2023-11-15
This commit is contained in:
6
CS1029/notes-2023-11-01
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CS1029/notes-2023-11-01
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Edge contractions: merge two vertices, removing an edge between them
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Representations: adjacency list
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adjacency matrix
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What does the determinant of an adjacency matrix mean?
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incidence matrix: vertices against edges, 1 where edge is connected to vertex
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1
CS1029/notes-2023-11-07
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CS1029/notes-2023-11-07
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1
CS1029/notes-2023-11-08
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CS1029/notes-2023-11-08
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1
CS1029/notes-2023-11-14
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CS1029/notes-2023-11-14
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58
CS1029/practical-2023-11-03/relations
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CS1029/practical-2023-11-03/relations
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1. a) {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 5) }
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Reflexive: \forall x. (x, x) \in R
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Antisymmetric (& not symmetric): all pairs have x <= y. forall x, y. x <= y, !(y <= x) || y == x
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Transitive
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b) {(1, 2), (1, 4), (2, 1), (2, 3), (3, 2), (3, 4), (4, 1), (4, 3)}
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Not reflexive: there are no pairs of (x, x)
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Symmetric - all pairs have their reverse represented
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Not antisymmetric: symmetric and anti-symmetric are mutually exclusive
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Not transitive: (1, 2) and (2, 3) - 1 + 3 is not odd
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2. {(item, quantity)}
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{(Name, {(key, value)})}
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{(Name, Address, {(Room type, price, {(key, value)})}, ...)}
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3. 105 305 306 505 705 707 905 906 909
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4. a) ab ac bc cb
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b) {(a, a), (a, b), (a, c), (b, b), (b, a), (b, c), (c, a), (c, b), (d, d)}
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5.a) {(1, 1), (1, 2), (1, 4), (2, 1), (2, 3), (3, 2), (3, 3), (3, 4), (4, 1), (4, 3), (4, 4)}
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Not reflexive: (2, 2) is not present
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Symmetric, therefore not anti-symmetric
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Not transitive: (1, 2) and (2, 3), but not (1, 3)
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b) 12 21 14 41 32 23 43 34
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Not reflexive
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Symmetric, therefore not antisymmetric
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Not transitive: 12 and 23 but not 13
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6. 1 1 0 0
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1 1 0 0
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1 0 1 1
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0 0 0 1
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Yes, it's reflexive
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8. {(a, b) | a divides b OR b divides a}
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9. No, it's not transitive. (a, b) & (b, d), but not (a, d)
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10. a) Not equivalence relation: missing transitivity
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(1, 3) and (3, 2), but not (1, 2)
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b) {0}, {1, 2}, {3}
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11. \forall n \in N_0:
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0 + 3n
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1 + 3n
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2 + 3n
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12. a) Y
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b) N: 0 is in both - not disjoint
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c) Y
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d) N: 0 is missing
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13. a) 00 11 22 33 44 55 12 21 34 43 35 53 45 54
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b) 00 11 22 33 44 55 01 10 23 32 45 54
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c) 00 11 22 33 44 55 01 10 02 20 12 21 34 43 35 53 45 54
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14. a) Y, trivially
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b) N: not antisymmetric ((2, 3) and (3, 2))
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c) N: not reflexive (no (3, 3))
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64
CS1029/practical-2023-11-10/graphs
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CS1029/practical-2023-11-10/graphs
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1. Undirected, unlooped, multi-edged: multigraph
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b) directed, looped, multi-edged: directed pseudo-multigraph
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2. ac bd
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b) cd cd dd ee ab bc
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3. {{paper}}
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4. vertices: 6
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edges: 6
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degree: a: 2 b: 4 c: 1 f: 3 e: 2 d: 0
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isolated: d
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pendant: c
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b) vertices: 5
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edges: 14
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degree: a: 6 b: 6 c: 6 d: 5 e: 3
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isolated: -
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pendant: -
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5. vertices: 4
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in-a : 2
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out-a: 2
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in-b: 3
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out-b: 4
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in-c: 2
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out-c: 1
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in-d: 1
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out-d: 1
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6. {ac} {bde}
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b) Not bipartite: 3-loop bcf would require 3 sets
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7. {{ paper }}
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8. a -> abcd
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b -> d
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c -> ab
