Page 352 - 35Linear Algebra

352 Sample Final Exam

(b) n.

(c) Yes.
(d) n × n.

(e) m × m.
T
(f) Yes. This relies on kerM = 0 because if M M had a non-trivial kernel,
T
then there would be a non-zero solution X to M MX = 0. But then
T
by multiplying on the left by X we see that ||MX|| = 0. This in turn
implies MX = 0 which contradicts the triviality of the kernel of M.
T
T
T
T T
(g) Yes because M M T = M (M ) = M M.
(h) Yes, all symmetric matrices have a basis of eigenvectors.
(i) No, because otherwise it would not be invertible.
(j) Since the kernel of L is non-trivial, M must have 0 as an eigenvalue.
(k) Since M has a zero eigenvalue in this case, its determinant must vanish.
I.e., det M = 0.

4. To begin with the system becomes

x
 
   
1 1 1 1 1
y
 
     
1 2 2 2   = 1
 z
1 2 3 3 1
w
Then
     
1 1 1 1 1 0 0 1 1 1 1
     
M = 1 2 2 2 = 1 1 0 0 1 1 1
1 2 3 3 1 0 1 0 1 2 2
   
1 0 0 1 1 1 1
= 1 1 0 0 1 1 1 = LU

 

1 1 1 0 0 1 1
 
a
So now MX = V becomes LW = V where W = UX =   (say). Thus
b
c
we solve LW = V by forward substitution

a = 1, a + b = 1, a + b + c = 1 ⇒ a = 1, b = 0, c = 0 .

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