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Section 22.4 Examples

Subsection 22.4.1 Computing a transition matrix

Let's work through an example similar to Discovery 22.6, but we'll make it a little more involved by choosing a different basis than the standard basis for \(\matrixring_2(\R)\text{.}\)

Example 22.4.1.

Here are two bases of \(\matrixring_2(\R)\text{:}\)

\begin{align*} \basisfont{B} \amp = \left\{ \begin{bmatrix} 1 \amp 1 \\ 0 \amp 0 \end{bmatrix}, \left[\begin{array}{rr} 1 \amp -1 \\ 0 \amp 0 \end{array}\right], \begin{bmatrix} 0 \amp 0 \\ 1 \amp 1 \end{bmatrix}, \left[\begin{array}{rr} 0 \amp 0 \\ -1 \amp 1 \end{array}\right] \right\} \text{,}\\ \\ \basisfont{B}' \amp = \left\{ \begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \end{bmatrix}, \begin{bmatrix} 2 \amp 1 \\ 0 \amp 0 \end{bmatrix}, \begin{bmatrix} 3 \amp 2 \\ 1 \amp 0 \end{bmatrix}, \begin{bmatrix} 4 \amp 3 \\ 2 \amp 1 \end{bmatrix} \right\} \text{.} \end{align*}

Following Procedure 22.3.1, we first need to compute coordinate vectors, which means determining linear combinations. For example, for the first vector in \(\basisfont{B}\text{,}\) we set up the vector equation

\begin{equation*} \begin{bmatrix} 1 \amp 1 \\ 0 \amp 0 \end{bmatrix} = k_1 \begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \end{bmatrix} + k_2 \begin{bmatrix} 2 \amp 1 \\ 0 \amp 0 \end{bmatrix} + k_3 \begin{bmatrix} 3 \amp 2 \\ 1 \amp 0 \end{bmatrix} + k_4 \begin{bmatrix} 4 \amp 3 \\ 2 \amp 1 \end{bmatrix} \end{equation*}

in the unknown scalars \(k_1,k_2,k_3,k_4\text{.}\) We could probably solve this vector equation by inspection, but let's work it out fully. Collapse the linear combination into

\begin{equation*} \begin{bmatrix} 1 \amp 1 \\ 0 \amp 0 \end{bmatrix} = \begin{bmatrix} k_1 + 2 k_2 + 3 k_3 + 4 k_4 \amp k_2 + 2 k_3 + 3 k_4 \\ k_3 + 2 k_4 \amp k_4 \end{bmatrix}\text{,} \end{equation*}

and then comparing entries on either side leads to linear system

\begin{equation*} \left\{ \begin{array}{rcrcrcrcr} k_1 \amp + \amp 2 k_2 \amp + \amp 3 k_3 \amp + \amp 4 k_4 \amp = \amp 1 \text{,} \\ \amp \amp k_2 \amp + \amp 2 k_3 \amp + \amp 3 k_4 \amp = \amp 1 \text{,} \\ \amp \amp \amp \amp k_3 \amp + \amp 2 k_4 \amp = \amp 0 \text{,} \\ \amp \amp \amp \amp \amp \amp k_4 \amp = \amp 0 \text{.} \end{array} \right. \end{equation*}

Use an augmented matrix to reduce the system:

\begin{equation*} \left[\begin{array}{cccc|c} 1 \amp 2 \amp 3 \amp 4 \amp 1 \\ 0 \amp 1 \amp 2 \amp 3 \amp 1 \\ 0 \amp 0 \amp 1 \amp 2 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \amp 0 \end{array}\right] \qquad \rowredarrow \qquad \left[\begin{array}{cccc|r} 1 \amp 0 \amp 0 \amp 0 \amp -1 \\ 0 \amp 1 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \amp 0 \end{array}\right]\text{.} \end{equation*}

The solution is

\begin{align*} k_1 \amp = -1 \text{,} \amp k_2 \amp = 1 \text{,} \amp k_3 \amp = 0 \text{,} \amp k_4 \amp = 0 \text{,} \end{align*}

giving us coordinate vector

\begin{equation*} \matrixOf{A_1}{B'} = \left[\begin{array}{r} -1 \\ 1 \\ 0 \\ 0 \end{array}\right] \end{equation*}

(where \(A_1\) represents the first vector in basis \(\basisfont{B}\)).