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d -> bcd
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9. | a b c d
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--+--------
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a | 1 1 1 1
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b | 0 0 0 1
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c | 1 1 0 0
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d | 0 1 1 1
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10. {{ paper }}
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11. v1 -> u1
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v2 -> u4
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v3 -> u2
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v4 -> u5
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v5 -> u3
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12. v1 -> u4
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v2 -> u3
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v3 -> u1
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v4 -> u2
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13. PSCL
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a) YNN4
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b) N---
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c) N---
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d) YYY5
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1
CS1032/notes-2023-11-02
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CS1032/notes-2023-11-02
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1
CS1032/notes-2023-11-03
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CS1032/notes-2023-11-03
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Pencil & eraser for final exam
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1
CS1032/notes-2023-11-09
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CS1032/notes-2023-11-09
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CS1032/practical-2023-11-01/head.py
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CS1032/practical-2023-11-01/head.py
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import sys, itertools
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def open_arg(_arg0, filename = "-"): # Default of - means stdin
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if filename == "-":
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return sys.stdin
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return open(filename, 'r')
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def head(file):
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for line in file.readlines()[:10]:
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print(line, end = "")
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if __name__ == '__main__':
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try:
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file = open_arg(*sys.argv)
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head(file)
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except FileNotFoundError:
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print("File not found!", file=sys.stderr)
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sys.exit(1)
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except:
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print("Some other error occured", file=sys.stderr)
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CS1032/practical-2023-11-01/tail.py
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CS1032/practical-2023-11-01/tail.py
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import sys, itertools
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def open_arg(_arg0, filename = "-"): # Default of - means stdin
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if filename == "-":
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return sys.stdin
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return open(filename, 'r')
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def tail(file):
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for line in file.readlines()[-10:]:
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print(line, end = "")
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if __name__ == '__main__':
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try:
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file = open_arg(*sys.argv)
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tail(file)
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except FileNotFoundError:
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print("File not found!", file=sys.stderr)
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sys.exit(1)
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except:
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print("Some other error occured", file=sys.stderr)
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70
CS1032/practical-2023-11-08/calc.py
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CS1032/practical-2023-11-08/calc.py
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# This function adds two numbers
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def add(x, y):
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return x + y
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# This function subtracts two numbers
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def subtract(x, y):
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return x - y
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# This function multiplies two numbers
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def multiply(x, y):
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return x * y
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# This function divides two numbers
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def divide(x, y):
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return x / y
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def avg(x, y):
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return (x + y) / 2
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def sci(x, y):
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return x * 10 ** y
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print("Select operation.")
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print("1.Add")
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print("2.Subtract")
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print("3.Multiply")
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print("4.Divide")
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print("5.Average")
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print("6.Scientific Notation")
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while True:
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# take input from the user
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choice = input("Enter choice(1/2/3/4/5/6): ")
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# check if choice is one of the four options
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if choice in ('1', '2', '3', '4', '5', '6'):
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try:
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num1 = float(input("Enter first number: "))
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num2 = float(input("Enter second number: "))
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except ValueError:
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print("Invalid input. Please enter a number.")
