General solutions

Here's the picture of xA x for the 2 × 2 matrix A = (v v ) (s ) (u ) 1 2 1 0 1 s u 0 2 2.

You see that A(u_1 + u_2) = 3v_1 + 0v_2 so the singular values are s_1 = 3 and s_2 = 0.

Solving A x = y when s_2 = 0.

First, you know that A x = y will have a solution only if y is a multiple of v_1.
In the graphic you can spot that y = -4.5v_1.  
Since s_1 = 3, you know that -4.5/3u_1 = -1.5u_1 is one solution to A x = y.
Second, you know that changing the amount of u_2 in x does not affect A x .  
So  x = -1.5u_1 + t u_2 is a solution for every value of t.

The general solution to A x = y is x = -1.5u_1 + t u_2.

Here's the picture of xA x for the 2 × 2 matrix A = (v v ) (s ) (u ) 1 2 1 0 1 s u 0 2 2.

(Q29) From the picture you can see that  s_1 = ___ and s_2 = ___ and y = (___v) _1.

(Q30) In terms of u_1and u_2, give the general solution to A x = y .
Answer: x = ___ u_1 + ___ u_2.

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Copyright © 2007 Todd Will
Created by Mathematica  (April 14, 2007)