Math 511: Linear Algebra

4.8 Applications of Vector Spaces

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4.8.1 Linear Differential Equations

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A linear differential operator or order $n$ is a linear transformation. We will study linear transformations in chapter 6. What makes a function linear?

Definition

A linear transformation is a function $L:V\rightarrow W$:

• whose domain is a vector space
• whose codomain is a vector space
• if $\mathbf{v}_1,\ \mathbf{v}_2\in V$ and $\alpha,\ \beta\in\mathbb{R}$, then

$$ L\left(\alpha\mathbf{v}_1 + \beta\mathbf{v}_2\right) = \alpha L(\mathbf{v}_1) + \beta L(\mathbf{v}_2) $$

Basically a linear transformation is a function from one vector space to another so that the image of a linear combination of vectors in the domain is that very same linear combination of the images of each vector in the column space. Every linear transformation from finite dimensional vector space $\mathbb{R}^n$ to finite dimensional vector space $\mathbb{R}^m$ can be represented by an $m\times n$ matrix $A$. We know from chapter 1 that

$$ L(\alpha\mathbf{v}_1 + \beta\mathbf{v}_2) = A(\alpha\mathbf{v}_1 + \beta\mathbf{v}_2) = \alpha A\mathbf{v}_1 + \beta A\mathbf{v}_2 = \alpha L(\mathbf{v}_1) + \beta L(\mathbf{v}_2) $$

Let us consider the vector space of functions on the real line with two derivatives $C''(\mathbb{R})$ (Remember $C^2(\mathbb{R})$ is the vector space of twice continuously differentiable functions). In fact $C^2(\mathbb{R})$ is an infinite dimensional subspace of $C''(\mathbb{R})$. We also need the vector space of functions on the real line $W$. A differentiable operator from $C''(\mathbb{R})$ to $W$ is a function whose inputs are a function (vector) in $C''(\mathbb{R})$ and whose output is a function(vector) in $W$.

Example 1

The derivative operator $D:C''(\mathbf{R})\rightarrow W$ is a function such that for every $f\in C''(\mathbf{R})$

$$ D(f) = f'(x) $$

The derivative operator is linear because if $f,\ g\in C''(\mathbb{R})$ and $\alpha,\ \beta\in\mathbb{R}$, then

$$ D(\alpha f + \beta g) = \alpha f'(x) + \beta g'(x) = \alpha D(f) + \beta D(g) $$

Definition

A linear differentiable operator $L:C^{(n)}(\mathbb{R})\rightarrow W$ is a linear transformation of the form

$$ L[y] = y^{(n)} + g_{n-1}(x)y^{(n-1)} + g_{n-2}(x)y^{(n-2)} + \dots + g_1(x)y' + g_0(x)y $$

It is linear because if $y_1,\ y_2\in C^{(n)}(\mathbb{R})$ and $\alpha,\ \beta\in\mathbb{R}$, then

$$ \begin{align*} L(\alpha y_1 + \beta y_2) &= \dfrac{d^{n}}{dx^n}(\alpha y_1 + \beta y_2) + g_{n-1}(x)\dfrac{d^{n-1}}{dx^{n-1}}(\alpha y_1 + \beta y_2) + \dots \\ \\ &\qquad\qquad + g_1(x)(\alpha y_1 + \beta y_2)' + g_0(x)(\alpha y_1 + \beta y_2) \\ \\ &= \alpha y_1^{(n)} + \beta y_2^{(n)} + g_{n-1}(x)\alpha y_1^{(n-1)} + g_{n-1}(x)\beta y_2^{(n-1)} + \dots \\ \\ &\qquad\qquad g_1(x)\alpha y_1' + g_1(x)\beta y_2' + g_0(x)\alpha y_1 + g_0(x)\beta y_2 \\ \\ &= \alpha(y_1^{(n)} + g_{n-1}(x)y_1^{(n-1)} + \dots + g_1(x)y_1' + g_0(x)y_1) + \\ \\ &\qquad\qquad\beta(y_2^{(n)} + g_{n-1}(x)y_2^{(n-1)} + \dots + g_1(x)y_2' + g_0(x)y_2) \\ \\ &= \alpha L[y_1] + \beta L[y_2] \end{align*} $$

