coordinates: Substitute (4) into (6) to express \(\mathbf{v}\) in terms of the old basis: \[ \mathbf{v} = k_{1}(a\mathbf{u}_{1} + b\mathbf{u}_{2}) + k_{2}(c\mathbf{u}_{1} + d\mathbf{u}_{2}) \] Rearranging terms: \[ \mathbf{v} = (k_{1}a + k_{2}c)\mathbf{u}_{1} + (k_{1}b + k_{2}d)\mathbf{u}_{2} \] Thus, the old coordinate vector for \(\mathbf{v}\) is: \[ [\mathbf{v}]_{B} = \begin{bmatrix} k_{1}a + k_{2}c \\ k_{1}b + k_{2}d \end{bmatrix} \]

Next, we take an arbitrary vector \(\mathbf{v}\) and express it using its coordinates in the new basis \(B'\). Then, we substitute the expressions for \(\mathbf{u}_1'\) and \(\mathbf{u}_2'\) (from the previous slide) into this equation. After regrouping the terms by \(\mathbf{u}_1\) and \(\mathbf{u}_2\), we get the coefficients that represent \(\mathbf{v}\) in the old basis \(B\). These coefficients form the old coordinate vector \([\mathbf{v}]_B\).

The Transition Matrix

The relationship found can be expressed as a matrix multiplication.

Using the new coordinate vector \([\mathbf{v}]_{B'} = \begin{bmatrix} k_{1} \\ k_{2} \end{bmatrix}\), the old coordinate vector can be written as: \[ [\mathbf{v}]_{B} = \left[ \begin{array}{ll}a & c \\ b & d \end{array} \right]\left[ \begin{array}{l}k_{1} \\ k_{2} \end{array} \right] = \left[ \begin{array}{ll}a & c \\ b & d \end{array} \right][\mathbf{v}]_{B^{\prime}} \]

The matrix \(P = \left[ \begin{array}{ll}a & c \\ b & d \end{array} \right]\) is called the transition matrix. Notice that the columns of \(P\) are \([\mathbf{u}_{1}']_{B}\) and \([\mathbf{u}_{2}']_{B}\).

The crucial observation is that the relationship between the old and new coordinate vectors can be written as a matrix-vector product. The matrix, which we call \(P\), has its columns composed of the coordinate vectors of the new basis vectors, expressed in terms of the old basis. This matrix \(P\) is our “transition matrix,” and it’s the key to changing bases.

Solution of the Change of Basis Problem (General Case)

This generalizes to \(n\)-dimensional spaces.

Solution of the Change of Basis Problem

If we change the basis for a vector space \(V\) from an old basis \(B = \{\mathbf{u}_{1}, \ldots , \mathbf{u}_{n}\}\) to a new basis \(B^{\prime} = \{\mathbf{u}_{1}^{\prime}, \ldots , \mathbf{u}_{n}^{\prime}\}\), then for each vector \(\mathbf{v}\) in \(V\), the old coordinate vector \([\mathbf{v}]_{B}\) is related to the new coordinate vector \([\mathbf{v}]_{B^{\prime}}\) by the equation: \[ [\mathbf{v}]_{B} = P[\mathbf{v}]_{B^{\prime}} \tag{7} \] where the columns of \(P\) are the coordinate vectors of the new basis vectors relative to the old basis; that is, the column vectors of \(P\) are: \[ [\mathbf{u}_{1}^{\prime}]_{B}, \quad [\mathbf{u}_{2}^{\prime}]_{B}, \ldots , \quad [\mathbf{u}_{n}^{\prime}]_{B} \tag{8} \]

So, in general, to convert coordinate vectors from a new basis \(B'\) to an old basis \(B\), we multiply by a transition matrix \(P\). The columns of this matrix \(P\) are formed by taking each vector from the new basis \(B'\) and expressing it as a coordinate vector with respect to the old basis \(B\). This is a fundamental formula in linear algebra.

