Inverse of A

Suppose A = (v    v    v ) (s           ) (u )       1    2    3    1   0    0     1                   ...      0     2   0     2                                s    u                     0    0     3    3 where {u_1, u_2, u_3} and {v_1, v_2, v_3} are orthonormal bases for ^3.

To solve A x = y is a three step process.
If A is invertible you can write x = A^(-1) y where A^(-1) does all three steps at once.
You can also write A^(-1) as a product M_3M_2M_1 where matrix M_i does step i.

Step 1:  Find v_i coords of y .  Since {v_1, v_2, v_3} is an orthonormal basis the v_i coords are c_i = y  v_i .      
Choose M_1 = (v )         1         v         2         v         3.  Then M_1y   = (v ) y = (v   y ) = (c )                       1        1           1      ...          2                      v        v   y      c                      3        3           3.

Step 2:  Divide the numbers (c )   1   c   2   c   3by the singular values.  
Choose M_2 = (1/s               )            1   0      0                1/s        0         2   0                       1/s        0      0         3.  
Then M_2(c ) = (1/s               ) (c ) = (c /s )      1        1   0      0       1      1  1     ...     2  2      c                    1/s    c      c /s      3     0      0         3    3      3  3.

Step 3:  Use the entries of (c /s )   1  1   c /s   2  2   c /s   3  3 as the u_i coords of x.
Choose M_3 = (u    u    u )         1    2    3.
Then M_3(c /s ) = (u    u    u ) (c /s ) = c_1/s_1u_1 + c_2/s_2u_2 + c_3/s_3u_3      1  1      1    ...   2                     2  2      c /s                     c /s      3  3                     3  3.

Combining all three steps, the solution to A x = y  is
    x = Underscript[Underscript[(u    u    u ) (1/s               ) (v ), ︸], A^(-1)] y     ...                       1/s    v                                             0      0         3    3.

If A is invertible, then the solution to A x = y is just x = A^(-1) y .  So it's a safe bet that the product (u    u    u ) (1/s               ) (v )   1    2    3      1   0      0       1               ...      2   0       2                                1/s    v                 0      0         3    3 is really just A^(-1).  

(Q43) Suppose A = (v    v    v ) (5   0   0) (u )       1    2    3                1                     0   ...               0   0   2    2                                  u                                  3.
Express A^(-1) as a product of three matrices.

Of course since step 2 divides by the singular values, this procedure only works if the none of the singular values is zero.    What happens if s_3 = 0???


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Copyright 2007 Todd Will
Created by Mathematica  (April 15, 2007)