notes and homework

1 files changed, 183 insertions, 0 deletions

diff --git a/li/matlab_hw.py b/li/matlab_hw.py
new file mode 100644
index 0000000..71b89b6
--- /dev/null
+++ b/li/matlab_hw.py
@@ -0,0 +1,183 @@
+import numpy as np
+import scipy.linalg
+
+print("Section 1.3: #30")
+
+coeff = (
+    [[ 1, 1, 1 ],
+     [ 1, 2, 2 ],
+     [ 2, 3, -4]]
+    )
+print(
+np.linalg.solve(
+    coeff,
+    [6, 11, 3])
+)
+
+print(
+np.linalg.solve(
+    coeff,
+    [7, 10, 3])
+)
+
+print("Section 1.3: #32")
+
+dimension = 3
+tot = [0]*3
+ct = 10000
+for x in range(ct):
+    P, L, U = scipy.linalg.lu(np.random.rand(dimension, dimension))
+    for i in range(dimension):
+        tot[i] += U[i][i]
+print([x/ct for x in tot]);
+
+print("Section 1.4: #21")
+
+A = [[.5, .5], [.5, .5]]
+B = [[1, 0], [0, -1]]
+C = np.matmul(A, B)
+print("A^2")
+print(np.linalg.matrix_power(A, 2))
+print("A^3")
+print(np.linalg.matrix_power(A, 3))
+print("A^k = A");
+print("B^2")
+print(np.linalg.matrix_power(B, 2))
+print("B^3")
+print(np.linalg.matrix_power(B, 3))
+print("B^{2k} = B^2. B^{2k+1} = B.")
+print("C^2")
+print(np.linalg.matrix_power(C, 2))
+print("C^3")
+print(np.linalg.matrix_power(C, 3))
+print("C^k = 0")
+
+print("Section 1.4: #59")
+
+print("A * v = [3, 4, 5]")
+print("v' * v = 50")
+print(f"v * A = {np.matmul([3,4,5], np.identity(3))}")
+
+print("Section 1.4: #60")
+
+print("Av = [4. 4. 4. 4.]")
+print("Bw = [10. 10. 10. 10.]")
+
+print("Section 1.6: #12")
+
+print("Section 1.6: #32")
+
+print("the inverse of the first matrix is")
+print(np.linalg.inv(5*np.identity(4) - np.ones((4,4))))
+print("(meaning a = .4, b = .2)")
+print("ones(4)*ones(4) =")
+print(np.matmul(np.ones((4,4)),np.ones((4,4))))
+print("so we can solve this like the following equation: (a*eye +\n\
+b*ones)(c*eye + d*ones) = ac*eye + (ad+bc+dimension*bd)*ones = eye")
+print("c = 1/a. d = -b/a/(a+dimension*b), giving for the next matrix")
+print("a = 1/6, d = 6/(1/6+5*-1)") # WRONG
+
+print("Section 1.6: #47")
+
+print("I'm not sure. This is Python. It says (in either case):")
+print("numpy.linalg.LinAlgError: Singular matrix")
+'''
+print(np.linalg.solve(np.ones((4, 4)), np.random.rand(4, 1)))
+print(np.linalg.solve(np.ones((4, 4)), np.ones((4, 1))))
+'''
+
+''' #68
+dimension = 500
+northwest = np.random.rand(dimension, dimension)
+for x in range(0, dimension):
+    for y in range(x+1, dimension):
+        northwest[y][x] = 0
+print(northwest)
+'''
+print("Section 1.6: #69")
+
+
+from time import perf_counter
+matrix = np.random.rand(500, 500)
+start = perf_counter()
+np.linalg.inv(matrix)
+first_time = perf_counter() - start
+print("500x500 inverse:", str(first_time) + "s")
+matrix = np.random.rand(1000, 1000)
+start = perf_counter()
+np.linalg.inv(matrix)
+second_time = perf_counter() - start
+print("1000x1000 inverse:", str(second_time) + "s")
+print("factor:", second_time / first_time)
+
+print("Section 1.6: #70")
+
+dimension = 1000
+I = np.identity(dimension)
+A = np.random.rand(dimension, dimension)
+U = np.random.rand(dimension, dimension)
+for x in range(0, dimension):
+    for y in range(x+1, dimension):
+        U[y][x] = 0
+
+start = perf_counter()
+np.linalg.inv(U)
+print("inverse of U:", str(perf_counter() - start) + "s")
+start = perf_counter()
+np.linalg.solve(U, I)
+print("U\I:", str(perf_counter() - start) + "s")
+start = perf_counter()
+np.linalg.inv(A)
+print("inverse of A:", str(perf_counter() - start) + "s")
+start = perf_counter()
+np.linalg.solve(A, I)
+print("A\I:", str(perf_counter() - start) + "s")
+
+#print("Section 1.6: #71")
+
+print("Section 2.2: #33")
+
+print("For the first matrix:")
+A = [[1, 3, 3], [2, 6, 9], [-1, -3, 3]]
+b = [[1, 5, 5]]
+# TODO
+print(scipy.linalg.null_space(A))
+print("For the second matrix:")
+A = [[1, 3, 1, 2], [2, 6, 4, 8], [0, 0, 2, 4]]
+b = [[1, 3, 1]]
+
+print("Section 2.2: #35")
+
+print("For the first system:")
+A = [[1, 2], [2, 4], [2, 5], [3, 9]]
+print(sympy.Matrix.rref(A)) # ew is there no automated way to do this ?
+
+print("Section 2.2: #36")
+
+print("(a)")
+A = np.array([[1, 2, 1], [2, 6, 3], [0, 2, 5]])
+print("Basis of the column space:")
+print(scipy.linalg.orth(A))
+print("Multiplying the rows by this gives zero (for this matrix, \
+equivalent to [0 0 0]^T)")
+print(scipy.linalg.null_space(A.transpose()))
+
+print("(b)")
+A = np.array([[1, 1, 1], [1, 2, 4], [2, 4, 8]])
+print("Basis of the column space:")
+print(scipy.linalg.orth(A))
+print("Multiplying the rows by this gives zero (for this matrix, \
+equivalent to [0 2 -1]^T)")
+print(scipy.linalg.null_space(A.transpose()))
+
+print("Section 2.3: #2")
+
+print("Section 2.3: #5")
+
+print("Section 2.3: #13")
+
+print("Section 2.3: #16")
+
+print("Section 2.3: #18")
+
+print("Section 2.3: #24")