Discover Linear Algebra

Example 45.4.1. From Discovery 45.2.

1. In Discovery 45.2.a, we considered \(\funcdef{T}{\matrixring_2(\R)}{\poly_2(\R)}\) defined by

   \begin{equation*} T\left(\begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix}\right) = (a + b) x^2 + (a + b + c) x - d \text{.} \end{equation*}

   Choosing the standard basis \(\basisfont{S}\) for domain space \(\matrixring_2(\R)\) and the choice \(\basisfont{S}' = \{ x^2, x, 1 \}\) of a standard basis for \(\poly_2(\R)\text{,}\) we create a chain of transformations

   \begin{equation*} \R^4 \xrightarrow{\invcoordmap{S}} \matrixring_2(\R) \xrightarrow{T} \poly_2(\R) \xrightarrow{\coordmap{S'}} \R^3 \end{equation*}

   where a typical vector \((a,b,c,d)\) in \(\R^4\) shuttles through the pipe by

   \begin{equation*} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} \mapsto \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix} \mapsto (a + b) x^2 + (a + b + c) x - d \mapsto \begin{bmatrix} a + b \\ a + b + c \\ -d \end{bmatrix}\text{.} \end{equation*}

   So the composite transformation \(\R^4 \to \R^3\) has input-output formulas

   \begin{equation*} \left\{\begin{array}{cclclclcl} y_1 \amp = \amp x_1 \amp + \amp x_2 \text{,} \\ y_2 \amp = \amp x_1 \amp + \amp x_2 \amp + \amp x_3 \text{,} \\ y_3 \amp = \amp \amp \amp \amp \amp \amp - \amp x_4 \text{.} \\ \end{array}\right. \end{equation*}

   Using these formulas as rows, we obtain matrix

   \begin{equation*} \matrixOf{T}{S'S} = \left[\begin{array}{rrrr} 1 \amp 1 \amp 0 \amp 0 \\ 1 \amp 1 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp -1 \end{array}\right]\text{.} \end{equation*}

   We should get the same result if we using Procedure 45.3.1. First, compute the images of the four basis vectors in \(\basisfont{S}\text{:}\)

   \begin{align*} T\left(\begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \end{bmatrix}\right) \amp = x^2 + x \text{,} \amp T\left(\begin{bmatrix} 0 \amp 1 \\ 0 \amp 0 \end{bmatrix}\right) \amp = x^2 + x \text{,}\\ T\left(\begin{bmatrix} 0 \amp 0 \\ 1 \amp 0 \end{bmatrix}\right) \amp = x \text{,} \amp T\left(\begin{bmatrix} 0 \amp 0 \\ 0 \amp 1 \end{bmatrix}\right) \amp = -1 \text{.} \end{align*}

   Relative to the chosen basis \(\basisfont{S}'\) of the codomain space \(\poly_2(\R)\text{,}\) we have coordinate vectors

   \begin{align*} \matrixOf{T(E_{11})}{S'} \amp = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \text{,} \amp \matrixOf{T(E_{12})}{S'} \amp = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \text{,}\\ \\ \matrixOf{T(E_{21})}{S'} \amp = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \text{,} \amp \matrixOf{T(E_{22})}{S'} \amp = \left[\begin{array}{r} 0 \\ 0 \\ -1 \end{array}\right]\text{.} \end{align*}

   Using these coordinate vectors as columns in a matrix, we obtain the same result for \(\matrixOf{T}{S'S}\) as before.

2. In Discovery 45.2.b, we considered \(\funcdef{T}{\poly_2(\R)}{\uppermatring_2(\R)}\text{,}\) for \(\uppermatring_2(\R)\) the space of \(2 \times 2\) upper triangular matrices, defined by

   \begin{equation*} T(a x^2 + b x + c) = \begin{bmatrix} a - b \amp b + c \\ 0 \amp c - a \end{bmatrix} \text{.} \end{equation*}

   Again, we take \(\basisfont{S} = \{ x^2, x, 1 \}\) as a choice of standard basis for \(\poly_2(\R)\text{,}\) and choose \(\basisfont{S}' = \{ E_{11}, E_{12}, E_{22} \}\) as a basis for \(\uppermatring_2(\R)\text{.}\) With these choices, we create a chain of transformations

   \begin{equation*} \R^3 \xrightarrow{\invcoordmap{S}} \poly_2(\R) \xrightarrow{T} \uppermatring_2(\R) \xrightarrow{\coordmap{S'}} \R^3 \end{equation*}

   where a typical vector \((a,b,c)\) in \(\R^3\) shuttles through the pipe by

   \begin{equation*} \begin{bmatrix} a \\ b \\ c \end{bmatrix} \mapsto a x^2 + b x + c \mapsto \begin{bmatrix} a - b \amp b + c \\ 0 \amp c - a \end{bmatrix} \mapsto \begin{bmatrix} a - b \\ b + c \\ c - a \end{bmatrix}\text{.} \end{equation*}

   So the composite transformation \(\R^3 \to \R^3\) has input-output formulas

   \begin{equation*} \left\{\begin{array}{ccrcccl} y_1 \amp = \amp x_1 \amp - \amp x_2 \text{,} \\ y_2 \amp = \amp \amp \amp x_2 \amp + \amp x_3 \text{,} \\ y_3 \amp = \amp - x_1 \amp \amp \amp + \amp x_3 \text{.} \\ \end{array}\right. \end{equation*}