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continue
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if choice == '1':
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print(num1, "+", num2, "=", add(num1, num2))
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elif choice == '2':
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print(num1, "-", num2, "=", subtract(num1, num2))
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elif choice == '3':
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print(num1, "*", num2, "=", multiply(num1, num2))
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elif choice == '4':
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print(num1, "/", num2, "=", divide(num1, num2))
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elif choice == '5':
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print(f"avg({num1}, {num2})", "=", avg(num1, num2))
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elif choice == '6':
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print(f"{num1}e{num2}", "=", sci(num1, num2))
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# check if user wants another calculation
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# break the while loop if answer is no
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next_calculation = input("Let's do next calculation? (yes/no): ")
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if next_calculation.lower().startswith('n'):
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break
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else:
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print("Invalid Input")
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@@ -8,8 +8,9 @@
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\date{}
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\date{}
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\author{}
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\author{}
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\renewcommand{\Re}[1]{\operatorname{\mathbb{R}e}(#1)}
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\newcommand{\paren}[1]{\left(#1\right)}
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\renewcommand{\Im}[1]{\operatorname{\mathbb{{I}}m}(#1)}
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\renewcommand{\Re}[1]{\operatorname{\mathbb{R}e}\paren{#1}}
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\renewcommand{\Im}[1]{\operatorname{\mathbb{{I}}m}\paren{#1}}
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\newcommand{\C}{\mathbb{C}}
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\newcommand{\C}{\mathbb{C}}
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\newcommand{\N}{\mathbb{N}}
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\newcommand{\N}{\mathbb{N}}
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\newcommand{\Z}{\mathbb{Z}}
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\newcommand{\Z}{\mathbb{Z}}
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@@ -18,8 +19,8 @@
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\newcommand{\conj}[1]{\overline{#1}}
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\newcommand{\conj}[1]{\overline{#1}}
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\renewcommand{\mod}[1]{\left|#1\right|}
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\renewcommand{\mod}[1]{\left|#1\right|}
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\newcommand{\abs}[1]{\left|#1\right|}
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\newcommand{\abs}[1]{\left|#1\right|}
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\newcommand{\paren}[1]{\left(#1\right)}
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\newcommand{\polar}[2]{#1\paren{\cos{\paren{#2}} + i\sin{\paren{#2}}}}
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\newcommand{\polar}[2]{#1\paren{\cos{\paren{#2}} + i\sin{\paren{#2}}}}
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\newcommand{\adj}[1]{\operatorname{adj}#1}
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\makeatletter
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\makeatletter
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\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
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\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
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@@ -219,7 +219,7 @@ $\det{A} = \prod^{N}_{i=0}~a_{ij} \forall~\text{ row-echelon }A$
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\end{description}
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\end{description}
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Note: $\det{A} = \det{A^T}~\forall~A$
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Note: $\det{A} = \det{A^T}~\forall~A$
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\section*{Matrix Multiplication}
|
\section*{Matrix Multiplication}
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LHS has columns $=$ rows of RHS
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LHS has columns $=$ rows of RHS \\
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It's the cartesian product
|
It's the cartesian product
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\[A\times B = (a_{i1}b_{j1} + a_{i2}b_{2j} + \ldots + a_{im}b_{mj})_{ij}\]
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\[A\times B = (a_{i1}b_{j1} + a_{i2}b_{2j} + \ldots + a_{im}b_{mj})_{ij}\]
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\begin{align*}
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\begin{align*}
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@@ -240,4 +240,119 @@ It's the cartesian product
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9 & 10 & 11 & 12 \\
|
9 & 10 & 11 & 12 \\
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\end{pmatrix}
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\end{pmatrix}
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\end{align*}
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\end{align*}
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\[A\vec{x} = \vec{b}\]
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where $A$ is the coefficient matrix, $\vec{x}$ is the variables, and $\vec{b}$ is the values of the equations of a linear equation system.
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\subsection*{Inverse Matrices}
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The identity matrix exists as $I_n$ for size $n$.
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|
\[AA^{-1} = I_n = A^{-1}A \quad \forall~\text{matrices }A \text{ of size } n\]
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Assume that $A$ has two distinct inverses, $B$ and $C$.
