Page 357 - 35Linear Algebra

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Finally, for λ = 3
2
 3 1   1 3   
− 1 0 1 − 0 1 0 −1 0
2 2 2 2
1 1 5 5




 −1 0  ∼  0 − 0  ∼  0 1 −1 0  .

 2 2 4 4
1 1 − 3 0 0 5 − 5 0 0 0 0 0
2 2 4 4
 
s
s
So we find the eigenvector   where s 6= 0 is arbitrary.
s
 
1
If the mistake X is in the direction of the eigenvector  −2 , then Y = 0.

1
I.e., the satellite returns to the origin O. For all subsequent orbits it will
again return to the origin. NASA would be very pleased in this case.
 
−1
If the mistake X is in the direction  0 , then Y = −X. Hence the

1
satellite will move to the point opposite to X. After next orbit will move
back to X. It will continue this wobbling motion indefinitely. Since this is a
stable situation, again, the elite engineers will pat themselves on the back.
 
1
1
Finally, if the mistake X is in the direction   , the satellite will move to a
1
3
point Y = X which is further away from the origin. The same will happen
2
for all subsequent orbits, with the satellite moving a factor 3/2 further away
from O each orbit (in reality, after several orbits, the approximations used
by the engineers in their calculations probably fail and a new computation
will be needed). In this case, the satellite will be lost in outer space and the
engineers will likely lose their jobs!
     
 1 0 0 
3
0
0
10. (a) A basis for B is      
,
,
1
0 0 1
 
(b) 3.
3
(c) 2 = 8.
3
(d) dimB = 3.
3
(e) Because the vectors {v 1 , v 2 , v 3 } are a basis any element v ∈ B can be
1
 
b
3
1
2
written uniquely as v = b v 1 +b v 2 +b v 3 for some triplet of bits  2  .
b
b 3
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