Definition> Linear Differential Equation

A Linear Differential Equation is the equation that results when we set the output of a linear differentiable operator to a particular vector in the codomain $f(x)\in W$.

$$ \begin{align*} L[y] &= f(x) \\ \\ y^{(n)} + g_{n-1}(x)y^{(n-1)} + g_{n-2}(x)y^{(n-2)} + \dots + g_1(x)y' + g_0(x)y &= f(x) \end{align*} $$

If the function $f(x)\equiv 0$, then we have the homogeneous linear differential equation

$$ y^{(n)} + g_{n-1}(x)y^{(n-1)} + g_{n-2}(x)y^{(n-2)} + \dots + g_1(x)y' + g_0(x)y = 0 $$

Example 2

Show that $y_1 = \cos(x)$ and $y_2 = \sin(x)$ are solutions to the homogeneous linear differential equation

$$ y'' + y = 0 $$

We differentiate each function twice and substitute the function and its second derivative into the differential equation.

$$ \begin{align*} y_1(x) &= \sin(x) \\ y_1'(x) &= \cos(x) \\ y_1''(x) &= -\sin(x) \\ y_1'' + y_1 &= -\sin(x) + \sin(x) = 0\ \Large{\color{green}{\checkmark}} \\ \\ y_2(x) &= \cos(x) \\ y_2'(x) &= -\sin(x) \\ y_2''(x) &= -\cos(x) \\ y_2'' + y_2 &= -\cos(x) + \cos(x) = 0\ \Large{\color{green}{\checkmark}} \\ \end{align*} $$

We know that the set of all solutions to the homogeneous linear differential equation is the null space of the differentiable operator. We also know from the study of the derivative operator $D$ that the dimension of the null space of a linear differentiable operator of order $n$ is ... $n$.

In example 4.8.2 we have a linear differential equation of order 2, hence the dimension of the subspace of solutions for this homogeneous equation is 2. We also know from computing the Wronskian that

$$ \begin{align*} \left|W(\cos, \sin)(x)\right| &= \begin{vmatrix} \cos(x) & \sin(x) \\ -\sin(x) & \cos(x) \end{vmatrix} = 1 \neq 0 \end{align*} $$

so $\cos(x)$ and $\sin(x)$ are two linearly independent vectors in the two dimensional subspace of solutions to the homogeneous differential equation.

Theorem 4.8.1 Solutions of a Linear Homogeneous Differential Equation

Every $n$-th order linear homogeneous differential equation

$$ y^{(n)} + g_{n-1}(x)y^{(n-1)} + g_{n-2}(x)y^{(n-2)} + \dots + g_1(x)y' + g_0(x)y = 0 $$

has $n$ linearly independent solutions. Moreover if $\left\{y_1, y_2, \dots, y_n\right\}$ is a set of linearly independent solutions, and every solution is of the form

$$ y = c_1y_1 + c_2y_2 + \dots + c_{n-1}y_{n-1} + c_ny_n $$

where $c_1$, $c_2$, $\dots$, $c_n$ are real numbers.

