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Discovery guide 22.1 Discovery guide

Our goal in this set of discovery activities is to figure out how to convert from describing vectors in terms of one basis to describing vectors in terms of another basis.

First, let's remind ourselves of the concept of coordinate vector.

Discovery 22.1.

Suppose \(V\) is a three-dimensional vector space, and \(\basisfont{B} = \{ \uvec{u}_1, \uvec{u}_2, \uvec{u}_3 \}\) is a basis for \(V\text{.}\)

(a)

If \(\uvec{v} = 5 \uvec{u}_1 + 3 \uvec{u}_2 + (-1)\uvec{u}_3 \text{,}\) then

\begin{equation*} \matrixOf{\uvec{v}}{B} = \begin{bmatrix} \underline{\hspace{0.909090909090909em}} \\ \underline{\hspace{0.909090909090909em}} \\ \underline{\hspace{0.909090909090909em}} \end{bmatrix}\text{.} \end{equation*}
(b)

If \(\matrixOf{\uvec{w}}{B} = \left[\begin{smallmatrix} -2 \\ 1 \\ 2 \end{smallmatrix}\right] \text{,}\) then \(\uvec{w} = \underline{\hspace{0.909090909090909em}}\uvec{u}_1 + \underline{\hspace{0.909090909090909em}}\uvec{u}_2 + \underline{\hspace{0.909090909090909em}}\uvec{u}_3 \text{.}\)

Now let's revisit how the concept of coordinate vector interacts with the vector operations.

Discovery 22.2.

Again, suppose \(V\) is a three-dimensional vector space, and \(\basisfont{B} = \{ \uvec{u}_1, \uvec{u}_2, \uvec{u}_3 \}\) is a basis for \(V\text{.}\)

(a)

If \(\matrixOf{\uvec{v}}{B} = \left[\begin{smallmatrix} 5 \\ 3 \\ -1 \end{smallmatrix}\right] \) and \(\matrixOf{\uvec{w}}{B} = \left[\begin{smallmatrix} -2 \\ 1 \\ 2 \end{smallmatrix}\right] \text{,}\) then

\begin{align*} \uvec{v} \amp = \underline{\hspace{0.909090909090909em}}\uvec{u}_1 + \underline{\hspace{0.909090909090909em}}\uvec{u}_2 + \underline{\hspace{0.909090909090909em}}\uvec{u}_3 \text{,} \\ \uvec{w} \amp = \underline{\hspace{0.909090909090909em}}\uvec{u}_1 + \underline{\hspace{0.909090909090909em}}\uvec{u}_2 + \underline{\hspace{0.909090909090909em}}\uvec{u}_3 \text{,} \end{align*}

and so

\begin{equation*} \uvec{v} + \uvec{w} = \underline{\hspace{0.909090909090909em}}\uvec{u}_1 + \underline{\hspace{0.909090909090909em}}\uvec{u}_2 + \underline{\hspace{0.909090909090909em}}\uvec{u}_3 \end{equation*}

and

\begin{equation*} \matrixOf{\uvec{v} + \uvec{w}}{B} = \begin{bmatrix} \underline{\hspace{0.909090909090909em}} \\ \underline{\hspace{0.909090909090909em}} \\ \underline{\hspace{0.909090909090909em}} \end{bmatrix}\text{.} \end{equation*}
(b)

Describe the pattern from Task a.

First, in symbols: \(\matrixOf{\uvec{v} + \uvec{w}}{B} = \underline{\hspace{4.545454545454546em}}\text{.}\)

And again in words: the coordinate vector of a sum is .

(c)

If \(\matrixOf{\uvec{v}}{B} = \left[\begin{smallmatrix} 5 \\ 3 \\ -1 \end{smallmatrix}\right] \text{,}\) then

\begin{equation*} \uvec{v} = \underline{\hspace{0.909090909090909em}}\uvec{u}_1 + \underline{\hspace{0.909090909090909em}}\uvec{u}_2 + \underline{\hspace{0.909090909090909em}}\uvec{u}_3 \text{,} \end{equation*}

and so

\begin{equation*} 7\uvec{v} = \underline{\hspace{0.909090909090909em}}\uvec{u}_1 + \underline{\hspace{0.909090909090909em}}\uvec{u}_2 + \underline{\hspace{0.909090909090909em}}\uvec{u}_3 \end{equation*}

and

\begin{equation*} \matrixOf{7 \uvec{v}}{B} = \begin{bmatrix} \underline{\hspace{0.909090909090909em}} \\ \underline{\hspace{0.909090909090909em}} \\ \underline{\hspace{0.909090909090909em}} \end{bmatrix}\text{.} \end{equation*}
(d)

Describe the pattern from Task c.

First, in symbols: \(\matrixOf{k \uvec{v}}{B} = \underline{\hspace{4.545454545454546em}}\text{.}\)

And again in words: the coordinate vector of a scalar multiple is .

(e)

Let's combine the patterns from Task a and Task c, without bothering with an explicit example.

