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Math 2370 – Fall 2008
Quiz #4
PICK AND SOLVE ONE OF THE FOLLOWING PROBLEMS:
Problem 1: Let θ be a real number. Show that the following two matrices are similar over
the field of complex numbers:
⎡cos θ
⎢ sin θ
⎣
− sin θ ⎤
,
cos θ ⎥⎦
⎡e iθ
⎢
⎣0
0 ⎤
⎥
e − iθ ⎦
Problem 2: Let T be a linear operator on IR 2 defined by T (a1 , a 2 ) = (− a 2 , a1 ) . Prove that
for every real number c the operator (T − cI ) is invertible (without the use of determinants or
eigenvalues).
Proof of 1: Label the matrices A and B. A and B are similar if there is an invertible matrix U
such that A = UBU −1 , i.e., AU = UB . This equation defines a system of linear equations for
the elements u ij of U:
u11 cos θ − u 21 sin θ = u11e iθ = u11 (cos θ + i sin θ )
u11 sin θ + u 21 cos θ = u 21e iθ = u 21 (cos θ + i sin θ )
u12 cos θ − u 22 sin θ = u12 e −iθ = u12 (cos θ − i sin θ )
u12 sin θ + u 22 cos θ = u 22 e −iθ = u 22 (cos θ − i sin θ )
which implies that − u 21 = iu11 and u 22 = iu12 . An example of an invertible matrix U that
obeys these relations is
i ⎤
⎡1
U =⎢
⎥
⎣− i − 1⎦
Proof of 2: It is sufficient to find the inverse of (T − cI ) or to show that the nullspace of
(T − cI ) is {0} for all c. The condition (T − cI )(a1 , a 2 ) = 0 implies that
− a 2 − ca1 = 0
a1 − ca 2 = 0
which has only the trivial solution for all c. Thus, the nullspace of (T − cI ) is {0} for all c
and (T − cI ) is invertible.