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4.8.2 Conic Sections and Rotation

The conic sections or quadratic curves obtained by intersection of the surface of a cone and a plane result in three four types of graphs

Figure 1, From Wikipedia

All of these conic sections are graphs of a quadratic equation of the form

$$ ax^2 + bxy + cy^2 + dx + ey + f = 0 $$

If $b=0$, then the graph of the equation can be obtained by completing the square.

$$ \begin{align*} ax^2 + cy^2 + dx + ey + f &= 0 \\ a\left( x^2 + \frac{d}{a}x \right) + c\left( y^2 + \frac{e}{c}y \right) &= -f \\ a\left( x^2 + \frac{d}{a}x + \frac{d^2}{4a^2} \right) + c\left( y^2 + \frac{e}{c}y + \frac{e^2}{4c^2} \right) &= \frac{d^2}{4a} + \frac{e^2}{4c} - f \\ \\ h &:= -\frac{d}{2a} \\ k &:= -\frac{e}{2c} \\ \\ a\left(x - h\right)^2 + c\left(y - k\right)^2 &= \frac{d^2}{4a} + \frac{e^2}{4c} - f \\ \frac{(x - h)^2}{c} + \frac{(y - k)^2}{a} &= \frac{d^2}{4a^2c} + \frac{e^2}{4ac^2} - \frac{f}{ac} \\ \end{align*} $$

We usually normalize the equation by dividing both sides of the last equation by the right-hand side to obtain

1. an ellipse

$$ \frac{(x-h)^2}{\alpha^2} + \frac{(y-k)^2}{\beta^2} = 1 $$

with axes of length $2\alpha$ and $2\beta$ centered at the point $(h,k)$

2. a circle if the axes of the ellipse are equall, $\alpha = \beta$.

$$ (x-h)^2 + (y-k)^2 = \alpha^2 $$

3. an hyperbola

$$ \frac{(x-h)^2}{\alpha^2} - \frac{(y-h)^2}{\beta^2} = \pm\,1 $$

4. a parabola

$$ (x-h)^2 = 4p(y-k) $$

or

$$ (y-k)^2 = 4p(x-h) $$

The axes of these curves are parallel to the coordinate axes because there is no $\ bxy\ $ term in the general formula.

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4.8.3 Rotation of Axes

Exercise 1

Find the standard coordinates of a point $(x,y)\in\mathbb{R}^2$ relative to the basis

$$ \Theta = \left\{\begin{bmatrix} \cos(\theta) \\ \sin(\theta) \end{bmatrix},\ \begin{bmatrix} -\sin(\theta) \\ \ \ \cos(\theta) \end{bmatrix} \right\} $$

for any angle $\theta$ in radians.

Follow Along

We need to find the transition matrix from the standard coordinate system to the non-standard coordinate system $\Theta$. We start with the simpler to construct transition matrix from the new coordinate system to this standard coordinate system

$$ T = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) &\ \ \cos(\theta) \end{bmatrix} $$
$T^{-1}$ is the transition matrix from the standard coordinate system to the new coordinate system.

$$ \begin{align*} T^{-1} &= \frac{1}{\left|T\right|}\text{adj}(T) \\ \\ &= \frac{1}{\cos^2(\theta) + \sin^2(\theta)}\ \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) &\ \ \cos(\theta) \end{bmatrix}^T \\ \\ &= \begin{bmatrix}\ \ \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix} \end{align*} $$
Now we can compute the coordinates of the vector $(x,y)$ in standard coordinates relative to the new coordinate system $\Theta$.

$$ \begin{bmatrix} x \\ y \end{bmatrix}_{\Theta} = T^{-1}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix}\ \ x\cos(\theta) + y\sin(\theta) \\ -x\sin(\theta) + y\cos(\theta) \end{bmatrix}_{\Theta} $$

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4.8.4 General Second-Degree Equation

Theorem 4.8.2

The general form of the second-degree equation

$$ ax^2 + bxy + cy^2 + dx + ey + f = 0 $$ can be re-written in the form

$$ \alpha u^2 + \gamma v^2 + \delta u + \epsilon v + \zeta = 0, $$

by rotating the axis counterclockwise through the angle $\theta$, where

$$ \cot(2\theta) = \frac{a-c}{b} $$

The coefficients of the equation are obtained from the substitutions

$$ x = u\cos(\theta) - v\sin(\theta),\qquad\text{and}\qquad y = u\sin(\theta) + v\cos(\theta) $$

