Flashcards in Raphael Deck (34)
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1
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.
2
Dot products
v= (v1, v2) w=(w1,w2)
v.w = v1w1 + v2w2
3
Length of a vector
sqrt(v.v) = sqrt(v1^2 + v2^2 + ...)
4
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.
5
Angle between a vector
a.b/|a||b| = cos theta
a.b=0 when v and w are perpendicular to each other
6
Cauchy Schwasz
Triangle Inequality
|v.w| <= ||v|| ||w||
||x +y|| <= ||x|| + ||y||
7
Elimination
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
8
Elimination matrix
Identity matrix with a non-zero entry -L in position ij
9
Matrix Pij for row exchange
Identity matrix with column swapped so the rows will exchange
10
Augmated matrix
Ax=b
[ A | b]
11
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
12
Block matrix
Can cut bigger matrices in to smaller blocks
13
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
14
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
15
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.
LDU = A
Using LU to solve Ax=b
sole Ly = b for y
solve Ux=y for x
16
Inverse matrix for A is 2x2
A = [a b]
[c d]
A^-1 = 1/(ad -bc) [d -b]
[ -c a]
17
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
18
Transposes
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
19
Symmetic matrices
S^T =S
(S^-1) = S
If S^T =S then LDU of S is LDL^T
20
Permutation matrix
Has the columns of I in any order
P^-1 = P^T
21
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
22
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
23
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
24
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
25
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
N(A)=N(U)=N(R)
if N(A) = 0
The column are independent, no combinations give 0 except the zero combination.
26
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
27
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
28
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
29
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)
30
