Matrices

Let $A~\in~\mathbb{M}_{nm}(\mathbb{F})$, $\bm{x}, \bm{y}~\in~\mathbb{F}^{m}$ and $\lambda \in \mathbb{F}^{1}$

Hence, one can multiple an $n \times m$ matrix by an vector of length $m$, the output will be a vector of length $n$.

The $i$th element of the output vector is the dot product between the $i$th row of $A$ and the vector $\bm{x}$.

Properties:

• $A(\bm{x} + \bm{y}) = A\bm{x} + A \bm{y}$
• $A(\lambda)\bm{x} = \lambda A \bm{x}$

Let $A,B~\in~\mathbb{M}_{nm}$

The same logic follows for matrix subtraction.

Let $A,B~\in~\mathbb{M}_{nm}(\mathbb{C})$, $C~\in~\mathbb{M}_{ml}(\mathbb{C})$.

The conjugate transpose of $A$, given by $A^{\dagger}~\in~\mathbb{M}_{mn}(\mathbb{C})$, is the matrix obtained by first transposing $A$, given by $A^{t}$, and then taking the complex conjugate of all its elements, given by $A^{*}$. This operations can be done is any order, such that $A^{\dagger} = (A^{t})^{*} = (A^{*})^{t}$.

Properties:

• $(AC)^{\dagger} = C^{\dagger}A^{\dagger}$
• $(A+B)^{\dagger} = A^{\dagger} + B^{\dagger} = B^{\dagger} + A^{\dagger}$.

Let $A,B~\in~\mathbb{M}_{nn}(\mathbb{F})$.

The matrix $A$ is invertible (non-singular) if there exists a matrix $A^{-1}~\in~\mathbb{M}_{nn}(\mathbb{F})$ such that

where $\mathbb{I}$ is the identity matrix.

Properties

• $(A^{-1})^{-1} = A$
• If $BA=\mathbb{I}$ for some $B$ then $B=A^{-1}$.
• If both $A$ and $B$ are invertible the $AB$ is invertible with $(AB)^{-1} = B^{-1}A^{-1}$.

Let $A~\in~\mathbb{M}_{nm}$ and $B~\in~\mathbb{M}_{ml}$ and $C=AB$

The elements $c_{ij}$ is given by

which is the dot product between the $i$th row of A and $j$th column of B. Elements Form.

Note the columns of $A$ must equal the rows of $B$ for matrix multiplication to be a valid operation. The output matrix $C$ is then a $n \times l$ matrix, $C~\in~\mathbb{M}_{nl}$. See the below figure for details.

Properties:

• $AB \neq BA$ in general

Let $A,B~\in~\mathbb{M}_{nm}$ such that $A,B \neq 0$, where 0 is the zero matrix

• $\exists~A,B$ such that $AB = 0$, meaning two non-zero matrices can be multiplied to give the zero matrix. Likewise, $\exists ~ A$ such that $A^{2}=AA=0$ for certain $A$.

Let $A,B~\in~\mathbb{M}_{nm}$ and $C~\in~\mathbb{M}_{ln}$

• $C(A+B) = CA + CB$

Let $A,B~\in~\mathbb{M}_{nm}$ and $C~\in~\mathbb{M}_{ml}$

• $(A+B)C = AC + BC$

Let $A~\in~\mathbb{M}_{nm}$, $B~\in~\mathbb{M}_{ml}$ and $C~\in~\mathbb{M}_{lk}$

• $A(BC) = (AB)C$

Let $A~\in~\mathbb{M}_{nm}$.

A is symmetric if $A=A^{t}$ where $(\cdot)^{t}$ is the transpose operation.

Let $A~\in~\mathbb{M}_{nm}$ with elements $a_{ij}$ ($i$th row and $j$th column)

$A$ is anti-symmetric if $a_{ij} = -a_{ji}$. For example

Let $A~\in~\mathbb{M}_{nm}$ with elements $a_{ij}$ ($i$th row and $j$th column)

$A$ is diagonal if $a_{ij} = 0$ if $i \neq j$.

Let $A~\in~\mathbb{M}_{nm}$ with elements $a_{ij}$ ($i$th row and $j$th column)

$A$ is positive definite if for all column vectors $\bm{x}$,

where $*$ is the conjugate transpose operation.

$A$ is positive semi-definite if for all column vectors $\bm{x}$,

Negative and negative semi-definite are defined the say way but with inequalities in the opposite direction.

A matrix is positive definite if it is both hermitian and has all positive eigenvalue - just having positive eigenvalues if not enough for a matrix to be positive definite or positive semi-definite.