Transition Matrices Notation

We often use specific notation to clarify the direction of the transition.

The matrix \(P\) in Equation (7) is called the transition matrix from \(B^{\prime}\) to \(B\), denoted by \(P_{B^{\prime} \rightarrow B}\). It can be expressed as: \[ P_{B^{\prime} \rightarrow B} = \left[ [\mathbf{u}_{1}^{\prime}]_{B} \mid [\mathbf{u}_{2}^{\prime}]_{B} \mid \dots \mid [\mathbf{u}_{n}^{\prime}]_{B} \right] \tag{9} \]

Similarly, the transition matrix from \(B\) to \(B^{\prime}\) is \(P_{B \rightarrow B^{\prime}}\), and its columns are: \[ P_{B \rightarrow B^{\prime}} = \left[ [\mathbf{u}_{1}]_{B^{\prime}} \mid [\mathbf{u}_{2}]_{B^{\prime}} \mid \dots \mid [\mathbf{u}_{n}]_{B^{\prime}} \right] \tag{10} \]

Tip

Memory Aid: The columns of the transition matrix from an old basis to a new basis are the coordinate vectors of the old basis relative to the new basis. (This statement in the text seems to have a typo. It should be: “The columns of the transition matrix from a new basis to an old basis are the coordinate vectors of the new basis relative to the old basis.”) Let’s stick to the definition: \(P_{B' \to B}\) takes coordinates in \(B'\) to \(B\). Its columns are the new basis vectors expressed in the old basis.

To avoid confusion, we use subscripts to denote the direction of the transition. \(P_{B' \to B}\) takes coordinates from \(B'\) and gives you coordinates in \(B\). Its columns are the vectors of \(B'\) expressed in terms of \(B\). Conversely, \(P_{B \to B'}\) takes coordinates from \(B\) and gives you coordinates in \(B'\), and its columns are the vectors of \(B\) expressed in terms of \(B'\). The memory aid in the textbook can be a bit confusing, so let’s rely on the formal definitions and the direction of the arrow.

EXAMPLE 1: Finding Transition Matrices

Consider bases \(B = \{\mathbf{u}_{1}, \mathbf{u}_{2}\}\) and \(B^{\prime} = \{\mathbf{u}_{1}^{\prime}, \mathbf{u}_{2}^{\prime}\}\) for \(R^{2}\), where:

\(\mathbf{u}_{1} = (1,0)\), \(\mathbf{u}_{2} = (0,1)\)

\(\mathbf{u}_{1}^{\prime} = (1,1)\), \(\mathbf{u}_{2}^{\prime} = (2,1)\)

(a) Find \(P_{B^{\prime} \rightarrow B}\) (from \(B'\) to \(B\)).

New basis vectors in terms of old basis \(B\):

\(\mathbf{u}_{1}^{\prime} = 1\mathbf{u}_{1} + 1\mathbf{u}_{2} \implies [\mathbf{u}_{1}^{\prime}]_{B} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\)

\(\mathbf{u}_{2}^{\prime} = 2\mathbf{u}_{1} + 1\mathbf{u}_{2} \implies [\mathbf{u}_{2}^{\prime}]_{B} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}\)

So, \(P_{B^{\prime} \rightarrow B} = \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix}\).

Let’s work through an example. We have two bases for \(R^2\). For part (a), we want to go from \(B'\) to \(B\). This means \(B'\) is our “new” basis, and \(B\) is our “old” basis. We need to express \(\mathbf{u}_1'\) and \(\mathbf{u}_2'\) in terms of \(\mathbf{u}_1\) and \(\mathbf{u}_2\). Since \(\mathbf{u}_1\) and \(\mathbf{u}_2\) are the standard basis vectors, this is straightforward. The coefficients become the columns of our transition matrix.

EXAMPLE 1: Finding Transition Matrices (Continued)

(b) Find \(P_{B \rightarrow B^{\prime}}\) (from \(B\) to \(B'\)).

