updating week 35

doc/src/week35/week35.do.txt

@@ -1322,7 +1322,7 @@ As an example, the above defective matrix can be decomposed as
 
 !bt
 \[
-\bm{X} = \frac{1}{\sqrt{2}}\begin{bmatrix}  1&  1 \\ 1& -1\\ \end{bmatrix} \begin{bmatrix}  4&  0 \\ 0& 0\\ \end{bmatrix}    \frac{1}{\sqrt{2}}\begin{bmatrix}  1&  -1 \\ 1& 1\\ \end{bmatrix}=\bm{U}\bm{\Sigma}\bm{V}^T, 
+\bm{X} = \frac{1}{\sqrt{2}}\begin{bmatrix}  1&  1 \\ 1& -1\\ \end{bmatrix} \begin{bmatrix}  2&  0 \\ 0& 0\\ \end{bmatrix}    \frac{1}{\sqrt{2}}\begin{bmatrix}  1&  -1 \\ 1& 1\\ \end{bmatrix}=\bm{U}\bm{\Sigma}\bm{V}^T, 
 \]
 !et
 
@@ -1378,7 +1378,7 @@ def SVDinv(A):
     SVD is numerically more stable than the inversion algorithms provided by
     numpy and scipy.linalg at the cost of being slower.
     '''
-    U, S, VT = np.linalg.svd(A)
+    U, S, VT = np.linalg.svd(A,full_matrices=True)
     print('test U')
     print( (np.transpose(U) @ U - U @np.transpose(U)))
     print('test VT')
@@ -1394,12 +1394,12 @@ def SVDinv(A):
 
 
 X = np.array([ [1.0,-1.0], [1.0,-1.0]])
+#X = np.array([[1, 2], [3, 4], [5, 6]])
+
 print(X)
-A = np.transpose(X) @ X
-print(A)
-C = SVDinv(A)
+C = SVDinv(X)
 # Print the difference between the original matrix and the SVD one
-print(C-A)
+print(C-X)
 !ec
 
 The matrix $\bm{X}$ has columns that are linearly dependent. The first