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|
\begin{align*}
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& \text{matrix multiplication is associative} \\
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\therefore~ & C(AB) = (CA)B \\
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\therefore~ & C I_n = I_n B \\
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\therefore~ & C = B \\
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& \text{
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|
As $B = C$, while $B$ and $C$ are assumed to be distinct, matrices have no more than one unique inverse by contradiction
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|
}
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|
\end{align*}
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Matrices are invertible $\iff \det{A} \neq 0$
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\[\det{AB} = \det{A}\det{B}\]
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\[\therefore~ \det{A}\det{A^{-1}} = \det{I_n} = 1\]
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\[\therefore~ \det{A} \neq 0 \]
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|
\begin{align*}
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|
\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}
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|
\end{align*}
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|
\subsubsection*{Computation thereof}
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|
\[\det{A} = \sum_{k = 1}^{n}~a_{ik}(-1)^{i+j}\det{A_{ij}} \quad \text{ for any $i$}\]
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|
\begin{description}
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|
\item[Matrix of Cofactors: $C$] determinants of minors \& signs of laplace expansion \\
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|
ie. $\sum A \odot C = \det{A}$
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|
\item[$\adj{A}$ Adjucate of $A$ =] $C^T$
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|
\end{description}
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|
\begin{align*}
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|
A & = \begin{pmatrix}
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|
1 & 0 & 1 \\
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|
-1 & 1 & 2 \\
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|
2 & 0 & 1
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|
\end{pmatrix} \\
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|
C(A) & = \begin{pmatrix}
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|
1 & 5 & -2 \\
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|
0 & -1 & 0 \\
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-1 & -3 & 1 \\
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|
\end{pmatrix}
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|
\end{align*}
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|
$$ A^{-1} = \frac{\adj{A}}{\det{A}} $$
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|
Gaussian elimination can also be used: augmented matrix with $I_n$ on the right,
|
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|
reduce to reduced row-echelon. If the left is of the form $I_n$, the right is
|
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|
the inverse. If there is a zero row, $\det{A} = 0$, and the $A$ has no inverse.
|
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|
\section*{Linear Transformations}
|
||||||
|
\begin{align*}
|
||||||
|
f: & ~ \R^n \to \R^m \\
|
||||||
|
f & (x_1, \cdots, x_n) = (f_1(x_1, \cdots, x_n), f_2(x_1, \cdots, x_n), \cdots, f_m(x_1, \cdots, x_n))
|
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|
\end{align*}
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|
$f$ is a linear transformation if \(\forall i.~f_i(x_1, \cdots, x_n)\) is a
|
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|
linear polynomial in $x_1, \cdots, x_n$ with a zero constant term
|
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|
\begin{align*}
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|
f(x_1,~ x_2) & = (x_1 + x_2,~ 3x_1 - x_2,~ 10x_2) \tag{is a linear transformation} \\
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|
g(x_1,~ x_2,~ x_3) & = (x_1 x_2,~ x_3^2) \tag{not a linear transformation} \\
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h(x_1,~ x_2) & = (3x_1 + 4,~ 2x_2 - 4) \tag{not a linear transformation} \\
|
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|
\end{align*}
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\[f: \R^n \to \R^m = \vec{x} \to A\vec{x} \]
