Third task

5 files changed, 253 insertions, 18 deletions

diff --git a/four.py b/four.py
index 4f7cd54..12074cf 100644
--- a/four.py
+++ b/four.py
@@ -23,8 +23,10 @@ lbdE = Matrix([
     [0, 0, 0, Poly([0, 1])]
 ])
 
-# uncomment to see characteristic poly
-#print((A - lbdE).det())
+print((A - lbdE).det())
+
+
+
 
 
 
@@ -45,11 +47,149 @@ A.make_D(0, -1)
 A.make_U(1, 2)
 A.make_D(1, Fraction(1, 4))
 
-gen_figure('fsr_A_T',
+gen_figure('V_1',
+f'''
+    {before}
+    {FOUR_SIMEQ}
+    {A.to_tex()}
+''')
+
+
+
+
+
+
+
+
+A = Matrix([
+    [-1, 1, 1, -1],
+    [-7, 4, 4, -3],
+    [4, -1, -2, 1],
+    [4, -1, -2, 1]
+])
+A = A@A
+
+before = A.to_tex()
+
+A.triangulate()
+
+
+A.make_D(0, -Fraction(1, 2))
+A.make_D(1, -6)
+
+gen_figure('V_0',
 f'''
     {before}
     {FOUR_SIMEQ}
     {A.to_tex()}
 ''')
 
-print(A)
\ No newline at end of file
+
+
+
+
+
+# solve 16-equation system
+# unknown matrix is (answer_to_system).reshape(4, 4),
+# so we 'flatten' it to solve for each unknown
+
+def make_row(vec, ans):
+    return [
+        [0, 0, 0, 0] * i +
+        vec +
+        [0, 0, 0, 0] * (3 - i) +
+        [ans[i]]
+        for i in range(4)
+    ]
+
+v_11 = [1, 3, 0, 0]
+v_12 = [1, 0, 3, 3]
+v_01 = [-1,-3, 1, 0]
+v_02 = [12, 10, 0, 3]
+e_3 = [0, 0, 1, 0]
+e_4 = [0, 0, 0, 1]
+
+rows = []
+rows += make_row(v_11, [0, 0, 0, 0])
+rows += make_row(v_12, [0, 0, 0, 0])
+rows += make_row(e_3, v_01)
+rows += make_row(e_4, v_02)
+
+sys = Matrix(rows)
+
+before = sys.to_tex()
+
+sys.triangulate(True)
+sys.make_D(1, -Fraction(1, 3))
+sys.make_D(5, -Fraction(1, 3))
+sys.make_D(9, -Fraction(1, 3))
+sys.make_D(13, -Fraction(1, 3))
+
+gen_figure('sys_orig',
+f'''
+    \\begingroup % keep the change local
+    \\setlength\\arraycolsep{{4pt}}
+    {before}
+    {FOUR_SIMEQ}
+    {sys.to_tex()}
+    \\endgroup
+''')
+
+ans = [row[-1] for row in sys.rows]
+reshape = [ans[4 * i:4 * i + 4] for i in range(4)]
+
+psi = Matrix(reshape)
+
+gen_figure('psi', 
+f'''
+\\psi' = {psi.to_tex()}
+''')
+
+
+
+
+
+
+inverse = Matrix([
+    [ 0,  Fraction(1, 3), 0, 0,   1, 0, 0, 0],
+    [ 1, -Fraction(1, 3), 0, 0,   0, 1, 0, 0],
+    [-3, -1, 1, 0,                0, 0, 1, 0],
+    [-3, -1, 0, 1,                0, 0, 0, 1]
+])
+
+before = inverse.to_tex()
+
+inverse.triangulate(True)
+inverse.make_D(1, 3)
+
+gen_figure('inverse',
+f'''
+    {before}
+    {FOUR_SIMEQ}
+    {inverse.to_tex()}
+''')
+
+
+
+
+C = Matrix([
+    [ 0,  Fraction(1, 3), 0, 0],
+    [ 1, -Fraction(1, 3), 0, 0],
+    [-3, -1, 1, 0],
+    [-3, -1, 0, 1]
+])
+
+inv = Matrix([
+    [1, 1, 0, 0],
+    [3, 0, 0, 0],
+    [6, 3, 1, 0],
+    [6, 3, 0, 1]
+])
+
+
+gen_figure('inverse_mul',
+f'''
+    \\psi = C^{{-1}}\\psi'C =
+    {((inv @ psi) @ C).to_tex()}
+'''
+)
diff --git a/gen_all.py b/gen_all.py
index 07a1c0f..3c42413 100644
--- a/gen_all.py
+++ b/gen_all.py
@@ -1,2 +1,3 @@
 import one
 import two
+import four
diff --git a/hw3.tex b/hw3.tex
index 2ec8de5..e064c25 100644
--- a/hw3.tex
+++ b/hw3.tex
@@ -117,18 +117,108 @@
         \rangle
     $
 
