Linear Algebra
Inverse of a matrix
Elementary Row Transformations
- \(R_i \leftrightarrow R_j\) interchange any tow rows
- \(R_i \rightarrow kR_i, k \ne 0\)
- \(R_i \rightarrow R_i + k R_j, k \ne 0\)
Rank of a matrix \(\rho(A)\)
Definition
Rank of a matrix is \(r\) iff
- it has at least one non zero minor of order \(r\)
- every minor of order higher than \(r\) is 0
Echelon form - upper triangular matrix
- number of non-zero rows in this form of matrix is equal to rank of matrix
Normal Form
Any of these forms
here rank of these matrix is \(n\)
Solution of Linear Equations
- \(B = 0\)
- homogeneous linear system of equations
- if \(\rho(A) = \text{no of unknowns} \implies\) unique trivial solution, all equal to 0
- if \(\rho(A) \ne \text{no of unknowns} \implies\) infinite no of non-trivial solutions
- \(B \ne 0\)
- non homogeneous linear system of equations
- augumented matrix \([A | B]\)
- if \(\rho([A|B]) = \rho(A)\)
- consistent
- \(\rho(A) = \text{no of unknowns}\) - unique solution
- \(\rho(A) \ne \text{no of unknowns}\) - infinite many solutions
- if \(\rho([A|B]) \ne \rho(A)\)
- inconsistents
- no solutions
- if \(\rho([A|B]) = \rho(A)\)
Linearly Dependent and Linearly Independent Vectors
Vectors \(X_1, X_2, \dots, X_n\) are said to be linearly dependent, iff \(\exists\) \(\lambda_1, \lambda_2 \dots, \lambda_n\) \(\lambda_i \ne 0\) and \(\sum_{i=0}^{n} \lambda_i X_i = 0\)
Vectors \(X_1, X_2, \dots, X_n\) are said to be linearly independent, iff \(\sum_{i=0}^{n} \lambda_i X_i = 0\) only when all \(\lambda_i = 0\)
Vectors, \(X_1, X_2, \dots, X_n\), are linearly independent if \(\rho( \begin{bmatrix} X_1 & X_2 & \dots & X_n \\ \end{bmatrix} ) = n\), else they are linearly dependent.
Characteristics Equations
Given a square matrix \(\textbf{A}\), \(|\textbf{A} - \lambda \textbf{I} | = 0\) is its characteristics equation.
Cayle Hamilton Theorem
Every square matrix satisfies its own characteristics equation
Eigen Values
- Solution to this equation gives eigen values.
Eigen Vectors
- corresponding to each eigen value
- the non zero solution to the equation
- \((\textbf{A} - \lambda \textbf{I}) \textbf{X} = 0\)
- give eigen vector
Diagonalization of a Matrix
- Steps to diagonalize a matrix
- find characteristics equation
- find eigen values \(\lambda_1, \lambda_2 \dots, \lambda_n\)
- find eigen vector \(X_1, X_2, \dots, X_n\)
- construct moadal matrix \(P = [X_1 X_2 \dots X_n]\)
- find \(P^{-1}\)
- finally \(D = P^{-1}AP = \operatorname{diag}(\lambda_1, \lambda_2 \dots, \lambda_n)\)
- property of diagonal matrix
- \(A^n = PD^nP^{-1}\)