Matrix Operations

Assume A , B , and C are matrices and $α$ and $β$ are constants.

A = [\begin{matrix} a & b & c \\ d & e & f \end{matrix}]

A = [\begin{matrix} g & h & i \\ j & k & l \end{matrix}]

Matrix Times Vector

Given a $m \times n$ matrix and a vector of size $n$

[\begin{matrix} a_{11} & \dots & a_{1 n} \\ ⋮ & ⋱ & ⋮ \\ a_{m 1} & \dots & a_{m n} \end{matrix}] \times [\begin{matrix} b_{1} \\ ⋮ \\ b_{n} \end{matrix}] = b_{1} [\begin{matrix} a_{11} \\ ⋮ \\ a_{m 1} \end{matrix}] + \dots + b_{n} [\begin{matrix} a_{1 n} \\ ⋮ \\ a_{m n} \end{matrix}]

Matrix Times Matrix

Dot Product Rule

Given two matrices $A$ and $B$ of $m \times n$ and $n \times p$ sizes respectively (Where the number of rows in A is equal to the number of columns in B).

The $(i, j)$ th entry of the resulting matrix is row $A_{i}$ $\cdot$ column $B_{j}$ (Row i of matrix A dotted with column j of matrix B).

Determinants of a Matrix

Definition 1: If A is a 1x1 matrix, then $A = [a]$ , then $det (A) = a$

Definition 2: If A is a 2x2 matrix, then $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ , then $det (A) = a d - b c$

Definition 3*: If A is an $n \times m$ matrix , then

A = [\begin{matrix} a_{11} & \dots & a_{1 n} \\ ⋮ & ⋱ & ⋮ \\ a_{m 1} & \dots & a_{m n} \end{matrix}] det (A) =

Determinants are recursive, you split a larger matrix into smaller ones to find the determinant for each and combine them accordingly.

To compute $det (A)$ , we can expand across any row/column
It is better to simplify (using row ops) before computing $det (A)$

This is because the determinant of A does not change when row ops are applied

There are 2 easy ways to tell if a the determinant of a matrix is 0:

An entire row or column is = 0
2 rows or columns are equal

Invertible Matrices

If A is a square matrix, a matrix B is the inverse of A if and only if
$A B = I \land B A = I$

i.e. if matrix A and B multiply communitatively and the result is the identity matrix A is the inverse of B and vice-versa.

Elementary Matrices

A matrix $A$ is elementary if the can be obtained from I using a single row operation.
e. g.

If $E = R_{1} \to 3 R_{1} + R_{2} = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

Inverse of an elementary matrix is just the inverse operation that was used to obtain the elementary matrix.
Can compute products/powers of elementary matrices by repeating the row operation n times.

e.g.

{| \begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix} |}^{10} = | \begin{matrix} 1 & 0 \\ 0 & 2^{10} \end{matrix} |

{| \begin{matrix} 1 & 0 \\ 2 & 1 \end{matrix} |}^{10} = | \begin{matrix} 1 & 0 \\ 2 * 10 & 1 \end{matrix} |

Importance:

Suppose A -> B takes one operation. If E = I correlated to this same row operation, then $B = E A$

i.e. performing row operations are just performing matrix multiplications by an elementary matrix in disguise.

This can be expanded express $B \to A$ in terms of elementary matrices.

e.g. Assuming we get B from A and $E_{1}$ is the first "row operation"

B = E_{n} . . . E_{2} * E_{1} * A \to A = (E_{n} * E_{n - 1} . . . E_{2} * E_{1})^{- 1} B

We can also factor out invertible matrices in terms of elementary matrices.

e.g. Assuming A is an invertible matrix and $E_{1}$ is the first "row operation"

E_{n} * E_{n - 1} . . . E_{2} * E_{1} * A = I \to (E_{n} * E_{n - 1} . . . E_{2} * E_{1})^{- 1} * I = A

