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Section 44.4 Examples

Subsection 44.4.1 Computing with compositions

Example 44.4.1. Computing the image of a composition.

Consider

\begin{align*} \amp\funcdef{T}{\poly_3(\R)}{\matrixring_2(\R)} \text{,} \amp \amp\funcdef{S}{\matrixring_2(\R)}{\R} \end{align*}

defined by

\begin{align*} T(a x^3 + b x^2 + c x + d) \amp = \begin{bmatrix} a + b \amp b + c \\ c + d \amp a - d \end{bmatrix} \text{,} \amp S(A) \amp = \trace A\text{.} \end{align*}

Then the composition is

\begin{equation*} \funcdef{ST}{\poly_3}{\R} \text{.} \end{equation*}

What is the image vector under the composition for

\begin{equation*} p(x) = 3 x^3 - 2 x^2 + x - 5 \text{?} \end{equation*}

Compute:

\begin{gather*} T\bigl(p(x)\bigr) = \begin{bmatrix} 3 + (-2) \amp (-2) + 1 \\ 1 + (-5) \amp 3 - (-5) \end{bmatrix} = \begin{bmatrix} 1 \amp -1 \\ -4 \amp 8 \end{bmatrix} \text{,}\\ \\ S\left( \begin{bmatrix} 1 \amp -1 \\ -4 \amp 8 \end{bmatrix} \right) = 1 + 8 = 9 \text{.} \end{gather*}

So

\begin{equation*} ST(3 x^3 - 2 x^2 + x - 5) = 9 \text{.} \end{equation*}
Example 44.4.2. Computing input-output formulas for a composition.

Consider

\begin{align*} \amp\funcdef{D}{\poly_3(\R)}{\poly_2(\R)} \text{,} \amp \amp\funcdef{E_3}{\poly_2(\R)}{\R} \text{,} \amp \amp\funcdef{T}{\R}{\R^3}\text{,} \end{align*}

where \(D\) is differentiation, \(E_3\) is evaluation at \(x = 3\text{,}\) and \(T\) is defined by

\begin{equation*} T(y) = (y, 0, -y) \text{.} \end{equation*}

We can compute an input-output formula for the composition \(S = T E_3 D\) by

\begin{align*} S(a x^3 + b x^2 + c x + d) \amp = T\Bigl(E _3\bigl(D(a x^3 + b x^2 + c x + d)\bigr)\Bigr) \\ \amp = T\bigl(E _3(3 a x^2 + 2 b x + c)\bigr) \\ \amp = T(27 a + 3 b + c) \\ \amp = (27 a + 3 b + c, 0, - 27 a - 3 b - c) \text{.} \end{align*}

Subsection 44.4.2 Checking and computing inverses

Example 44.4.3. Determining whether a transformation is invertible.

Consider \(\funcdef{T}{\R^4}{\matrixring_{2\times 3}(\R)}\) defined by

\begin{equation*} T(a,b,c,d) = \begin{bmatrix} a + d \amp b + c \amp a + d \\ b - c \amp a - d \amp b - c \end{bmatrix} \text{.} \end{equation*}

Does \(T\) have an inverse transformation \(funcdef{T}{\im T}{\R^4}\text{?}\) We can use the kernel to investigate:

\begin{gather*} \phantom{\implies} T(a,b,c,d) = \zerovec \\ \implies \begin{bmatrix} a + d \amp b + c \amp a + d \\ b - c \amp a - d \amp b - c \end{bmatrix} = \begin{bmatrix} 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \end{bmatrix}\\ \implies \left\{ \begin{array}{ccccl} a \amp + \amp d \amp = \amp 0 \text{,} \\ b \amp + \amp c \amp = \amp 0 \text{,} \\ a \amp + \amp d \amp = \amp 0 \text{,} \\ b \amp - \amp c \amp = \amp 0 \text{,} \\ a \amp - \amp d \amp = \amp 0 \text{,} \\ b \amp - \amp c \amp = \amp 0 \text{.} \end{array}\right. \end{gather*}

We could use row reducing to solve this homogeneous system, but it should be clear that we cannot have both \(a = d\) and \(a = - d\) simultaneously unless both \(a,d\) are zero, and similarly for \(b\) and \(c\text{.}\)

Therefore, since \(\ker T = \{\zerovec\}\text{,}\) \(T\) is one-to-one and an inverse \(\funcdef{\inv{T}}{\im T}{\R^4}\) exists.