A similar procedure can be used to compute coordinate vectors for the other vectors in \(\basisfont{B}\text{.}\) Here are the results:

\begin{align*} \matrixOf{A_2}{B'} \amp = \left[\begin{array}{r} 3 \\ -1 \\ 0 \\ 0 \end{array}\right] \text{,} \amp \matrixOf{A_3}{B'} \amp = \left[\begin{array}{r} 1 \\ -1 \\ -1 \\ 1 \end{array}\right] \text{,} \amp \matrixOf{A_4}{B'} \amp = \left[\begin{array}{r} -1 \\ 3 \\ -3 \\ 1 \end{array}\right] \end{align*}

(where again \(A_2,A_3,A_4\) represent the remaining vectors in \(\basisfont{B}\)).

Combine these four coordinate vectors into the transition matrix:

\begin{equation*} \ucobmtrx{B}{B'} = \left[\begin{array}{rrrr} -1 \amp 3 \amp 1 \amp -1 \\ 1 \amp -1 \amp -1 \amp 3 \\ 0 \amp 0 \amp -1 \amp -3 \\ 0 \amp 0 \amp 1 \amp 1 \end{array}\right]\text{.} \end{equation*}

Let's test this transition matrix against an example coordinate vector. We'll start with a \(\basisfont{B}\)-coordinate vector

\begin{equation*} \matrixOf{Y}{B} = \left[\begin{array}{r} 2 \\ -1 \\ 3 \\ -2 \end{array}\right] \text{,} \end{equation*}

which corresponds to vector

\begin{equation*} Y = 2 \begin{bmatrix} 1 \amp 1 \\ 0 \amp 0 \end{bmatrix} + (-1) \left[\begin{array}{rr} 1 \amp -1 \\ 0 \amp 0 \end{array}\right] + 3 \begin{bmatrix} 0 \amp 0 \\ 1 \amp 1 \end{bmatrix} + (-2) \left[\begin{array}{rr} 0 \amp 0 \\ -1 \amp 1 \end{array}\right] = \begin{bmatrix} 1 \amp 3 \\ 5 \amp 1 \end{bmatrix}\text{.} \end{equation*}

Now convert to a \(\basisfont{B}'\)-coordinate vector:

\begin{equation*} \matrixOf{Y}{B'} = \ucobmtrx{B}{B'} \matrixOf{Y}{B} = \left[\begin{array}{rrrr} -1 \amp 3 \amp 1 \amp -1 \\ 1 \amp -1 \amp -1 \amp 3 \\ 0 \amp 0 \amp -1 \amp -3 \\ 0 \amp 0 \amp 1 \amp 1 \end{array}\right] \left[\begin{array}{r} 2 \\ -1 \\ 3 \\ -2 \end{array}\right] = \left[\begin{array}{r} 0 \\ -6 \\ 3 \\ 1 \end{array}\right]\text{,} \end{equation*}

which corresponds to vector

\begin{equation*} Y = 0 \begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \end{bmatrix} + (-6) \begin{bmatrix} 2 \amp 1 \\ 0 \amp 0 \end{bmatrix} + 3 \begin{bmatrix} 3 \amp 2 \\ 1 \amp 0 \end{bmatrix} + 1 \begin{bmatrix} 4 \amp 3 \\ 2 \amp 1 \end{bmatrix} = \begin{bmatrix} 1 \amp 3 \\ 5 \amp 1 \end{bmatrix}\text{.} \end{equation*}

As this linear combination results in the same vector \(Y\) as before, the coordinate conversion was correct.

Subsection 22.4.2 Computing a transition matrix for \(\R^n\)

Now let's use Procedure 22.3.2 to compute a transition matrix for \(\R^4\text{.}\)

Example 22.4.2.

Here are two bases of \(\R^4\text{:}\)

\begin{align*} \basisfont{B} \amp = \left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \left[\begin{array}{r} 1 \\ -1 \\ 0 \\ 0 \end{array}\right], \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \end{bmatrix}, \left[\begin{array}{r} 0 \\ 0 \\ -1 \\ 1 \end{array}\right] \right\} \text{,}\\ \\ \basisfont{B}' \amp = \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 3 \\ 2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 4 \\ 3 \\ 2 \\ 1 \end{bmatrix} \right\} \text{.} \end{align*}