Eigenvalues

Eigenvalues are paired with eigenvectors

To find an eigenvalue we use the formula: $$\det(A-\lambda I)=0$$

This comes from the equivalence $A x = λ x = (λ I) x$ where $A$ is a square invertible matrix, and $x$ is our eigenvector

Removing the $x$ vector from our equation moving everything over to one side we get $A - (λ I) = 0$ . The only way this is true is if the determinant of the LHS is equal to 0. Therefore we get $det (A - λ I) = 0$

Eigenvectors

Eigenvectors are paired with eigenvalues

To find an eigenvalue we first have to find its corresponding eigenvalue or $λ$ .

From there we sub in the value of the eigenvalue to the equation

Theorems

#determinants
You cannot distribute determinants to matrices i.e $det (A + B) \neq det (A) + det (B)$

If A is a $n \times n$ matrix:

$det (A^{T}) = det (A)$
$det (k A) = k^{n} det (A)$

#determinants
If A, B are both $n \times n$ :

Prof. Hernandez calls it the "beautiful theorem"

$det (A B) = det (A) det (B)$

This provides a few things consequentially:

$det (A B) = det (A) det (B) = det (B) det (A) = det (B A)$
$det (A_{1} . . . A_{n}) = det (A_{1}) . . . det (A_{n})$
If $A_{1} . . . A_{n}$ the previous formula becomes $det (A^{n}) = det (A)^{n}$

#determinants
Suppose that A is $n \times n$ and also that A is invertible:

$det (A) \neq 0$
$det (A^{- 1}) = det (A)^{- 1} = \frac{1}{det (A)}$
$det (A^{T}) = det (A)$

If A is a square matrix, then the following statements are equivalent

A is invertible
The only solution to $A x = 0$ is $x = 0$
RREF of $A$ is $I$
A is a product of elementary matrices
The determinant of $A \neq 0$

$1 \to 2 \to 3 \to 4 \to 5 \to 1$

#determinants
If A is a square matrix and is invertible:

If a multiple of a row is added to another row $det (A) = det (A)$
If a row is multiplied by a scalar $k$ then $det (A) = k det (A)$
If a row is swapped with another then $det (A) = - det (A)$
If any two rows or columns are identical then $det (A) = 0$
If there exists a row or column of zeroes then $det (A) = 0$

#diagonalizeable
A square matrix $A$ is diagonalizable if we can factor $A$ as $A = P D P^{- 1}$ where P=invertible and D=diagonal

Ideas:

Diagonal matrices are easy to deal with (compute det, eigenvalue, eigenvector, etc.)
If A is diagonalizable then A inherits those properties from D

#diagonalizeable
A square matrix $A$ is diagonalizable if and only if we can build an invertible matrix P where columns are eigenvectors. In this case if $P = [x_{1}, x_{2}, . . . x_{n}]$ which is also invertible

Properties

$α$ and $β$ are both constants while $A$ , $B$ , and $C$ are all matrices.

A + B = B + A
#associative-property
(A + B) + C = A + (B + C)
#communitaive-property
$α$ (A + B) = $α$ A + $α$ B
#distributive-property
$α + β$ (A + B) = $α$ A + $β$ B
#distributive-property
( $α β$ )(A) = $α$ ( $β$ A)
$(A^{T})^{T} = A$
$(A B)^{T} = B^{T} A^{T}$ order matters
$I$ is the identity matrix and has the same properties as the constant 1
$I A = A I$ for all matrices
Inverted Matrices
- $B = A^{- 1}$ exists exactly when $B = A^{- 1} = \frac{1}{A}$ which should be $A B = B A = I$
- $C = (A B)^{- 1} = B^{- 1} A^{- 1} = I$ order matters
- $(A^{k})^{- 1} = (A^{- 1})^{k}$ k can be a Transpose i.e. $(A^{T})^{- 1} = (A^{- 1})^{T}$