Ma1513 cheat sheet PDF

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MA1513 – Linear Algebra with Differential Equations Ch1- Linear Systems and Matrix Algebra Number of solutions No solution One solution Infinite solutions Last column is Every column except Non-pivot row pivot last is pivot present (except last) Solving matrix equations (a) inverse of 2x2 matrix (b) inverse of a nxn matrix using GJE to make the LHS an I matrix A

( )

a b , c d x=ad − bc d −b −1 x A = x a −c x x =

rows)

[ | ] [

Ax=0

= 1 : Unit Vector,

Let

u∙ v ¿ ¿ −1 θ=cos ¿ u ∙ v=0 : orthogonal

vectors Standard Basis Vectors

trivial and non-trivial

R

For

singular

matrix

(due

u,v

not

(

)

| |

3

:

det ( A ) = product of diagonal entri

det ( A ) =0

If det(A) = 0, A is singular (columns of A are not linearly indep.), vice versa Properties −1 n

det ( cA ) =c de det ( AB )= det ( A ) 1 −1 det ( A T )=det det ( A ) = det ( A

A A =I −1 ( AT ) = ( A−

Rank Rank of A – number of non-zero rows in the row-echelon form of A (b) rank of identity matrix (a) rank of zero matrix (all

zero

e 1=( 1, 0,0 ), e 2= ( 0, 1, 0) , e3=(0, 0,1 )

rank ( I n )=n

- (no zero

span { e1 , e2 , e 3 }= R

3

S= { u1 , u2 … , uk }

be a subset of vector space

is a basis for V if: span { u 1 , u2 … , uk} =V u1 ,u 2 … , uk are linearly indep.

Given matrix A, Row space Non-zero rows in the REF form of A

If

R

n

is a subset of

n

that satisfies the

in

V

R

conditions: 1. V can be expressed in linear span form there is a set of vectors

such

2.

Column space Reduce A to REF form, column space is the corresponding pivot column in REF form in A *exclude redundant vectors

Nullspace Solve

Ax=0

Ax=0 , nullity ( A ) n n dim( V )≠ R ≠ n ,V ≠ R

Dimension Theorem for Matrices (Rank-Nullity Theorem) If A is a matrix with n columns, then rank(A) + nullity(A) = n Coordinate Vectors

that

V =span {vectors } cu ∈ V

and

* every subspace of

R

n

is a subset of

R

angle between them) and new y’=

n

but not vice

versa Given span{a, b}, solve the solution to find the plane -> Let t=__ (particular sol) Solution space (ss) of a homogeneous system n Let , its solution set is a subspace of (ss)

R

θ ( cos sin θ ) α (cos sin α )

(where

Given a standard basis, x’=

2. V satisfies the closure properties

Ax=0

V

S 1.

aka

Subspaces

u+ v ∈ V

are scalar multiples

dim for ss of

The set of all linear combinations.

-

u,v

Dim of nullspace of A :

Linear Span

A subspace V of

(d) triangular matrix

scalar

(a) Basis

to

n ×n matrix a b c A= d e f det ( A ) =a e f −b h i g h i

( )

rows)

||u||= √ u ∙u

||u||= √ u ∙u ||u||∙∨|v|∨¿

4.

, at least

multiples, after GE, no non-pivot column *lie in the same plane only if the vectors are linearly dependent Basis and Dimension

)

a b A= c d det ( A ) =ad−b

rank ( 0) =0

3.

||u−v||= √(u−v)∙ (u−v)

(b) Dimension

=

det ( A ) =0

one of the vectors can be written as a linear combination of the other 2 vector s

n

2.

* singular x non-singular det(AB)=det(A)det(B)=0) Determinants (a) 2x2 matrix (b)

(c)contains same rows/columns

R

Linearly Dependent Scalars that are non-zero

det ( A ) ≠ 0

R

such that

*usually written in column

R

u ∙ v=u1 v 1 +u2 v 2 + … + u n v n

(

Null space of A

span {c 1 , c 2 , …Vectors u Au=0

if a linear system has n variables, its solutions are n-vectors, is a n . subspace of

1.

solution

x=0

I

n Ch2- Linear Combinations and Linear Spans n Solution Set as a subspace of

Length, Distance and angles in

x= A b ¿

Column space

span {r 1 ,r 2 ,

Linear Independence Linearly Independent 1 Only possible scalars are vector 0

0

 A is non-singular RREF of A is

−1

trivial

≠

det(A)

