Modulus

Essentially the length of the vector from the origin to $(a, b)$. $$|w| = |a + bi| = \sqrt{a^2 + b^2}$$

Argument

The line from the origin to (a, b) has an angle $\theta$. This is called the argument and is denoted by arg $\theta$. The principal argument Arg $\theta$ (with the initial letter capitalized) is the value in the interval $(−\pi, \pi]$

Polar form

Given $r = |w|$ and $\theta = arg\:w$ $$w = r\:cos\:\theta + i\:r\:sin\:\theta$$

Multiplication

$$ |wz| = |w||z| $$ $$ arg(wz) = arg(w) + arg(z) $$

Division

$$ \left|\frac{z}{w}\right| = \frac{|z|}{|w|} $$ $$ arg\left(\frac{z}{w}\right) = arg(z) - arg(w) $$

Complex form

Euler's formula: $$ e^{ i\theta } = \cos \theta + i\sin \theta $$

Based on that formula, we can write a complex number $w$ in complex form. $$r = |w|$$ $$\theta = Arg\;w$$ $$w = r\,e^{i\theta + 2 \pi i k} = r\,e^{i(\theta + 2 \pi k)}$$ where $k = 0, \;\pm 1, \;\pm 2, \;\dots$

Roots

When finding the $n$th roots of a complex number $w$, we can find $n$ roots. These $n$ roots are distinct complex numbers. For example, let's find $\sqrt[3]{w}$ where $w = 4 + 4i$. We expect to find 3 roots.

First, write $w$ in complex form: $$ r = |w| = \sqrt{4^2 + 4^2} = \sqrt{32} $$ $$ \theta = arg(w) = tan^{-1}\left(\frac{4}{4}\right) = \frac{\pi}{4} $$ $$ w = \sqrt{32} \: e^{i(\pi/4 + 2 \pi k)} $$ where $k = 0, \;\pm 1, \;\pm 2, \;\dots$

Next comes the most important step: raise both sides to the power of $\frac{1}{3}$: $$ w^{\frac{1}{3}} = \left(\sqrt{32} \: e^{i(\pi/4 + 2 \pi k)} \right)^{\frac{1}{3}} $$ $$ \sqrt[3]{w} = \sqrt{32}^{\frac{1}{3}} \: e^{\frac{1}{3} i(\pi/4 + 2 \pi k)} $$ Finally insert $k = 0, 1, 2$ to get the three 3rd roots of $w$ respectively.
Note: If you try to insert $k = 3$, you'll see that it gives the same root as $k = 0$

Complex functions

$$ \cos{z} = \frac{e^{iz} + e^{-iz}}{2} $$ $$ \sin{z} = \frac{e^{iz} - e^{-iz}}{2i} $$

Linear, homogenous equations with constant coefficients

$$ y''+py'+qy=0 $$ Solve $\lambda^2+p\lambda+q=0$

Inhomogeneous equations

The inhomogenous linear equation $$ y''+py'+qy=f $$ Has the general solution: $$ y=y_p+C_1y_1+C_2y_2 $$ Where $y_1$ and $y_2$ are solutions to the associated homogenous equation $$ y''+py'+qy=0$$ and $y_p$ is a particular solution to the inhomogenous equation $$ y''+py'+qy=f$$

In order to find $y_p$, you can use one of the following methods:

• Undetermined coefficients
• Variation of parameters

Harmonic oscillator

$$E: y'' + 2cy' + w_0^2 y = cos(wt)$$ $$E_h: y'' + 2cy' + w_0^2 y = 0$$

Step 1:
Solve the homogenous equation by solving the characteristic equation

$$\Delta = 4c^2 - 4w_0^2$$

if $\Delta < 0$: Underdamped (Oscillations. Goes to halt slowly.)

if $\Delta = 0$: Critically damped (the fastest way the system goes to halt)

if $\Delta > 0$: Overdamped (an exponential function that slowly approaches 0)

Step 2:
Compute $y_p$ by using undetermined coefficients

$$ y_p = A \:cos(wt) + B \:sin(wt) $$

General solution: $$y = y_h + y_p$$ where $y_h$ tend to 0. It is the transient term. $y_p$ will keep oscillating. It's called the steady state of the system.

Normalization of vectors

When we normalize a vector we are making the length of the vector $1$ (for the zero vector, with length equal to zero, this is not possible). This is an important operation in regards to steady-state vectors and computer graphics. A normalized vector is also called a unit vector.

$$ \vec{\hat{v}} = \frac{\vec{v}}{\begin{Vmatrix} \vec{v} \end{Vmatrix}} $$ Where $\begin{Vmatrix}\vec{v}\end{Vmatrix}$ is the length of the vector: $$ \begin{Vmatrix}\vec{v}\end{Vmatrix} = \sqrt{v_1^2 + v_2^2 + \dots + v_n^2}$$

Terminology

For example, when a text says "Suppose a 4 x 7 matrix A (...)", that means that m = 4, n = 7. Can't remember the order? Rule of thumb: Just remember the word "man" and that m comes before n. m is the height of the matrix, and n is the width of the matrix.

