Matrix

Basics

Identity Matrix

The identity matrix, $I$, is defined as the matrix that satisfies the condition

$IA=A\!$

For any m-by-n matrix $A$.

For example the identity matrix in R ³

$\mathbf {I} _{3}={\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}}$

Vectors

A three diemensional vector

${\overrightarrow {v}}_{3}=v_{1}{\hat {i}}+v_{2}{\hat {j}}+v_{3}{\hat {k}}\!$

has the matrix representation

${\overrightarrow {v}}_{3}={\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\end{bmatrix}}$

Or more generally, an n-diemensional vector has the matrix form

${\overrightarrow {v}}_{n}={\begin{bmatrix}v_{1}\\v_{2}\\\vdots \\v_{n}\end{bmatrix}}$

Determinants

The determinant of a 2-by-2 matrix

$A={\begin{bmatrix}a&b\\c&d\end{bmatrix}}$

is

$det{\mathit {A}}={\begin{vmatrix}a&b\\c&d\\\end{vmatrix}}=ad-bc$

Eigenvalue Analysis

Let

${\mathit {A}}={\begin{bmatrix}2&3\\3&-6\end{bmatrix}}$

We must find all scalars $\lambda$ such that the matrix equation

$(A-\lambda I)\mathbf {x} =\mathbf {0}$

so we subtract $A$ by $\lambda I$

$A-\lambda I={\begin{bmatrix}2&3\\3&-6\end{bmatrix}}-{\begin{bmatrix}\lambda &0\\0&\lambda \end{bmatrix}}={\begin{bmatrix}2-\lambda &3\\3&-6-\lambda \end{bmatrix}}$

So the eigenvalues of $A$ are the solutions of the equation

$det(A-\lambda I)={\begin{vmatrix}2-\lambda &3\\3&-6-\lambda \end{vmatrix}}=0$

This gives us

$(2-\lambda )(-6-\lambda )-(3)(3)=\lambda ^{2}+4\lambda -21=0\!$

Solving this polynomial we find that the eigenvalues of $A$ are

$\lambda =3,-7$\!