Page 107 - DMTH502_LINEAR_ALGEBRA

Unit 6: Computation Concerning Subspaces

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Thus  =  –  . Similarly we obtain    and  = 2 – 2 +  . Notes
1 1 2 2 2 3 1 2 3
Hence
 1 0 2 
 
P   1 1  2 

 0 0 1 
 
Now let us see how we would answer the questions by the second method which we described.
We form the 4 × 3 matrix B with column vectors  ,  ,  :
1 2 3
2
 1 0  
 2 2 0 
B =  
 2 0  4
 
 1 1 3 
We inquire for which y , y , y , y the system BX = Y has a solution.
1 2 3 4
1 2
1 0 0 y 1 y 2 y 4
1 0 2 y 1 1 0 2 y 1 1 0 2 y 1 3 3
1
2 2 0 y 0 2 4 y 2y 0 0 6 y 2y 0 0 1 (2y y )
2 2 1 2 4 4 2
6
2 0 4 y 3 0 0 0 y 3 2y 1 0 1 5 y 4 y 1 5 2
1 1 3 y 0 1 5 y y 0 0 0 y 2y 0 1 0 y 1 y 2 y 4
4 4 1 3 1 6 3
0 0 0 y 3 2y 1
Thus the condition that the system BX = Y have a solution is y = 2y . So  = (b , b , b , b ) is in W
3 1 1 2 3 4
if and only if b – 2b . If  is in W, then the coordinates (x , x , x ) in the ordered basis { ,  ,  }
3 1 1 2 3 1 2 3
can be read off from the last matrix above. We obtain once again the formulas (1) for those
coordinates
The questions (c) and (d) are now answered as before.

Example 2: We consider the 5 × 5 matrix

 1 2 0 3 0
 1 2  1  1 0 
 

A = 0 0 1 4 0
 
 2 4 1 10 1 
 0 0 0 0 1 
 
and the following problems concerning A
(a) Find an invertible matrix P such that PA is a row-reduced echelon matrix R.

(b) Find a basis for the new row space W of A.
(c) Say which vectors (b , b , b , b , b ) are in W.
1 2 3 4 5
(d) Find the coordinate matrix of each vector (b , b , b , b , b ) in W in the ordered basis chosen
1 2 3 4 5
in (b).
(e) Write each vector (b , b , b , b , b ) in W as a linear combination of the rows of A.
1 2 3 4 5
(f) Give an explicit description of the vector space V of all 5 × 1 column matrices X such that
AX = 0.

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