Old basis vectors in terms of new basis \(B'\):

We need to find \(c_1, c_2\) such that \(\mathbf{u}_{1} = c_{1}\mathbf{u}_{1}^{\prime} + c_{2}\mathbf{u}_{2}^{\prime}\).

\((1,0) = c_{1}(1,1) + c_{2}(2,1) \implies \begin{cases} c_1 + 2c_2 = 1 \\ c_1 + c_2 = 0 \end{cases}\)

Solving gives \(c_1 = -1, c_2 = 1\). So, \([\mathbf{u}_{1}]_{B^{\prime}} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}\).

Similarly for \(\mathbf{u}_{2}\):

\((0,1) = d_{1}(1,1) + d_{2}(2,1) \implies \begin{cases} d_1 + 2d_2 = 0 \\ d_1 + d_2 = 1 \end{cases}\)

Solving gives \(d_1 = 2, d_2 = -1\). So, \([\mathbf{u}_{2}]_{B^{\prime}} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}\).

Thus, \(P_{B \rightarrow B^{\prime}} = \begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix}\).

For part (b), we want to go from \(B\) to \(B'\). Now, \(B\) is the “new” basis, and \(B'\) is the “old” basis. This means we need to express \(\mathbf{u}_1\) and \(\mathbf{u}_2\) in terms of \(\mathbf{u}_1'\) and \(\mathbf{u}_2'\). This typically involves solving a system of linear equations for each basis vector. Once we find the coefficients, they form the columns of \(P_{B \to B'}\).

Interactive Example: Verify Transition Matrix \(P_{B \rightarrow B'}\)

Let’s use Python to solve the systems and construct \(P_{B \rightarrow B'}\).

Using numpy, we can automate the process of solving the linear systems. We form a matrix A whose columns are the vectors of \(B'\). Then, we solve \(A\mathbf{c} = \mathbf{u}_1\) and \(A\mathbf{d} = \mathbf{u}_2\) to find the coordinates. The output confirms our manual calculations for \(P_{B \to B'}\), demonstrating the power of computational tools for these tasks.

EXAMPLE 2: Computing Coordinate Vectors

Let \(B\) and \(B^{\prime}\) be the bases from Example 1. Use an appropriate formula to find \([\mathbf{v}]_{B}\) given that \([\mathbf{v}]_{B^{\prime}} = \begin{bmatrix} -3 \\ 5 \end{bmatrix}\).

Solution: We want to find \([\mathbf{v}]_{B}\) from \([\mathbf{v}]_{B'}\). This means we need the transition matrix \(P_{B^{\prime} \rightarrow B}\). From Example 1(a), \(P_{B^{\prime} \rightarrow B} = \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix}\).

Using the formula \([\mathbf{v}]_{B} = P_{B^{\prime} \rightarrow B}[\mathbf{v}]_{B^{\prime}}\): \[ [\mathbf{v}]_{B} = \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} -3 \\ 5 \end{bmatrix} = \begin{bmatrix} (1)(-3) + (2)(5) \\ (1)(-3) + (1)(5) \end{bmatrix} = \begin{bmatrix} 7 \\ 2 \end{bmatrix} \]

So, the coordinate vector of \(\mathbf{v}\) relative to basis \(B\) is \(\begin{bmatrix} 7 \\ 2 \end{bmatrix}\).

Now, let’s see how to use these transition matrices. If we’re given the coordinates of a vector \(\mathbf{v}\) in the \(B'\) basis, and we want its coordinates in the \(B\) basis, we simply multiply \([\mathbf{v}]_{B'}\) by the transition matrix \(P_{B' \to B}\). As you can see, a simple matrix multiplication gives us the required coordinate vector.

Interactive Example: Compute \([\mathbf{v}]_{B}\)

Let’s use Python to perform the matrix multiplication.