-%\dmquestion{4}
-%    Найдем хар. многочлен $\varphi$:
-%    \[
-%        \mathrm{det} (\varphi - \lambda E)
-%        \overset{\texttt{four.py}}=
-%        \lambda^4 - \lambda^3 + \lambda^2 =
-%        \lambda^2(\lambda - 1)^2
-%    \]
-%
-%    $
-%        V^1 = \ker ((\varphi - E)^2)
-%    $, так как у каждого собственного значения степень 2
-%    \input{figures/fig_four_fsr_A_T}
+\dmquestion{4}
+    Найдем хар. многочлен $\varphi$:
+    \[
+        \mathrm{det} (\varphi - \lambda E)
+        \overset{\texttt{four.py}}=
+        \lambda^4 - \lambda^3 + \lambda^2 =
+        \lambda^2(\lambda - 1)^2
+    \]
+
+    $
+        V^1 = \ker ((\varphi - E)^2)
+    $, так как у каждого собственного значения степень 2
+    \input{figures/fig_four_V_1}
+
+    ФСР:
+    \[
+        v_{11} = (1, 3, 0, 0), \quad v_{12} = (1, 0, 3, 3)
+    \]
+
+    $
+        V^0 = \ker(\varphi^2)
+    $
+    \input{figures/fig_four_V_0}
+
+    ФСР:
+    \[
+        v_{01} = (-1, -3, 1, 0), \quad v_{02} = (12, 10, 0, 3)
+    \]
+
+    Дополним $v_{11}$ и $v_{12}$ до базиса $\mathbb{R}^4$:
+
+    \[
+        \begin{bmatrix}
+            1 & 3 & 0 & 0\\
+            1 & 0 & 3 & 3
+        \end{bmatrix}
+        \simeq
+        \begin{bmatrix}
+            1 & 3 & 0 & 0\\
+            0 & -3 & 3 & 3
+        \end{bmatrix}
+        \simeq
+        \begin{bmatrix}
+            1 & 3 & 0 & 0\\
+            0 & 1 & -1 & -1
+        \end{bmatrix}
+    \]
+
+    Дополним стандартными векторами $e_3$ и $e_4$
+
+    Мы хотим, чтобы $\Image \psi = V^0, \quad \ker \psi = V^1$.
+    Для этого, отправим $v_{11}$ и $v_{12}$ в $0$, а $e_3$ и $e_4$ в $v_{01}$ и $v_{02}$. Тогда у нас будет однозначно определенный оператор, надо будет лишь перейти обратно к стандартному базису. Пусть $\psi'$ - матрица $\psi$ в базисе $(v_{11}, v_{12}, e_3, e_4)$.
+
+    \[
+        \begin{cases}
+            \psi'v_{11} = 0\\
+            \psi'v_{12} = 0\\
+            \psi'e_3 = v_{01}\\
+            \psi'e_4 = v_{02}\\
+        \end{cases}
+    \]
+
+    \input{figures/fig_four_sys_orig}
+
+    \input{figures/fig_four_psi}
+
+    Теперь надо перевести $\psi'$ в стандартный базис:
+    \[
+        \begin{cases}
+            e_1 = v_{12} - 3e_3 - 3e_4\\
+            e_2 = \frac{1}{3}v_{11} - \frac{1}{3}v_{12} - e_3 - e_4\\
+            e_3 = e_3\\
+            e_4 = e_4
+        \end{cases}
+    \]
+
+    Выпишем разложение в столбцы матрицы перехода:
+    \[
+        C = (v_{11}, v_{12}, e_3, e_4) \to e =
+        \begin{bmatrix}
+             0 &  \frac{1}{3} & 0 & 0\\ 
+             1 & -\frac{1}{3} & 0 & 0\\
+            -3 & -1           & 1 & 0\\
+            -3 & -1           & 0 & 1
+        \end{bmatrix}
+    \]
+
+    и найдем обратную:
+    \input{figures/fig_four_inverse}
+
+    \[
+        C^{-1} =
+        \begin{bmatrix}
+            1 & 1 & 0 & 0\\
+            3 & 0 & 0 & 0\\
+            6 & 3 & 1 & 0\\
+            6 & 3 & 0 & 1
+        \end{bmatrix}
+    \]
+
+    Тогда
+    \input{figures/fig_four_inverse_mul}
+
 
 \end{document}
diff --git a/intro.tex b/intro.tex
index f4f9d67..eab3965 100644
--- a/intro.tex
+++ b/intro.tex
@@ -53,6 +53,7 @@
   \mathrel{\vbox{\baselineskip.65ex\lineskiplimit0pt\hbox{.}\hbox{.}\hbox{.}}}%
 }
 
+\setcounter{MaxMatrixCols}{20}
 \newcommand{\question}[2]{
     \bigskip\bigskip
     \begin{cframed}
diff --git a/libsolve.py b/libsolve.py
index e9a0677..92f2b78 100644
--- a/libsolve.py
+++ b/libsolve.py
@@ -300,7 +300,7 @@ class Matrix:
             parts.append(' & '.join(map(str, row.lst)))
         return '\\begin{bmatrix}\n' + '\\\\\n'.join(parts) + '\n\\end{bmatrix}'
 
-    def triangulate(self):
+    def triangulate(self, swap=False):
         for i in range(self.shape[1]):
             for j in range(i, self.shape[0]):
                 assert(self[j][i].deg() == 0)
@@ -315,6 +315,9 @@ class Matrix:
                             continue
                         coef = -Fraction(self[k][i].vec[0]) / Fraction(self[j][i].vec[0])
                         self[k] = self[k] + self[j] * coef
+
+                    if swap:
+                        self.make_U(j, i)
                     break
 
 class Permutation: