From 1982e0b0abe8b94cbc6c3701ba3e6e193616d117 Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Wed, 6 Oct 2021 15:18:32 -0400
Subject: interim changes to matlab hw

---
 li/matlab_hw.py | 67 +++++++++++++++++++++++++++++++++++++++++++++++++++------
 1 file changed, 61 insertions(+), 6 deletions(-)

diff --git a/li/matlab_hw.py b/li/matlab_hw.py
index 71b89b6..3473934 100644
--- a/li/matlab_hw.py
+++ b/li/matlab_hw.py
@@ -133,24 +133,47 @@ start = perf_counter()
 np.linalg.solve(A, I)
 print("A\I:", str(perf_counter() - start) + "s")
 
-#print("Section 1.6: #71")
+print("Section 1.6: #71")
+
+def proof(dim):
+    L = np.identity(dim) - np.diag([x/(1+x) for x in range(1,dim)], -1)
+    print(L)
+    Linv = np.linalg.inv(L)
+    print("L^{-1}:")
+    print(Linv)
+    right = True
+    for j in range(1,dim+1):
+        for i in range(1,j+1):
+            if not np.isclose(Linv[j-1][i-1], i/j):
+                print("the assertion in the book is wrong at",i,j)
+                right = False
+    if right:
+        print("The book is right")
+
+proof(4)
+proof(5)
 
 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
+b = [1, 5, 5]
+print("null space:")
 print(scipy.linalg.null_space(A))
+print("particular solution:")
+print(np.linalg.lstsq(A, b, rcond=None)[0])
 print("For the second matrix:")
 A = [[1, 3, 1, 2], [2, 6, 4, 8], [0, 0, 2, 4]]
-b = [[1, 3, 1]]
+b = [1, 3, 1]
+print("null space:")
+print(scipy.linalg.null_space(A))
+print("particular solution:")
+print(np.linalg.lstsq(A, b, rcond=None)[0])
 
 print("Section 2.2: #35")
-
+print("TODO")
 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")
 
@@ -172,12 +195,44 @@ print(scipy.linalg.null_space(A.transpose()))
 
 print("Section 2.3: #2")
 
+print("Largest possible number of independent vectors is the dimension \
+of the space spanned by these vectors, or the number of vectors in the \
+basis, which in this case is:")
+v = [[ 1,  1,  1,  0,  0,  0],
+     [-1,  0,  0,  1,  1,  0],
+     [ 0, -1,  0, -1,  0,  1],
+     [ 0,  0, -1,  0, -1, -1]]
+print(len(scipy.linalg.orth(v)))
+
 print("Section 2.3: #5")
 
+print("(a)")
+
+v = [[1, 2, 3],
+     [3, 1, 2],
+     [2, 3, 1]]
+if len(scipy.linalg.orth(v)) == 3: print("they are independent")
+else: print("they are not independent")
+
+print("(b)")
+
+v = [[ 1,  2, -3],
+     [-3,  1,  2],
+     [ 2, -3,  1]]
+if len(scipy.linalg.orth(v)) == 3: print("they are independent")
+else: print("they are not independent")
+
 print("Section 2.3: #13")
+# probably out of spirit of the matlab hw
+print("This one can be done easily without a calculator because the \
+        echelon matrix has already been computed.")
+print("The dimensions of all of these is 2, and the row spaces are the \
+        same.")
 
 print("Section 2.3: #16")
+print("TODO")
 
 print("Section 2.3: #18")
+print("TODO")
 
 print("Section 2.3: #24")
-- 
cgit