# Matrix

## Basics

A square matrix of dimension $n$ is a table of $n\times n$ elements $a_{ij}$ given by

$$A = \begin{pmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \cdots & a_{2n} \\ \vdots & \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & a_{n3} & \cdots & a_{nn} \\ \end{pmatrix}$$

Suppose, we have an other $n\times n$ matrix $B$ with elements $b_{ij}$.

• Matrix elements $a_{ij}$ are (possibly complex) numbers.
• The sum of two matrices $A$ and $B$ is a matrix $C$ with matrix elements given by $c_{ij} = a_{ij} + b_{ij}$. Or, in matrix form, $C = A + B$.
• Their differences gives a matrix $D$ with matrix elements given by $d_{ij} = a_{ij} - b_{ij}$. Or, in matrix form, $D = A - B$
• Sum is commutative. i.e. $A + B = B + A$ but difference is not. i.e. $A - B \neq B - A$
• Product gives a matrix $E$ with matrix elements given by $e_{ij} = \sum_{k = 1}^{n} a_{ik} b_{kj}$. Or, in matrix form, $E = AB$
• Product is not commutative. i.e. $AB \neq BA$. To be precise, call as non-commutative. But it is associative. i.e. $A(BC) = (AB)C$.
• $a_{ij}, b_{ij}, c_{ij}, d_{ij},$ and $e_{ij}$ are ordinary numbers, meaning it does not matter in which order they are written. But, in matrix form may matter.
• To justify why product is not commutative. Let $e_{ij} = \sum_{k = 1}^{n} a_{ik} b_{kj}$ for $E = AB$ and $e'_{ij} = \sum_{k = 1}^{n} b_{ik} a_{kj}$ for $E' = BA$. Even though order does not matter but the elements are of different row and column.
• Commutator of two matrices $A$ and $B$ is given by $[A, B] := AB - BA$
• The product of a (possibly complex) number $x$ with a matrix $A$ is a matrix $B$ with elements: $b_{ij} = x a_{ij}$. In matrix form, $B = x A$.
• $B^{2} = x^{2} A^{2}$ can be written as in matrix component form as $\sum_{k} b_{ik} b_{kj} = x^{2} \sum_{k} a_{ik} a_{kj}$.

## Notes

• Matrix $L$ is not invertible if it has a zero mode i.e. $L e^{(0)} = 0$ where $e^{(0)} = (1, 1, \ldots, 1)^{T}$ is a constant eigenvector