Linear transformations

A matrix can be regarded as a transformation that transforms a vector. For example, a transformation $T$ can double the length of a 2D vector. Such a transformation $T$ and its standard matrix $A$ would look like this:

$$ T(x) = 2x $$ $$ A = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} $$

Inverse of a matrix

The inverse of a matrix $A$ is denoted $A^{-1}$ and is the matrix that one can multiply A with to get the identity matrix.

$$ A^{-1}A = I $$

For example, consider the following matrix

$$ A = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1\\ 0 & -1 & 2 \end{bmatrix}. $$

To find the inverse of this matrix, one takes the following matrix augmented by the identity, and row reduces it as a 3 by 6 matrix: $$ [\;A\;|\;I\;] = \left[ \begin{array}{rrr|rrr} 2 & -1 & 0 & 1 & 0 & 0\\ -1 & 2 & -1 & 0 & 1 & 0\\ 0 & -1 & 2 & 0 & 0 & 1 \end{array} \right]. $$

By performing row operations, one can check that the reduced row echelon form of the this augmented matrix is:

$$ [\;I\;|\;B\;] = \left[ \begin{array}{rrr|rrr} 1 & 0 & 0 & \frac{3}{4} & \frac{1}{2} & \frac{1}{4}\\[3pt] 0 & 1 & 0 & \frac{1}{2} & 1 & \frac{1}{2}\\[3pt] 0 & 0 & 1 & \frac{1}{4} & \frac{1}{2} & \frac{3}{4} \end{array} \right]. $$

By this point we can see that B is the inverse of A.

A matrix is invertible if, and only if, its determinant is nonzero.

LU factorization

With this technique, you can write $A = LU$ where $L$ is a lower triangular matrix with ones on the diagonal and $U$ is an upper triangular matrix (echelon form). Do this only if you can reduce A without row exchanges.

First, write down the augmented matrix: $$ [\;A\;|\;I\;] $$ Row reduce it until the left side is in echelon form (not reduced echelon form). You should now have $$ [\;U\;|\;L^{-1}\;] $$ Now, to find L, use your favorite technique for finding the inverse of a matrix.

Why LU factorization?
Because when $A = LU$, and you want to solve $Ax = b$ for $x$, then you can find $x$ by solving these two equations: $$Ux = y$$ $$Ly = b$$ Each of these two equations are relatively simple, and computers are able to solve them faster than solving the more complex equation $Ax = b$

Determinants

The determinant of a 2x2 matrix is simple to compute. Here's an example: $$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 4 & 9 \\ 3 & 2 \end{bmatrix} $$ $$ det\: A = ad - bc = 4 * 2 - 9 * 3 = -19 $$

Calculating the determinant of a larger matrix is harder. Look up "cofactor expansion" in your textbook or on the internet. Moreover, here are some useful theorems related to determinants:

• If $A$ is a triangular matrix, then $det\: A$ is the product of the entries on the main diagonal of $A$.

• Adding one row of $A$ to another row of $A$ will not change $det\: A$.

• A square matrix $A$ is invertible if and only if $det\: A \neq 0$.

Eigenvalues and eigenvectors

Eigenvalues and eigenvectors apply to square matrices.

Column Space and Null Space of a Matrix

The column space of a matrix A is the set Col A of all linear combinations of A. The vector $\textbf{b}$ is a linear combination of A if the equation $A\textbf{x} = \textbf{b}$ has a solution.

If $V$ is a vector space, then $dim\: V$ is the number of vectors in a basis for V.

The rank of a matrix $A$ is the dimension of the vector space generated (or spanned) by its columns. In other words, $rank \:A$ equals the number of pivot columns (in the echelon form of $A$). Rank A = dim Col A

The null space of a matrix A is the set Nul A of all solutions of the homogeneous equation $A\textbf{x} = \textbf{0}$

Orthogonality

A set of vectors is orthogonal if the vectors in the set are orthogonal (perpendicular) to each other. Two vectors $u$ and $v$ are orthogonal to each other if $u \cdot v = 0$. In other words, $u^T v = 0$

Symmetric matrices

A symmetric matrix is a square matrix $A$ such that $A^T = A$.

For any symmetric matrix $A$, this is true: $A^T A = A A^T$

If $A$ is symmetric, then any two eigenvectors from different eigenspaces are orthogonal.