In words first this time: the coordinate vector of a linear combination is .

And again in symbols: \(\matrixOf{k_1 \uvec{v}_1 + k_2 \uvec{v}_2 + \dotsb + k_n \uvec{v}_n}{B} = \underline{\hspace{4.545454545454546em}}\text{.}\)

Before we continue our investigation of coordinate vectors, let's revisit Discovery 21.2 to expose a new pattern of matrix multiplication.

Discovery 22.3.

Think of an \(m\times 3\) matrix \(A\) as being made out of three column vectors from \(\R^m\text{:}\)

\begin{equation*} A = \begin{bmatrix} | \amp | \amp | \\ \uvec{a}_1 \amp \uvec{a}_2 \amp \uvec{a}_3 \\ | \amp | \amp | \end{bmatrix}\text{.} \end{equation*}
(a)

Do you remember what happens when we compute \(A\uvec{e}_1\text{?}\) \(A\uvec{e}_2\text{?}\) \(A\uvec{e}_3\text{?}\)

Hint

Recall Discovery 21.1.

(b)

Suppose we want to compute \(A\uvec{x}\text{,}\) where \(\uvec{x} = (5,3,-1)\) (but as a column vector). Use what you remembered in Task a to fill in the following.

Since

\begin{equation*} \left[\begin{array}{r} 5 \\ 3 \\ -1 \end{array}\right] = 5 \uvec{e}_1 + 3 \uvec{e}_2 + (-1) \uvec{e}_3\text{,} \end{equation*}

then

\begin{equation*} A \left[\begin{array}{r} 5 \\ 3 \\ -1 \end{array}\right] = A (5\uvec{e}_1 + 3\uvec{e}_2 + (-1)\uvec{e}_3) = 5 \underline{\hspace{0.909090909090909em}} + 3 \underline{\hspace{0.909090909090909em}} + (-1) \underline{\hspace{0.909090909090909em}}\text{.} \end{equation*}
(c)

Describe the pattern from Task b:

Matrix \(A\) times column vector \(\uvec{x}\) can be expressed as a linear combination of , where the coefficients in the linear combination are .

In particular,

\begin{equation*} A = \begin{bmatrix} | \amp | \amp | \\ \uvec{a}_1 \amp \uvec{a}_2 \amp \uvec{a}_3 \\ | \amp | \amp | \end{bmatrix}, \quad \uvec{x} = \begin{bmatrix} k_1 \\ k_2 \\ k_3 \end{bmatrix} \qquad \implies \qquad A\uvec{x} = \underline{\hspace{2.272727272727273em}} + \underline{\hspace{2.272727272727273em}} + \underline{\hspace{2.272727272727273em}}\text{.} \end{equation*}

Now we will combine Discovery 22.2 and Discovery 22.3.

Discovery 22.4.

Once again, suppose \(V\) is a three-dimensional vector space, we have a basis

\begin{equation*} \basisfont{B} = \{ \uvec{u}_1, \uvec{u}_2, \uvec{u}_3 \} \end{equation*}

for \(V\text{,}\) and \(\uvec{w}\) is a vector in \(V\) for which we know the coordinate vector relative to \(\basisfont{B}\text{,}\)

\begin{equation*} \matrixOf{\uvec{w}}{B} = \left[\begin{array}{r} 5 \\ 3 \\ -1 \end{array}\right] \text{.} \end{equation*}

But this time, suppose we have a second basis \(\basisfont{B}'\) for \(V\text{,}\) and we don't know the coordinate vector of \(\uvec{w}\) relative to \(\basisfont{B}'\text{.}\)

Our goal is to figure out how to obtain \(\matrixOf{\uvec{w}}{B'}\) from \(\matrixOf{\uvec{w}}{B}\text{.}\)

(a)

Repeat Task 22.1.b: \(\matrixOf{\uvec{w}}{B} = \left[\begin{smallmatrix} 5 \\ 3 \\ -1 \end{smallmatrix}\right] \) means

\begin{gather} \uvec{w} = \underline{\hspace{0.909090909090909em}}\uvec{u}_1 + \underline{\hspace{0.909090909090909em}}\uvec{u}_2 + \underline{\hspace{0.909090909090909em}}\uvec{u}_3\text{.}\label{equation-change-of-basis-discovery-coord-vec-to-lincomb}\tag{\(\star\)} \end{gather}
(b)

Apply the pattern from Task 22.2.e to (\(\star\)) to obtain an expression for \(\matrixOf{\uvec{w}}{B'}\) as a linear combination of coordinate vectors.

(c)

Remember that the coordinate vectors in your linear combination from Task b are vectors in \(\R^n\text{,}\) so they can be thought of as column vectors (as we have been doing in this discovery guide).