Following Procedure 22.3.2, we form an augmented matrix with the vectors of \(\basisfont{B}'\) on the left and the vectors of \(\basisfont{B}\) on the right, and then reduce:

\begin{align} \amp \left[\begin{array}{rrrr|rrrr} 1 \amp 2 \amp 3 \amp 4 \amp 1 \amp 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 2 \amp 3 \amp 1 \amp -1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 2 \amp 0 \amp 0 \amp 1 \amp -1 \\ 0 \amp 0 \amp 0 \amp 1 \amp 0 \amp 0 \amp 1 \amp 1 \end{array}\right]\label{equation-change-of-basis-examples-cob-by-row-red}\tag{\(\star\)}\\ \notag\\ \rowredarrow \qquad \amp \left[\begin{array}{rrrr|rrrr} 1 \amp 0 \amp 0 \amp 0 \amp -1 \amp 3 \amp 1 \amp -1 \\ 0 \amp 1 \amp 0 \amp 0 \amp 1 \amp -1 \amp -1 \amp 3 \\ 0 \amp 0 \amp 1 \amp 0 \amp 0 \amp 0 \amp -1 \amp -3 \\ 0 \amp 0 \amp 0 \amp 1 \amp 0 \amp 0 \amp 1 \amp 1 \end{array}\right]\text{.}\notag \end{align}

Our transition matrix is now on the right, and we can extract it as

\begin{equation*} \ucobmtrx{B}{B'} = \left[\begin{array}{rrrr} -1 \amp 3 \amp 1 \amp -1 \\ 1 \amp -1 \amp -1 \amp 3 \\ 0 \amp 0 \amp -1 \amp -3 \\ 0 \amp 0 \amp 1 \amp 1 \end{array}\right]\text{.} \end{equation*}

Let's test this transition matrix against an example coordinate vector. We'll start with a \(\basisfont{B}\)-coordinate vector

\begin{equation*} \matrixOf{\uvec{v}}{B'} = \left[\begin{array}{r} 2 \\ -1 \\ 3 \\ -2 \end{array}\right] \text{,} \end{equation*}

which corresponds to vector

\begin{equation*} \uvec{v} = 2 \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + (-1) \left[\begin{array}{rr} 1 \\ -1 \\ 0 \\ 0 \end{array}\right] + 3 \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \end{bmatrix} + (-2) \left[\begin{array}{rr} 0 \\ 0 \\ -1 \\ 1 \end{array}\right] = \begin{bmatrix} 1 \\ 3 \\ 5 \\ 1 \end{bmatrix}\text{.} \end{equation*}

Now convert to a \(\basisfont{B}'\)-coordinate vector:

\begin{equation*} \matrixOf{\uvec{v}}{B'} = \ucobmtrx{B}{B'} \matrixOf{\uvec{v}}{B} = \left[\begin{array}{rrrr} -1 \amp 3 \amp 1 \amp -1 \\ 1 \amp -1 \amp -1 \amp 3 \\ 0 \amp 0 \amp -1 \amp -3 \\ 0 \amp 0 \amp 1 \amp 1 \end{array}\right] \left[\begin{array}{r} 2 \\ -1 \\ 3 \\ -2 \end{array}\right] = \left[\begin{array}{r} 0 \\ -6 \\ 3 \\ 1 \end{array}\right]\text{,} \end{equation*}

which corresponds to vector

\begin{equation*} \uvec{v} = 0 \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + (-6) \begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + 3 \begin{bmatrix} 3 \\ 2 \\ 1 \\ 0 \end{bmatrix} + 1 \begin{bmatrix} 4 \\ 3 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ 5 \\ 1 \end{bmatrix}\text{.} \end{equation*}

As this linear combination results in the same vector \(\uvec{v}\) as before, the coordinate conversion was correct.

Remark 22.4.3.

You may have noticed that Example 22.4.1 and Example 22.4.2 are essentially the same example, but using different methods to compute the transition matrix. This comparison provides evidence for the claim in Statement 2 of Remark 22.3.3 that Procedure 22.3.2 can be used in other spaces besides \(\R^n\text{.}\) In Example 22.4.1, we could have taken the usual standard basis

\begin{equation*} \basisfont{S} = \left\{ \begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \end{bmatrix}, \begin{bmatrix} 0 \amp 1 \\ 0 \amp 0 \end{bmatrix}, \begin{bmatrix} 0 \amp 0 \\ 1 \amp 0 \end{bmatrix}, \begin{bmatrix} 0 \amp 0 \\ 0 \amp 1 \end{bmatrix} \right\} \end{equation*}

of \(\matrixring_2(\R)\) as our intermediate basis, and we would have ended up reducing the same \(4 \times 8\) augmented matrix to compute \(\ucobmtrx{B}{B'}\) as we did in Example 22.4.2. (Look back at (\(\star\)), and notice the how the columns of that initial \(4 \times 8\) augmented matrix are essentially the vectors of \(\basisfont{B}'\) and \(\basisfont{B}\) written as columns instead of as \(2 \times 2\) matrices.)