Page 269 - Matrix Analysis & Applied Linear Algebra

264 Chapter 4 Vector Spaces

Solution: X is invariant because Tq 1 = q 1 +3q 2 and Tq 2 =2q 1 +4q 2 insure
that for all α and β, the images
T(αq 1 + βq 2 )=(α +2β)q 1 +(3α +4β)q 2
lie in X. The desired matrix Q is constructed by extending {q 1 , q 2 } to a basis
4
B = {q 1 , q 2 , q 3 , q 4 } for . If the extension technique described in Solution 2
of Example 4.4.5 is used, then
1 0
   
and q 4 =   ,
 0   0 
0 0
q 3 =  
0 1
and
2 −110
 

 −1 200 
Q = q 1 q 2 q 3  .
0 −100
q 4 = 
0 001
Since the first two columns of Q span a space that is invariant under T, it
follows from (4.9.9) that Q −1 TQ must be in block-triangular form. This is easy
to verify by computing
 
0 −1 −2 0
  12 0 −6
Q −1  0 0 −1 0  and Q −1 TQ =  34 0 −14  .


1 2 3 0   00 −1 −3 
= 
0 0 0 1 00 4 14
In passing, notice that the upper-left-hand block is

12
* +
T = .
34
/X {q 1 ,q 2 }
Example 4.9.3
Consider again the matrices of Example 4.9.2:
−1 −1 −1 −1 2 −1
     
 0 −5 −16 −22   −1  and  2 
 ,
 ,
 .
0 3 10 14 q 1 =  0 q 2 =  −1
T = 
4 8 12 14 0 0
There are infinitely many extensions of {q 1 , q 2 } to a basis B = {q 1 , q 2 , q 3 , q 4 }
4
for —the extension used in Example 4.9.2 is only one possibility. Another
extension is
0 0
   
and  .
 −1   0 
2  q 4 =  −1
q 3 = 
−1 1