Proof:

$$ \begin{align*} ax^2 + bxy + cy^2 + dx + ey + f &= 0 \\ \\ a(u\cos(\theta) - v\sin(\theta))^2 + \quad b(u\cos(\theta) - v\sin(\theta))(u\sin(\theta) + v\cos(\theta)) + \qquad\qquad \\ \\ \qquad c(u\sin(\theta) + v\cos(\theta))^2 + d(u\cos(\theta) - v\sin(\theta)) \qquad \\ \\ \qquad\qquad e(u\sin(\theta) + v\cos(\theta)) + f &= 0 \\ \\ \\ a(u^2\cos^2(\theta) - 2uv\cos(\theta)\sin(\theta) + v^2\sin^2(\theta)) + \qquad\qquad \\ \\ \quad b(u^2\cos(\theta)\sin(\theta) + uv\cos^2(\theta) - uv\sin^2(\theta) - v^2\cos(\theta)\sin(\theta)) + \qquad\quad \\ \\ c(u^2\sin^2(\theta) + 2uv\cos(\theta)\sin(\theta) + v^2\cos^2(\theta)) + \qquad \\ \\ d(u\cos(\theta) - v\sin(\theta)) + e(u\sin(\theta) + v\cos(\theta)) + f &= 0 \\ \\ \\ a(u^2\cos^2(\theta) - uv\sin(2\theta) + v^2\sin^2(\theta)) + \qquad\qquad\quad \\ \\ b(u^2\cos(\theta)\sin(\theta) + uv\cos(2\theta) - v^2\cos(\theta)\sin(\theta)) + \qquad\qquad \\ \\ c(u^2\sin^2(\theta) + uv\sin(2\theta) + v^2\cos^2(\theta)) + d(u\cos(\theta) - v\sin(\theta)) + \qquad \\ \\ e(u\sin(\theta) + v\cos(\theta)) + f &= 0 \\ \\ \\ u^2\left(a\cos^2(\theta) + b\cos(\theta)\sin(\theta) + c\sin^2(\theta)\right) + \qquad\qquad \\ \\ uv\left(-a\sin(2\theta) + b\cos(2\theta) +c\sin(2\theta)\right) + \qquad\quad \\ \\ v^2\left(a\sin^2(\theta) - b\cos(\theta)\sin(\theta) + c\cos^2(\theta)\right) + \qquad \\ \\ u\left(d\cos(\theta) + e\sin(\theta)\right) + v\left(e\cos(\theta) - d\sin(\theta)\right) + f &= 0 \end{align*} $$

If one defines $\theta$ by

$$ \cot(2\theta) = \frac{a-c}{b} $$

then one obtains

$$ \begin{align*} b\cos(2\theta) &= (a - c)\sin(2\theta) \\ \\ 0 &= -a\sin(2\theta) + b\cos(2\theta) + c\sin(2\theta) \end{align*} $$

Substituting yields

$$ u^2\left(a\cos^2(\theta) + b\cos(\theta)\sin(\theta) + c\sin^2(\theta)\right) + v^2\left(a\sin^2(\theta) - b\cos(\theta)\sin(\theta) + c\cos^2(\theta)\right) + \\ u\left(d\cos(\theta) + e\sin(\theta)\right) + v\left(e\cos(\theta) - d\sin(\theta)\right) + f = 0 $$

$$ \begin{align*} \alpha &= a\cos^2(\theta) + b\cos(\theta)\sin(\theta) + c\sin^2(\theta) \\ \beta &= 0 \\ \gamma &= a\sin^2(\theta) - b\cos(\theta)\sin(\theta) + c\cos^2(\theta) \\ \delta &= d\cos(\theta) + e\sin(\theta) \\ \epsilon &= e\cos(\theta) - d\sin(\theta) \\ \zeta &= f \end{align*} $$

This completes the proof. $\tombstone$

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4.8.5 Rotation of a Conic Section

Example 3

Perform a rotation of axes to eliminate the $xy$-term from

$$ 5x^2 - 6xy + 5y^2 + 14\sqrt{2}x - 2\sqrt{2}y + 18 = 0 $$

and sketch the graph in the $uv$-plane.