|
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|
\[\exists \text{ a matrix $A$ of dimension $n$x$m$ } \forall\text{ linear transforms } f \]
|
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|
\[\forall \text{ matrices $A$ of dimension $n$x$m$ } \exists \text{ a linear transform $f$ of dimension $n$x$m$ such that } f(\vec{x}) = A\vec{x} \]
|
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|
Function composition of linear translations is is just matrix multiplication:
|
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|
\begin{align*}
|
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|
f(\vec{x}) & = A\vec{x} \\
|
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|
g(\vec{y}) & = B\vec{y} \\
|
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|
(f\cdot g)(\vec{x}) & = g(f(\vec{x})) = BA\vec{x}
|
||||||
|
\end{align*}
|
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|
A function \(f: \R^n \to \R^m\) is a linear transformation iff:
|
||||||
|
\begin{enumerate}
|
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|
\item $f(\vec{x} + \vec{y}) = f(\vec{x}) + f(\vec{y}) \quad \forall~\vec{x},~\vec{y} \in \R^n $
|
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|
\item $f(r\vec{x}) = r\cdot f(\vec{x}) \quad \forall~\vec{x} \in \R^n, r \in \R $
|
||||||
|
\end{enumerate}
|
||||||
|
\subsection*{Building the matrix of a linear transform}
|
||||||
|
\[ f(\vec{x}) = f(x_1\vec{e}_1 + x_2\vec{e}_2) = f(x_1\vec{e}_1) + f(x_2\vec{e}_2) = x_1f(\vec{e}_1) + x_2f(\vec{e}_2) \]
|
||||||
|
\[ A = \begin{pmatrix} f(\vec{e}_1) & f(\vec{e}_2) \end{pmatrix} \]
|
||||||
|
\begin{align*}
|
||||||
|
& \vec{e}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}
|
||||||
|
\\ & \vec{e}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}
|
||||||
|
\\ & \vdots
|
||||||
|
\\ & \forall \vec{x}.~ \vec{x} = \sum_{i}^{n}~\vec{e}_i x_i
|
||||||
|
\end{align*}
|
||||||
|
\subsection*{Composition}
|
||||||
|
\[ \paren{f \cdot g}\paren{\vec{x}} = f(g(\vec{x})) = AB\vec{x} \]
|
||||||
|
where: $f(\vec{x}) = A\vec{x}$, $g(\vec{x}) = B\vec{x}$
|
||||||
|
\subsection*{Geometry}
|
||||||
|
\begin{description}
|
||||||
|
\item[rotation of $x$ by $\theta$ anticlockwise] \( = R_\theta = \begin{pmatrix} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta} \end{pmatrix} \)
|
||||||
|
\item[reflection about a line at angle $\alpha$ from the $x$-axis] \( = T_\alpha = R_{\alpha}T_0R_{-\alpha}\) where \( T_0 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \)
|
||||||
|
\item[scaling by $\lambda \in \R$] \( = S_\lambda = \lambda I_n\)
|
||||||
|
\item[Skew by $\alpha$ in $x$ and $\gamma$ in $y$] \( \begin{pmatrix} \alpha & 0 \\ 0 & \gamma \end{pmatrix}\)
|
||||||
|
\end{description}
|
||||||
|
The image of the unit square under the linear transform $A$ is a parallelogram of $(0, 0)$, $(a_{11}, a_{21})$, $(a_{12}, a_{22})$, $(a_{11} + a_{12}, a_{21} + a_{22})$, with area $ \abs{\det{A}} $
|
||||||
|
\subsection*{Inversion}
|
||||||
|
Inversion of a linear transformation is equivalent to inversion of its representative matrix
|
||||||
|
\subsection*{Eigen\{values, vectors\}}
|
||||||
|
\[ \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} a \\ 0 \end{pmatrix} = a\vec{e}_1\]
|
||||||
|
\[ \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ b \end{pmatrix} = b\vec{e}_2\]
|
||||||
|
\[ T_\alpha \vec{x} = \vec{x} \text{ for $\vec{x}$ along the line of transformation }\]
|
||||||
|
\begin{description}
|
||||||
|
\item[Eigenvector (of some transformation $f$)] A non-zero vector $\vec{x}$ such that $f(\vec{x}) = \lambda\vec{x}$ for some value $\lambda$
|
||||||
|
\item[Eigenvalue] $\lambda$ as above
|
||||||
|
\end{description}
|
||||||
|
\[ \forall \text{ eigenvectors of $A$ } \vec{x}, c \in R, \neq 0 .~ c\vec{x} \text{ is an eigenvector with eigenvalue } \lambda\]
|
||||||
|
|
||||||
|
\[ \forall A: \text{$n$x$n$ matrix}.\quad P_A\paren{\lambda} = \det{\paren{A - \lambda I_n}} \tag{characteristic polynomial in $\lambda$}\]
|
||||||
|
Eigenvalues of $A$ are the solutions of $P_A\paren{\lambda} = 0$
|
||||||
|
\begin{align*}
|
||||||
|
& A\vec{x} = \lambda\vec{x} & x \neq 0\\
|
||||||
|
\iff & A\vec{x} - \lambda\vec{x} = 0 \\
|
||||||
|
\iff & (A - \lambda I_n)\vec{x} = 0 \\
|
||||||
|
\iff & \det{\paren{A - \lambda I_n}} = 0 \\
|
||||||
|
& \quad \text{ or $\paren{A - \lambda I_n}$ is invertible and $x = 0$ }
|
||||||
|
\end{align*}
|
||||||
|
\[ P_{R\theta}(\lambda) = \frac{2\cos{\theta} \pm \sqrt{-4\lambda^2\sin^2{\theta}}}{2}\]
|
||||||
|
\[ R_\theta \text{ has eigenvalues }\iff \sin{\theta} = 0 \]
|
||||||
\end{document}
|
\end{document}
|
||||||
Reference in New Issue
Block a user