If inverse exists, A is non-singular / invertible Setting Parameters If variables are dependent on each other, doesn’t matter which one it is - usually use the last non-zero column to be the parameter (REF form) Solutions of Linear Systems A is non-singular A is singular one solution ( no / infinite sol

Ax = b

Row space , A is full

rank Rank and Determinant If A is full rank and rank(A) = n, RREF of A has n pivot columns 

a b c1 0 0 1 0 0 d e f 0 1 0 ⟶ 0 1 0 g h i 0 0 1 0 0 1

( )

rank ( A )= min{m , n }

A is a mxn matrix. If

θ

is the

*you’re finding the

‘coordinate’ of new axis w.r.t the old one Given a non-standard basis

{( a , b ) , ( c , d )}

coordinates

the

of

v

c 1 ( e1 ) +c 2 (e 2 ) +…

w.r.t

basis

:

, find the new rewrite

as

Projection of a vector onto a subspace of Find

p

that

Vector

(u-p)

is

R

n

orthogonal

subspace

v

:

(u− p ) ∙ v=0 ¿|u−p|∨≤∨|u−v|∨¿

p is the projection if and only

Projection and Linear Approximation Given that p is a projection of v onto a plane, v-p is orthogonal to the plane (works for any plane that contains the origin) Least Squares solution n in that minimizes ( Vector

u

Au

¿| Ax− b|∨¿

R

T

and

‘Mapping’- mapping from

R

to

R

T :R →R

u∈R

defined

, Domain of

T (u ) =Au

by

A=(T ( e1) T ( e2 ) … T (e n ) )

CounterC rotation

Reflection about yaxis

(

Enlargement (A>1)

for

)

( ) j

( λI − A) x=0

Write as

Shearing parallel to xaxis

th column of A

.

det( λI − A )=0

λ

is all the diagonal entries

, solve

Solution space of the homogeneous system (subspace of )

Finding Eigenvectors:

( λI − A) x=0

2. Get the general solution and pull out the parameter 3. Remaining vector is the eigenvector of A

R

n

by

y2 ' ( t )

all

�,

eigenvalues

(

λ1 0

y '(t )= y (t )

2 1 Mass Spring System Undampened

Dampened where c is the damping constant, c>0

then

By

−1

A=PD P λ1 , λ2 , … with

of

A:

v1 , v2 , … 0 P=(v 1 v 2 ) and λ2

)

S=S λ ∪ S λ ∪ … ∪ S λ

Let

1

2

P= {u , u … ,u

1 2 n Population (Long Run) n −1 n 0 infinity Power of Matrices m 1

}

where

|S|

is

S= { u1 , u2 … , un }

λ ¿ m ¿ 0 λ2 0 ¿ ¿ ⋱ ¿ ¿ m m A =P ( ¿ λ n ¿ ) P−1

Ch4- System of Differential Equations ODE to SDE '' ' ODE:

a y + b y +c=0 y 1( t )= y and y 2( t )= y '

F

ODE:

−k y m

''

the

total

( )(

, then the

First Order Linear SDE '

y ( t )= A y (t )

find

n

where n tends to

)( ) y1 y2

y (t)

Constant Coeff : Entries of A are real numbers Variable Coeff: Entries of A are functions of t Superposition Principle

x (t)

If 1 linear SDE

and

x 2(t)

y ( t )= A y (t ) , c 1 x 1 (t ) +c 2 x 2 ( t ) scalars c 1 and c 2 '

λ

'

m y + c y +ky =0 0 1 y1 ' = −k −c y2 ' m m

*eigenvectors as the

k

, and

F2 =−cy ' '' m y =¿ F1 +F 2=−ky −cy '

Law,

'' 1 Harmonic Oscillation:

y '' =

diagonalizes A.

n

Hooke’s

F1=−ky

m y =¿

corresponding

number of basis vectors from all the eigenspaces

1. Let

the subject :

(diagonalizable)

x =P D P x = A x 0

- If 0 is an eigenvalue, A is singular Eigenvectors and Eigenspaces

1. Sub one eigenvalue into

the

and

for at least one � , then |S| < r1 + r2 + … + r ��

If |S|>n, A is diagonalizable - if

Eigenvalues

- If A is a triangular matrix,

of

square matrix

( 1a 10)

gives the

multiplicity

rows Let A be an nxn matrix, if A has n linearly independent eigenvectors/n distinct eigenvalues, A is diagonalizable.

1 a 0 1

( 10 −10 ) T (e j)

i

D=

Images of standard basis Multiplying

dim E λ...