This Python code snippet confirms the calculation. Defining the transition matrix and the coordinate vector, then using numpy’s matrix multiplication operator (@), we quickly get the coordinate vector in the old basis \(B\). This is a common operation in engineering to switch between different representations of the same physical quantity or state.

Invertibility of Transition Matrices

Transition matrices are always invertible, and their inverse has a special meaning.

If \(B\) and \(B^{\prime}\) are bases for \(V\): \[ [\mathbf{v}]_{B} = P_{B^{\prime} \rightarrow B}[\mathbf{v}]_{B^{\prime}} \quad \text{and} \quad [\mathbf{v}]_{B^{\prime}} = P_{B \rightarrow B^{\prime}}[\mathbf{v}]_{B} \tag{11} \]

Substituting the second into the first: \[ [\mathbf{v}]_{B} = P_{B^{\prime} \rightarrow B}(P_{B \rightarrow B^{\prime}}[\mathbf{v}]_{B}) = (P_{B^{\prime} \rightarrow B} P_{B \rightarrow B^{\prime}})[\mathbf{v}]_{B} \] This implies that \((P_{B^{\prime} \rightarrow B} P_{B \rightarrow B^{\prime}})\) must be the identity matrix \(I\). Thus, \(P_{B^{\prime} \rightarrow B}\) and \(P_{B \rightarrow B^{\prime}}\) are inverses of each other.

THEOREM 4.6.1

If \(P\) is the transition matrix from a basis \(B^{\prime}\) to a basis \(B\) for a finite-dimensional vector space \(V\), then \(P\) is invertible and \(P^{- 1}\) is the transition matrix from \(B\) to \(B^{\prime}\). \[ P_{B^{\prime} \rightarrow B}^{-1} = P_{B \rightarrow B^{\prime}} \]

An important property of transition matrices is their invertibility. If you can go from basis \(B'\) to \(B\), you should be able to go back from \(B\) to \(B'\). This means the transition matrices are inverses of each other. This is formally stated in Theorem 4.6.1. We can see this intuitively because transforming coordinates from \(B\) to \(B'\) and then back to \(B\) should result in the original coordinates, implying their product is the identity matrix.

Interactive Example: Verify Inverse Relationship

Let’s verify that \(P_{B^{\prime} \rightarrow B}\) and \(P_{B \rightarrow B^{\prime}}\) are inverses.

Using numpy, we can multiply the two transition matrices we found earlier. As expected, their product is the identity matrix. Furthermore, we can compute the inverse of \(P_{B' \to B}\) directly using np.linalg.inv and confirm that it matches \(P_{B \to B'}\). This numerical verification reinforces Theorem 4.6.1 and builds confidence in our calculations.

Efficient Method for Computing Transition Matrices for \(R^n\)

For \(R^n\), there’s an efficient procedure using augmented matrices. This avoids solving multiple systems separately.

A Procedure for Computing \(P_{B\rightarrow B^{\prime}}\) (Old to New)

Step 1. Form the augmented matrix \([B^{\prime} \mid B]\). (Here \(B'\) is the new basis, \(B\) is the old basis. Columns of \(B'\) and \(B\) are the basis vectors).

Step 2. Use elementary row operations to reduce the matrix in Step 1 to reduced row echelon form.

Step 3. The resulting matrix will be \([I \mid P_{B \rightarrow B^{\prime}}]\).

Step 4. Extract the matrix \(P_{B \rightarrow B^{\prime}}\) from the right side of the matrix in Step 3.

This procedure is summarized by the diagram: row operations \([ \text{new basis} \mid \text{old basis} ] \quad \to \quad [ I \mid \text{transition from old to new} ]\)

When working with \(R^n\), solving \(n\) separate linear systems to find the coordinates of each basis vector can be tedious. This efficient procedure streamlines the process by using an augmented matrix and row operations. The key is to set up the augmented matrix correctly: the “new basis” vectors on the left, and the “old basis” vectors on the right. After row reduction, the right side will directly give you the transition matrix from the old basis to the new basis. This method is incredibly useful in practical computations.

