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Which of the following statements are true/false? Justify your answer

(i) If #T:V->W# is a one-one linear transformation between two finite dimensional vector spaces #V# and #W# then #T# is invertible.

(ii) Every unitary operator is invertible.

(i) If #T:V->W# is a one-one linear transformation between two finite dimensional vector spaces #V# and #W# then #T# is invertible.

(ii) Every unitary operator is invertible.

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Feb 10, 2018

Answer:

# \ #

# "(i) The statement is False." #

# "(However, it can be salvaged. If" \ dim(V) = dim(W), \ "then it is True.)"#

# "(ii) The statement is True." #

Explanation:

# \ #

# "(i) Proof, by counter-example, that (i) is False." #

# "Let:" \qquad \qquad \qquad \qquad V = \ RR^{1} \quad "and" \quad W = \ RR^{2}. #

# "Let:" \qquad \qquad \qquad \qquad T: V rarr W, \qquad T: r |-> (r,0). #

# \qquad "a)" \quad \ T \ \ "is linear [note how important it is that the second" #
# "coordinate of elements in" \ W \ "is 0]:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \quad \ T( \alpha r + \beta s ) \ = \ ( \alpha r + \beta s, 0 ) #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ ( \alpha r, 0 ) \ + \( \beta s, 0 )#

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ \alpha (r, 0) \ + \ \beta (s, 0)#

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ \alpha T(r) \ + \ \beta T(s). #

# \qquad \qquad "Thus:" \qquad \qquad \qquad \qquad \qquad \ \ T \ \ "is linear." #

# \qquad "b)" \quad \ T \ \ "is one-one [injective]:" #

# \qquad \qquad \qquad \quad "Suppose:" \qquad \qquad \quad \ T( r ) \ = \ T( s ). #

# \qquad \qquad \qquad \qquad \qquad \qquad :. \qquad \qquad \quad (r, 0) \ = \ (s, 0). #

# :. \quad "equating first components:" \qquad \qquad r \ = \ s. #

# \qquad \qquad "Thus:" \qquad \qquad \qquad \quad \ \ T \ \ "is one-one [injective]." #

# \qquad "c)" \quad \ T \ \ "is not invertible [is not (one-one and onto)]:" #

# \qquad \qquad \qquad \quad "Let:" \qquad \qquad \quad \ w \ = \ (0, 1). #

# \qquad \qquad \qquad \quad "Clearly:" \qquad \qquad \quad \ w \in RR^2 \ = \ W. #

# \qquad \qquad \qquad \quad "Suppose:" \qquad \qquad \quad \ \exists \ r \in RR^1: qquad T( r ) \ = \ w. #

# \qquad \qquad \qquad \qquad \qquad \qquad :. \qquad \qquad \quad (r, 0) \ = \ (0, 1). #

# :. \quad "equating second components:" \qquad \qquad 0 \ = \ 1. #

# \qquad \qquad \qquad \qquad \qquad \qquad :. \qquad \qquad \quad rArr lArr. #

# \qquad \qquad "Thus:" \qquad \qquad \qquad \quad \ \ T \ \ "is not onto [surjective]." #

# \qquad \qquad "Thus:" \qquad \qquad \qquad \quad \ \ T \ \ "is not invertible ." #

# \qquad "d)" \quad "Having established (a), (b), and (c), the proof of the" #
# "falsity of (i) is complete." \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad square #

# \ #

# "(ii) Proof that (ii) is True." #

# "This is pretty much by definition." #

# \qquad \qquad \qquad \quad "Let:" \qquad \qquad \quad \ U \quad "be a unitary matrix". #

# "Then by definition:" \qquad \quad \ U U^H \ = \ I;" #
# \qquad \qquad "where" \ \ U^H \ "is the Hermitian Transpose of" \ \ U. #

# \qquad \qquad "Thus:" \qquad\quad \ \ T \ \ "is invertible. (Moreover," \ U^{-1} \ = \ U^H. ")" \quad \ square #

# \ #

# \ #

# "Quick proof of offered salvage of (i)" #
# \qquad \qquad "[some small steps I leave as exercises !!!]:" #

# "Suppose:" #
# \qquad \ \ T: V rarr W, \quad T \ \ "is one-one, and" \ dim(V) = dim(W). #

# "We want to show:" \qquad T \ \ "is invertible." #

# "Since" \ \ T \ \ "is one-one:" \qquad \qquad ker(T) \ = \ \{ 0 \}. #

# "As" \ \ T \ \ "is linear:" \qquad \quad V / { ker(T) } \ ~~\ Im(T) \ sube \ W. #

# \qquad :. \qquad \qquad \qquad \qquad \qquad \qquad V / { \{ 0 \} } \ ~~\ Im(T) \ sube \ W. #

# \qquad :. \qquad \qquad \qquad \qquad \qquad \qquad \quad V \ ~~\ Im(T) \ sube \ W. #

# \qquad :. \qquad \qquad \qquad \qquad \quad \ dim(V) \ = \ dim(Im(T)). #

# "As given:" dim(V) = dim(W), \ "we have:" #

# \qquad :. \qquad \qquad \qquad \qquad \quad \ dim(W) = \ dim(Im(T)). #

# :. \ Im(T) \ \ "is a subspace of" \ \ W, "and has same dimension as" \ W. #

# \qquad :. \qquad \qquad \qquad \qquad \qquad \quad \ Im(T) \ = \ \ W. #

# "Thus:" \qquad T \ \ "is onto [surjective]." #

# "By the given here:" \qquad T \ \ "is one-one [injective]." #

# "Thus we have:" #

# \qquad \qquad \qquad T \ \ "is one-one and onto [injective and surjective]." #

# "Hence:" \ \ T \ \ "is bijective, and so invertible." \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad square #

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