In Task 22.3.c, we established a new pattern for matrix-times-column-vector. Using this pattern backwards, what matrix \(P\) and what column vector \(\uvec{x}\) could be used to turn your linear combination for \(\matrixOf{\uvec{w}}{B'}\) from Task b into the matrix equation

\begin{equation*} \matrixOf{\uvec{w}}{B'} = P \uvec{x} \;\text{?} \end{equation*}

(Recognize the numbers in the vector \(\uvec{x}\) from earlier in this activity?)

The matrix in Discovery 22.4 that allows us to convert from coordinate vectors relative to one basis to coordinate vectors relative to another basis is called the transition matrix from basis \(\basisfont{B}\) to \(\basisfont{B}'\text{,}\) and is denoted \(\ucobmtrx{B}{B'}\text{.}\)

Notationally, the conversion process can be written

\begin{gather} \ucobmtrx{B}{B'} \matrixOf{\uvec{v}}{B} = \matrixOf{\uvec{v}}{B'}\text{.}\label{equation-change-of-basis-discovery-cob-matrix-convert}\tag{\(\star\star\)} \end{gather}
Discovery 22.5.

Summarize the pattern discovered in Discovery 22.4: the columns of a transition matrix \(\ucobmtrx{B}{B'}\) are .

Discovery 22.6.

Let's work out a transition matrix in a simple example.

Here are two bases of \(\matrixring_2(\R)\text{,}\) the 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\}\text{,} \end{equation*}

and another basis

\begin{equation*} \basisfont{B} = \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{equation*}
(a)

This vector space is four-dimensional, not three-dimensional like the abstract vector space used in our development through examples in the previous activities of this discovery guide. What size of matrix should \(\ucobmtrx{S}{B} \) be here?

(b)

Use the pattern you described in Discovery 22.5 to compute \(\ucobmtrx{S}{B} \text{.}\) (You can probably compute the columns of \(\ucobmtrx{S}{B} \) by inspection, without any lengthy calculations.)

(c)

As a test that you have the right transition matrix, write down \(\matrixOf{\uvec{v}}{S}\) for vector

\begin{equation*} \uvec{v} = \begin{bmatrix} 1 \amp 2 \\ 3 \amp 4 \end{bmatrix} \text{.} \end{equation*}

(Again, this can by done by inspection.)

Then compute \(\ucobmtrx{S}{B}\matrixOf{\uvec{v}}{S}\text{,}\) which by (\(\star\star\)) should be equal to \(\matrixOf{\uvec{v}}{B}\text{.}\)

Finally, check that \(\ucobmtrx{S}{B}\matrixOf{\uvec{v}}{S} = \matrixOf{\uvec{v}}{B}\) is correct by using the components of this column vector as coefficients in a linear combination (similarly to Task 22.1.b), which, if your calculations have been carried out correctly, should return you full circle to the \(2 \times 2\) matrix \(\uvec{v}\text{.}\)

Discovery 22.7.

In each task below, use (\(\star\star\)) to guide your thinking, instead of using the pattern from Discovery 22.5.

(a)

For a single basis \(\basisfont{B}\text{,}\) what form do you think the transition matrix \(\ucobmtrx{B}{B}\) will take?

(b)

Suppose you have three bases, \(\basisfont{B}\text{,}\) \(\basisfont{B}'\text{,}\) and \(\basisfont{B}''\text{,}\) of a particular space \(V\text{.}\) What is the relationship between the three transition matrices \(\ucobmtrx{B}{B'}\text{,}\) \(\ucobmtrx{B'}{B''}\text{,}\) and \(\ucobmtrx{B}{B''}\text{?}\)

(c)

For bases \(\basisfont{B}\) and \(\basisfont{B}'\text{,}\) what is the relationship between transition matrices \(\ucobmtrx{B}{B'}\) and \(\ucobmtrx{B'}{B}\text{?}\) Does your answer make sense both in terms of (\(\star\star\)) and in terms of your answers to Task a and Task b?

Discovery 22.8.

In this discovery activity, we'll consider transition matrices involving the standard basis of \(\R^n\text{,}\)

\begin{equation*} \basisfont{S} = \{ \uvec{e}_1, \uvec{e}_2, \dotsc, \uvec{e}_n \} \text{.} \end{equation*}
(a)

Temporarily suppose that \(n = 3\text{.}\)

How do you write the vector \(\uvec{v} = (5, 3, -1)\) as a linear combination of the standard basis vectors of \(\R^3\text{?}\)

And so, what is \(\matrixOf{\uvec{v}}{S}\text{?}\)

(b)

Suppose \(\basisfont{B}\) is another basis of \(\R^n\text{.}\) Considering what you learned in Task a, what do the columns of the transition matrix \(\ucobmtrx{B}{S}\) look like?

(d)

The takeaway from Task b is that computing a transition matrix where the second basis is the standard basis \(\basisfont{S}\) of \(\R^n\) is pretty simple.

Combine the pattern of Task b with what you learned in Discovery 22.7 to develop a simpler process to compute a transition matrix \(\ucobmtrx{B}{B'}\) for bases of \(\R^n\text{.}\)