EXAMPLE 3: Example 1 Revisited (Efficient Method)

Using the bases from Example 1: \(B = \{\mathbf{u}_{1}=(1,0), \mathbf{u}_{2}=(0,1)\}\) \(B^{\prime} = \{\mathbf{u}_{1}^{\prime}=(1,1), \mathbf{u}_{2}^{\prime}=(2,1)\}\)

(a) Find \(P_{B^{\prime} \rightarrow B}\) (from \(B'\) to \(B\)).

Here, \(B'\) is the old basis; \(B\) is the new basis. So, we want to find \(P_{B' \to B}\). According to the general formula \(P_{B' \to B} = [[\mathbf{u}_1']_B \mid [\mathbf{u}_2']_B]\). Since \(B\) is the standard basis, \([\mathbf{u}_{1}^{\prime}]_{B} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\) and \([\mathbf{u}_{2}^{\prime}]_{B} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}\). Thus, \(P_{B^{\prime}\rightarrow B} = \begin{bmatrix} 1 & 2\\ 1 & 1 \end{bmatrix}\). (This was trivial as \(B\) is standard basis). The procedure (new basis | old basis) would be \([B \mid B'] = \left[ \begin{array}{l l l}{1} & 0 & {1} & {2}\\ {0} & {1} & {1} & {1} \end{array} \right]\). Since the left side is already \(I\), \(P_{B^{\prime}\rightarrow B}\) is the right side.

Let’s apply the efficient method. For part (a), finding \(P_{B' \to B}\), we set up the augmented matrix with the new basis (\(B\)) on the left and the old basis (\(B'\)) on the right. Since \(B\) is the standard basis, the left block is already the identity matrix, and the right block immediately gives us the transition matrix. This matches our earlier result.

EXAMPLE 3: Example 1 Revisited (Efficient Method Continued)

(b) Find \(P_{B \rightarrow B^{\prime}}\) (from \(B\) to \(B'\)).

Here, \(B\) is the old basis; \(B'\) is the new basis. We want to find \(P_{B \rightarrow B^{\prime}}\). Form the augmented matrix \([B^{\prime} \mid B]\): \[ \left[ \begin{array}{ll|ll}1 & 2 & 1 & 0\\ 1 & 1 & 0 & 1 \end{array} \right] \] Reduce to reduced row echelon form: \[ \left[ \begin{array}{ll|cc}1 & 0 & -1 & 2\\ 0 & 1 & 1 & -1 \end{array} \right] \] So the transition matrix is \(P_{B\rightarrow B^{\prime}} = \begin{bmatrix} - 1 & 2\\ 1 & -1 \end{bmatrix}\), which matches our previous result.

For part (b), we are finding \(P_{B \to B'}\). So, \(B'\) is the new basis, and \(B\) is the old basis. We construct the augmented matrix \([B' \mid B]\). Then, we perform row operations to transform the left block (representing \(B'\)) into the identity matrix. The right block will then automatically become our desired transition matrix \(P_{B \to B'}\). This is a much faster way than solving two separate systems of equations.

Interactive Example: Efficient Method for \(P_{B \rightarrow B'}\)

Let’s use Python to perform the row operations for \(P_{B \rightarrow B'}\).

While numpy doesn’t have a direct “row reduce augmented matrix” function like some symbolic tools, we can achieve the same result. The transition matrix \(P_{B \to B'}\) is effectively the inverse of the matrix formed by \(B'\) vectors, multiplied by the matrix formed by \(B\) vectors. The code calculates this, confirming the result obtained by manual row operations. This highlights how computational tools simplify complex matrix algebra.

Transition to the Standard Basis for \(R^n\)

A special case arises when one of the bases is the standard basis.