The angle of rotation is given by

$$ \begin{align*} \cot(2\theta) &= \frac{a - c}{b} = \frac{5 - 5}{-6} = 0 \\ \frac{\cos(2\theta)}{\sin(2\theta)} &= 0 \\ \cos(2\theta) &= 0 \\ 2\theta &= \frac{\pi}{2} + 2n\pi,\ n\in\mathbb{Z} \\ \theta &= \frac{\pi}{4} \end{align*} $$

By substituting

$$ \begin{align*} x &= u\cos(\theta) - v\sin(\theta) = \frac{u - v}{\sqrt{2}} \\ y &= u\sin(\theta) + v\cos(\theta) = \frac{u + v}{\sqrt{2}} \end{align*} $$

into the original equation one obtains

$$ 5\left(\frac{u - v}{\sqrt{2}}\right)^2 - 6\left(\frac{u - v}{\sqrt{2}}\right)\left(\frac{u + v}{\sqrt{2}}\right) + 5\left(\frac{u + v}{\sqrt{2}}\right)^2 + 14\sqrt{2}\left(\frac{u - v}{\sqrt{2}}\right) - 2\sqrt{2}\left(\frac{u + v}{\sqrt{2}}\right) + 18 = 0 $$

$$ \frac{5}{2}(u^2 - 2uv + v^2) - 3(u^2 - v^2) + \frac{5}{2}(u^2 + 2uv + v^2) + 14(u-v) - 2(u+v) + 18 = 0 $$

$$ \begin{align*} 2u^2 + 8v^2 + 12u - 16v + 18 &= 0 \\ u^2 + 4v^2 + 6u - 8v + 9 &= 0 \\ u^2 + 6u + 9 + 4\left(v^2 - 2v + 1\right) + 9 &= 9 + 4 \\ \\ (u + 3)^2 + 4(v - 1)^2 &= 4 \\ \\ \frac{(u+3)^2}{4} + \frac{(v-1)^2}{1} &= 1 \end{align*} $$

Figure 2

The ellipse has a major axis of 2 parallel to the $u$-axis and minor axis of 1 parallel to the $v$-axis. Now if we apply the transition matrix

$$ T = \begin{bmatrix} \cos\left(\frac{\pi}{4}\right) & -\sin\left(\frac{\pi}{4}\right) \\ \sin\left(\frac{\pi}{4}\right) &\ \ \cos\left(\frac{\pi}{4}\right) \end{bmatrix} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & -1 \\ 1 &\ \ 1\end{bmatrix} $$

to every point on the ellipse in the $uv$-plane, then we have the graph of the original equation in the $xy$-plane.

Figure 3

Each of the antipodal points at the ends of the major axis in the $uv$-plane become

$$ \begin{align*} \begin{bmatrix} -5 \\ \ \ 1 \end{bmatrix}_{\theta} &= \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & -1 \\ 1 &\ \ 1\end{bmatrix}\begin{bmatrix} -5 \\ \ \ 1 \end{bmatrix} = \begin{bmatrix} \frac{-6}{\sqrt{2}} \\ \frac{-4}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} -3\sqrt{2} \\ -2\sqrt{2} \end{bmatrix} \\ \begin{bmatrix} -1 \\ \ \ 1 \end{bmatrix}_{\theta} &= \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & -1 \\ 1 &\ \ 1\end{bmatrix}\begin{bmatrix} -1 \\ \ \ 1 \end{bmatrix} = \begin{bmatrix} \frac{-2}{\sqrt{2}} \\ \frac{0}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} -\sqrt{2} \\ \ \ 0 \end{bmatrix} \end{align*} $$

in the $xy$-plane.

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