   Using these formulas as rows, we obtain matrix

   \begin{equation*} \matrixOf{T}{S'S} = \left[\begin{array}{rrrr} 1 \amp -1 \amp 0 \\ 0 \amp 1 \amp 1 \\ -1 \amp 0 \amp 1 \end{array}\right]\text{.} \end{equation*}

   We should get the same result if we using Procedure 45.3.1. First, compute the images of the three basis vectors in \(\basisfont{S}\text{:}\)

   \begin{align*} T(x^2) \amp = \left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp -1 \end{array}\right] \text{,} \amp T(x) \amp = \left[\begin{array}{rr} -1 \amp 1 \\ 0 \amp 0 \end{array}\right] \text{,} \amp T(1) \amp = \begin{bmatrix} 0 \amp 1 \\ 0 \amp 1 \end{bmatrix} \text{.} \end{align*}

   Relative to the chosen basis \(\basisfont{S}'\) of the codomain space \(\uppermatring_2(\R)\text{,}\) we have coordinate vectors

   \begin{align*} \matrixOf{T(x^2)}{S'} \amp = \left[\begin{array}{r} 1 \\ 0 \\ -1 \end{array}\right] \text{,} \amp \matrixOf{T(x)}{S'} \amp = \left[\begin{array}{r} -1 \\ 1 \\ 0 \end{array}\right] \text{,} \amp \matrixOf{T(1)}{S'} \amp = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}\text{.} \end{align*}

   Using these coordinate vectors as columns in a matrix, we obtain the same result for \(\matrixOf{T}{S'S}\) as before.

3. In Discovery 45.2.c, we considered the differentiation operator \(\funcdef{D}{V}{V}\) on the space \(V = \Span \{ e^x \sin x, e^x \cos x \}\text{.}\) The provided spanning set for \(V\) is linearly independent (check!). Using this basis on both the “input” and “output” ends, we create a chain of transformations

   \begin{equation*} \R^2 \xrightarrow{\invcoordmap{B}} V \xrightarrow{D} V \xrightarrow{\coordmap{B}} \R^2 \end{equation*}

   where a typical vector \((a,b)\) in \(\R^2\) shuttles through the pipe by

   \begin{equation*} \begin{bmatrix} a \\ b \end{bmatrix} \mapsto a e^x \sin x + b e^x \cos x \mapsto (a - b) e^x \sin x + (a + b) \cos x \mapsto \begin{bmatrix} a - b \\ a + b \end{bmatrix}\text{.} \end{equation*}

   So the composite transformation \(\R^2 \to \R^2\) has input-output formulas

   \begin{equation*} \left\{\begin{array}{cclcl} y_1 \amp = \amp x_1 \amp - \amp x_2 \text{,} \\ y_2 \amp = \amp x_1 \amp + \amp x_2 \text{.} \end{array}\right. \end{equation*}

   Using these formulas as rows, we obtain matrix

   \begin{equation*} \matrixOf{T}{B} = \left[\begin{array}{rr} 1 \amp -1 \\ 1 \amp 1 \end{array}\right]\text{.} \end{equation*}

   We should get the same result if we using Procedure 45.3.1. First, compute the images of the two basis vectors in \(\basisfont{S}\text{:}\)

   \begin{align*} D(e^x \sin x) \amp = e^x \sin x + e^x \cos x \text{,} \\ D(e^x \cos x) \amp = - e^x \sin x + e^x \cos x \text{.} \end{align*}

   Relative to the basis \(\basisfont{B}\) of \(V\text{,}\) we have coordinate vectors

   \begin{align*} \matrixOf{D(e^x \sin x)}{B} \amp = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \text{,} \amp \matrixOf{D(e^x \cos x)}{B} \amp = \left[\begin{array}{r} -1 \\ 1 \end{array}\right]\text{.} \end{align*}

   Using these coordinate vectors as columns in a matrix, we obtain the same result for \(\matrixOf{D}{B}\) as before.

Example 45.4.2. Computing an image vector.

Consider \(f(x) = 3 e^x \sin x - e^x \cos x\text{,}\) with

Then we can compute

Converting this coordinate vector back into a linear combination, we find

Example 45.4.3. Computing the image vector of a composition.

Again consider \(f(x) = 3 e^x \sin x - e^x \cos x\text{.}\) What if we would like to compute the second derivative of \(f(x) = 3 e^x \sin x - e^x \cos x\text{?}\) We could of course just multiply the resulting coordinate vector from Example 45.4.2 by \(\matrixOf{D}{B}\) again. But let's proceed as if we had not already calculated \(D(f)\text{.}\)

Since the second derivative of \(f\) is

we can compute

Then from this we can compute

Converting this coordinate vector back into a linear combination, we find

Example 45.4.4. Computing the image vector of the inverse of an operator.

Once again consider \(f(x) = 3 e^x \sin x - e^x \cos x\text{.}\) Since \(\matrixOf{D}{B}\) is invertible, so must be operator \(D\text{.}\) And if \(D\) represents differentiation, then \(\inv{D}\) must represent antidifferentiation, so computing \(\inv{D}(f)\) will be the antiderivative of \(f(x)\text{.}\)

First compute

With this, we can compute

Converting this coordinate vector back into a linear combination, we find that the antiderivative of \(f\) is