THEOREM 4.6.2

Let \(B^{\prime} = \{\mathbf{u}_{1},\mathbf{u}_{2},\ldots ,\mathbf{u}_{n}\}\) be any basis for the vector space \(R^{n}\) and let \(S = \{\mathbf{e}_{1},\mathbf{e}_{2},\ldots ,\mathbf{e}_{n}\}\) be the standard basis for \(R^{n}\). If the vectors in these bases are written in column form, then: \[ P_{B^{\prime}\rightarrow S} = [\mathbf{u}_{1}\mid \mathbf{u}_{2}\mid \dots \mid \mathbf{u}_{n}] \tag{15} \] That is, the transition matrix from \(B'\) to the standard basis \(S\) is simply the matrix whose columns are the basis vectors of \(B'\).

This means if \(A = [\mathbf{u}_{1}\mid \mathbf{u}_{2}\mid \dots \mid \mathbf{u}_{n}]\) is any invertible \(n\times n\) matrix, then \(A\) can be viewed as the transition matrix from the basis \(\{\mathbf{u}_{1},\mathbf{u}_{2},\ldots ,\mathbf{u}_{n}\}\) for \(R^{n}\) to the standard basis for \(R^{n}\).

Theorem 4.6.2 is a very convenient shortcut. If you’re transitioning to the standard basis \(S\), the transition matrix \(P_{B' \to S}\) is simply the matrix formed by placing the vectors of \(B'\) as its columns. This is because expressing a vector like \(\mathbf{u}_1'\) in terms of the standard basis \(\mathbf{e}_1, \mathbf{e}_2, \ldots\) just means its components are the coefficients themselves. This means any invertible matrix can be seen as a transition matrix from some basis to the standard basis.

Interactive Example: Transition to Standard Basis

Let’s use a non-standard basis for \(R^3\) and find its transition matrix to the standard basis.

Consider the basis: \(\mathbf{u}_{1} = (1,2,1)\), \(\mathbf{u}_{2} = (2,5,0)\), \(\mathbf{u}_{3} = (3,3,8)\)

The transition matrix \(P_{B \rightarrow S}\) would simply be: \[ P_{B\rightarrow S} = \left[ \begin{array}{lll}1 & 2 & 3\\ 2 & 5 & 3\\ 1 & 0 & 8 \end{array} \right] \]

Here, we define a basis \(B\) for \(R^3\). According to Theorem 4.6.2, the transition matrix from this basis \(B\) to the standard basis \(S\) is simply the matrix formed by taking \(\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\) as its columns. The Python code demonstrates this construction and then uses this transition matrix to convert a vector from \(B\) coordinates to standard coordinates. This is a very common operation in engineering, for example, transforming sensor readings from a device’s local coordinate system to a global reference frame.

Summary

• Coordinate Maps: Connect vectors in a general space \(V\) to coordinate vectors in \(R^n\).
• Change of Basis Problem: Relates coordinate vectors of a fixed vector in different bases (\([\mathbf{v}]_{B}\) vs. \([\mathbf{v}]_{B'}\)).
• Transition Matrix \(P_{B' \rightarrow B}\): Converts coordinates from new basis \(B'\) to old basis \(B\). Its columns are the new basis vectors expressed in the old basis.
• Invertibility: \(P_{B' \rightarrow B}^{-1} = P_{B \rightarrow B'}\).
• Efficient Method for \(R^n\): Use augmented matrix \([B' \mid B] \to [I \mid P_{B \rightarrow B'}]\).
• Standard Basis Shortcut: \(P_{B' \rightarrow S}\) is simply the matrix whose columns are the vectors of \(B'\).

In summary, we’ve learned how to navigate between different coordinate systems in vector spaces using change of basis. We defined coordinate maps, understood the change of basis problem, and derived the concept of transition matrices. These matrices allow us to convert coordinate vectors from one basis to another, and we saw that they are always invertible, with the inverse matrix performing the reverse conversion. We also covered an efficient method for \(R^n\) using augmented matrices and a useful shortcut for transitions involving the standard basis. These tools are indispensable for working with complex systems in various engineering disciplines.