Orthogonale Diagonalisierung

Gegeben: Matrix $A \in M( n \times n)$
Gesucht: Diagonalisierende Matrix $Q$ sodass $Q^T A Q = D$ wobei $D$ die Eigenwerte von A in der Hauptdiagonale enthält



In [9]:

    
# input: 
A = matrix([
    [1, -1, 2],
    [-1, 1, 2],
    [2, 2, -2]
])
#



In [10]:

    
V = []
EW = []
i = 0
for x in A.eigenvectors_left():
    for y in x[1]:
        EW.append(x[0])
        show(LatexExpr(r"\lambda_{} = ".format(i)), x[0])
        V.append(vector([val for val in y]))
        show(LatexExpr(r"v_{} = ".format(i)), y)
    i += 1









    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\lambda_0 = -4$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}v_0 = \left(1,\,1,\,-2\right)$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\lambda_1 = 2$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}v_1 = \left(1,\,0,\,\frac{1}{2}\right)$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\lambda_1 = 2$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}v_1 = \left(0,\,1,\,\frac{1}{2}\right)$



In [11]:

    
# optional input: EW, V if chosen in different order
#EW = []
V = [
    vector([-1, -1, 2]),
    vector([2,0,1]),
    vector([-1,1,0]),
]

Orthonormalisieren mit Gram-Schmidt Verfahren

$$ w_1 = \frac{1}{\| v_1 \|} v_1 $$$$ u_i = v_i - \sum_{k=1}^{i-1} \langle v_i, w_k \rangle w_k $$$$ w_i = \frac{1}{\| u_i \|} u_i $$



In [12]:

    
W = []
U = []
W.append((1/(V[0].norm())) * V[0])
show(LatexExpr("v_0 ="), V[0])
show(LatexExpr("w_0 ="), W[0])
print
for i, v in enumerate(V[1:]):
    show(LatexExpr("v_{} =".format(i+1)), v)
    s = sum([(v.dot_product(w) * w) for w in W])
    u = v - s
    show(LatexExpr("u_{} =".format(i+1)), u)
    U.append(u)
    w = (1/u.norm()) * u
    show(LatexExpr("w_{} =".format(i+1)), w)
    W.append(w)
    print









    




 $\newcommand{\Bold}[1]{\mathbf{#1}}v_0 = \left(-1,\,-1,\,2\right)$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}w_0 = \left(-\frac{1}{6} \, \sqrt{6},\,-\frac{1}{6} \, \sqrt{6},\,\frac{1}{3} \, \sqrt{6}\right)$ 






    









    




 $\newcommand{\Bold}[1]{\mathbf{#1}}v_1 = \left(2,\,0,\,1\right)$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}u_1 = \left(2,\,0,\,1\right)$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}w_1 = \left(\frac{2}{5} \, \sqrt{5},\,0,\,\frac{1}{5} \, \sqrt{5}\right)$ 






    









    




 $\newcommand{\Bold}[1]{\mathbf{#1}}v_2 = \left(-1,\,1,\,0\right)$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}u_2 = \left(-\frac{1}{5},\,1,\,\frac{2}{5}\right)$ 






    




 $\newcommand{\Bold}[1]{\mathbf{#1}}w_2 = \left(-\frac{1}{6} \, \sqrt{\frac{6}{5}},\,\frac{5}{6} \, \sqrt{\frac{6}{5}},\,\frac{1}{3} \, \sqrt{\frac{6}{5}}\right)$



In [13]:

    
Q = matrix(W).T
show(Q)









    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} -\frac{1}{6} \, \sqrt{6} & \frac{2}{5} \, \sqrt{5} & -\frac{1}{6} \, \sqrt{\frac{6}{5}} \\ -\frac{1}{6} \, \sqrt{6} & 0 & \frac{5}{6} \, \sqrt{\frac{6}{5}} \\ \frac{1}{3} \, \sqrt{6} & \frac{1}{5} \, \sqrt{5} & \frac{1}{3} \, \sqrt{\frac{6}{5}} \end{array}\right)$



In [14]:

    
D = Q.T * A *Q
show(D)









    




 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} -4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right)$



In [15]:

    
assert D.ncols() == D.nrows(), "not a n x n matrix!"
for i in range(D.ncols()):
    assert D[i,i] == EW[i], "D[{1},{1}] != EW[{1}]".format(i)