Example 44.4.4. Computing inverse images.

Let's continue with the invertible transformation \(T\) from Example 44.4.3. What is the inverse image

\begin{equation*} \inv{T}\left(\left[\begin{array}{crc} 6 \amp 1 \amp 6 \\ 5 \amp -4 \amp 5 \end{array}\right]\right) \text{?} \end{equation*}

First, recall that \(\inv{T}\) is only defined on \(\im T\text{.}\) Is the matrix used as input to \(\inv{T}\) above actually in \(\im T\text{?}\) Attempting to compute its inverse image will answer the question for us — if we are able to come to a solution, then it must be in \(\im T\text{,}\) and if we find that there is no solution, then it must not be in \(\im T\text{.}\) As usual, attempting to solve for the inverse image will lead to a linear system, and we know how to determine when a system has solutions or not (Proposition 2.5.6).

The inverse image we would like to calculate (if it exists) will be precisely the \(\R^4\) vector that satisfies

\begin{align*} T(a,b,c,d) \amp = \left[\begin{array}{crc} 6 \amp 1 \amp 6 \\ 5 \amp -4 \amp 5 \end{array}\right] \amp \amp \Longleftrightarrow \amp \begin{bmatrix} a + d \amp b + c \amp a + d \\ b - c \amp a - d \amp b - c \end{bmatrix} \amp = \left[\begin{array}{crc} 6 \amp 1 \amp 6 \\ 5 \amp -4 \amp 5 \end{array}\right]\text{,} \end{align*}

from which we obtain linear system

\begin{align*} \left\{ \begin{array}{ccccrl} a \amp + \amp d \amp = \amp 6 \amp \text{,} \\ b \amp + \amp c \amp = \amp 1 \amp \text{,} \\ a \amp + \amp d \amp = \amp 6 \amp \text{,} \\ b \amp - \amp c \amp = \amp 5 \amp \text{,} \\ a \amp - \amp d \amp = \amp -4 \amp \text{,} \\ b \amp - \amp c \amp = \amp 5 \amp \text{.} \end{array}\right. \end{align*}

This system can be converted to an augmented matrix and reduced, leading to one unique solution

\begin{align*} a \amp = 1 \text{,} \amp b \amp = 3 \text{,} \amp c \amp = -2 \text{,} \amp d \amp = 5 \text{,} \end{align*}

hence

\begin{equation*} \inv{T}\left(\left[\begin{array}{crc} 6 \amp 1 \amp 6 \\ 5 \amp -4 \amp 5 \end{array}\right]\right) = (1,3,-2,5) \text{.} \end{equation*}

Subsection 44.4.3 Checking surjectivity

The easiest way to determine surjectivity of a transformation is to use the Dimension Theorem to establish the dimension of the image.

Example 44.4.5. Determining whether a transformation is surjective.

In Example 44.4.3, we found that a transformation \(\funcdef{T}{\R^4}{\matrixring_{2\times 3}(\R)}\) had trivial kernel. Using the Dimension Theorem, we must then have

\begin{equation*} \dim (\im T) = \dim (\R^4) - \dim (\ker T) = 4 - 0 = 4 \text{,} \end{equation*}

and so

\begin{equation*} \dim (\im T) \lt \dim \bigl(\matrixring_{2 \times 3}(\R)\bigr) \text{.} \end{equation*}

Since \(\im T\) is a subspace of the codomain space \(\matrixring_{2 \times 3}(\R)\text{,}\) Statement 3 of Proposition 20.5.8 tells us that it is not possible for \(T\) to be surjective.

Subsection 44.4.4 Defining isomorphisms by choice of bases

Example 44.4.6. Creating an isomorphism by sending a domain space basis to a codomain space basis.

Let's create an isomorphsm \(\funcdef{T}{\R^4}{\matrixring_2(\R)}\) by sending the standard basis for \(\R^4\) to some nonstandard basis of \(\matrixring_2(\R)\text{:}\)

\begin{align*} T(1,0,0,0) \amp = \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix} \text{,} \amp T(0,0,1,0) \amp = \left[\begin{array}{rc} 0 \amp 1 \\ -1 \amp 0 \end{array}\right] \text{,}\\ T(0,1,0,0) \amp = \begin{bmatrix} 0 \amp 1 \\ 1 \amp 0 \end{bmatrix} \text{,} \amp T(0,0,0,1) \amp = \left[\begin{array}{cr} 1 \amp 0 \\ 0 \amp -1 \end{array}\right] \text{.} \end{align*}

We can determine an input-ouput formula for \(T\) by expanding an arbitrary input as a linear combination of the standard basis of \(\R^4\text{,}\) and using the linearity of \(T\text{:}\)

\begin{align*} T(a,b,c,d) \amp = a \, T(1,0,0,0) + b \, T(0,1,0,0) + c \, T(0,0,1,0) + d \, T(0,0,0,1) \\ \amp = a \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \end{bmatrix} + b \begin{bmatrix} 0 \amp 1 \\ 1 \amp 0 \end{bmatrix} + c \left[\begin{array}{rc} 0 \amp 1 \\ -1 \amp 0 \end{array}\right] + d \left[\begin{array}{cr} 1 \amp 0 \\ 0 \amp -1 \end{array}\right]\\ \amp = \begin{bmatrix} a + d \amp b + c \\ b - c \amp a - d \end{bmatrix} \text{.} \end{align*}

Similarly to Example 44.4.3, it is easy to check that this transformation is one-to-one by verifying that the kernel is trivial. Then, using the Dimension Theorem, we must have

\begin{equation*} \dim (\im T) = \dim (\R^4) - \dim (\ker T) = 4 - 0 = 4 = \dim \bigl(\matrixring_2(\R)\bigr) \text{,} \end{equation*}

from which we can conclude that \(T\) must be surjective as well. Together, trivial kernel and full image tells us that \(T\) is indeed an isomorphism.

Example 44.4.7. A geometric example: rotation in \(\R^3\).

Rotation around an axis should be an isomorphism of \(\R^3\) onto \(\R^3\text{,}\) as every vector in \(\R^3\) can be realized as the rotated image of one unique “input” vector.

Let \(\funcdef{T}{\R^3}{\R^3}\) represent counter-clockwise rotation by \(\pi/2\) around the axis through the origin and parallel to the vector

\begin{equation*} \uvec{n} = (2,-1,3) \text{,} \end{equation*}

where “counter-clockwise” is considered as one looks “downward” along \(\uvec{n}\) back toward the origin. Let's set up an orthogonal basis of \(\R^3\) that will help describe \(T\text{,}\) beginning with the axis \(\uvec{n}\text{.}\) It will be easiest to analyze rotation around \(\uvec{n}\) using vectors in the plane normal to \(\uvec{n}\text{.}\) So let's guess a vector

\begin{equation*} \uvec{v}_1 = (1,2,0) \end{equation*}

that is parallel to this plane (which we can verify by computing \(\dotprod{\uvec{n}}{\uvec{v}_1} = 0\)). Now, since \(T\) rotates by \(\pi/2\text{,}\) it will be convenient to take another vector in the normal plane that is also orthogonal to \(\uvec{v}_1\text{.}\) We could use Gram-Schmidt to do this, but instead we'll use the cross product to be able to use the right-hand rule to control the direction (“positive” or “negative”) of the result. Use the right-hand rule to convince yourself that

\begin{equation*} \uvec{v}_2 = \crossprod{\uvec{n}}{\uvec{v}_1} = (-6, 3, 5) \end{equation*}

will be counter-clockwise (relative to \(\uvec{n}\) as described above) from \(\uvec{v}_1\text{.}\)

Finally, we need to adjust \(\uvec{v}_1, \uvec{v}_2\) to be the same length, as vectors shouldn't change length as they rotate. However, instead of normalizing both to unit vectors, we can compute that \(\uvec{v}_2\) is \(\sqrt{14}\) times as long as \(\uvec{v}_1\text{,}\) so let's redefine \(\uvec{v}_1\) to be

\begin{equation*} \uvec{v}_1 = (\sqrt{14}, 2 \sqrt{14}, 0) \text{.} \end{equation*}

We now have an orthogonal basis \(\basisfont{B} = \{ \uvec{n}, \uvec{v}_1, \uvec{v}_2 \} \) of \(\R^3\text{.}\) Since \(T\) is uniquely defined by the collection of image vectors \(T(\basisfont{B})\text{,}\) we know \(T\) completely by knowing the images

\begin{align*} T(\uvec{n}) \amp = \uvec{n} \text{,} \amp T(\uvec{v}_1) \amp = \uvec{v}_2 \text{,} \amp T(\uvec{v}_2) \amp = - \uvec{v}_1\text{.} \end{align*}