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)