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linear combinations

Cu + Dv
for two vectors u and v, linear combinations are on a plane
if u and v are on the same line, the space of all the linear combinations is the line.


Dot products

v= (v1, v2) w=(w1,w2)
v.w = v1w1 + v2w2


Length of a vector

sqrt(v.v) = sqrt(v1^2 + v2^2 + ...)


Unit vector

length 1, u.u=1
we can divide any vector v by its length to get a unit vector in the same direction as v.


Angle between a vector

a.b/|a||b| = cos theta
a.b=0 when v and w are perpendicular to each other


Cauchy Schwasz
Triangle Inequality

|v.w| <= ||v|| ||w||

||x +y|| <= ||x|| + ||y||



Multiplier = coefficient/pivot = Lij, where I is the row of coefficient and j is the column of the pivot

If the second pivot is zero then there is no solutions
solve x,y and z for 3 simultaneous equations


Elimination matrix

Identity matrix with a non-zero entry -L in position ij


Matrix Pij for row exchange

Identity matrix with column swapped so the rows will exchange


Augmated matrix

[ A | b]


Rules for matrix operations

- Can add matrices of the same size
- Multiply A and B if A is mxn and B is nxr
- (AB)C = A(BC)
- A( B+C) = AB + AC
- AB is not equal to BA


Block matrix

Can cut bigger matrices in to smaller blocks


Inverse matrices( Properties)

A^(-1)A =I
Ax=b, x=A^(-1)b
(AB)^(-1) = B^-1A^-1

-The inverse exists if and only iff elimination produces n pivots for A nxn - full set
-The inverse is unique
-Ax=0, A does not have an inverse


Calculating inverse matrices by elimination

Gauss- Jordan solves all three systems using the augmented matrix
1. Eliminate both sides to upper triangle form - circle pivots
2. Eliminate back up to a diagonal matrix of the left hand side of the augmented matrix
3. divide each row by the pivots so the identity matrix is on the LHS and A^(-1) on the RHS
The product of the pivots = determinants


LDU decomposition

Elimination takes A to a upper triangle. (U for LU)
L is a lower triangle matrix, the identity matrix with the multiplies from elimination (l21, l31 ...) in there ij positions.
D is a diagonal matrix with the pivots on the diagonal
U - take the upper triangle matrix from elimination and divide each row by the pivots.

Using LU to solve Ax=b
sole Ly = b for y
solve Ux=y for x


Inverse matrix for A is 2x2

A = [a b]
[c d]
A^-1 = 1/(ad -bc) [d -b]
[ -c a]


Inverse of a diagonal matrix

A has d1,d2,d3 ...... on the diagonals and zeros elsewhere
A^-1 has 1/d1, 1/d2, 1/d3 on the diagonal and zeros elsewhere



Transfers the rows of a matrix A into a new matrix A^T
if A is a m x n matrix , A^T is a n x m matrix

(A+B)^T = A^T + B^T
(AB)^T = B^TA^T
(A^-1)^T = (A^T)^-1


Symmetic matrices

S^T =S
(S^-1) = S
If S^T =S then LDU of S is LDL^T


Permutation matrix

Has the columns of I in any order
P^-1 = P^T


Vector spaces
Real vector space
Subspace of a vector space

Vector space - The Space Rn consists of all the column vectors v with n components. E.g. R2 is the xy space, R3 is the xyz space and R1 is a line

Real vector space - Set of 'vectors' together with rules for vector addition and multiplication by real numbers

Subspace of a vector space - A set of vectors (including 0) such that if v and w are in the subspace
- v + w is in the subspace
- cv is in the subspace


The column space of A

Consists of all linear combinations of the columns. These are the possible vectors Ax, denote C(A)
C(A) is a subspace of Rm
C(a^T) is a subset of Rn


The Nullspace

N(A), consists of all the solutions to Ax=0. These solutions are vectors in Rn.
N(A) is a subspace of Rn
N(A^T) is a subset of Rm


Special solutions

Tp describe N(A) as the span of some vectors, we need two special solutions.
1. Set free variables as 1 or 0 for each.
2. Solve u [x1,x2,x3,x4] =0
s1 = , s2=
3. N(u) = CS1 + DS2
If A is m x n, with m>n, there is at least one free variable and so at least one free special solution
Free columns have no pivots


Reduced row echelon form

Continue to simplify a matrix after elimination to produce
1) Zeros above the pivots
2) 1's in the pivots
for m x n matrix

if N(A) = 0
The column are independent, no combinations give 0 except the zero combination.


The rank of A

Is the number of pivots. For rank r, A has r independent rows and r independent columns.
Rank is the dimension of C(A)
n - r is the dimension of N(A) for A is m x n


The complete solution to Ax=b

X = Particular solution + (n-r) special solutions

For particular solution use the augmented matrix and set both free variable to 0


Full column rank

Rank r = n
All columns have pivots
No free variables/ special solutions
N(A) = 0
If Ax=b, has a solution it is unique


Full row rank

rank r = m
All rows have pivots, rref has no 0 rows
Ax=b has a solution for every b
C(A) = Rm
n-r =n-m, special solutions in N(A)


Linear independence

The column of Matrix A are linearly independent when the only solution to Ax=0 is x=0

The sequence of vectors v1, v2, v3 ..... vn is linearly independent if the only combination x1v1 + x2v2 ....+xnvn =0 is when xi=0

The column are independent exactly when r=n. There are n pivots, no free variable N(A) = 0

If mm.