## (i) If $T : V \to W$ 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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3
Feb 10, 2018

$\setminus$

$\text{(i) The statement is False.}$

$\text{(However, it can be salvaged. If" \ dim(V) = dim(W), \ "then it is True.)}$

$\text{(ii) The statement is True.}$

#### Explanation:

$\setminus$

$\text{(i) Proof, by counter-example, that (i) is False.}$

$\text{Let:" \qquad \qquad \qquad \qquad V = \ RR^{1} \quad "and} \setminus \quad W = \setminus {\mathbb{R}}^{2} .$

$\text{Let:} \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad T : V \rightarrow W , \setminus q \quad T : r \mapsto \left(r , 0\right) .$

$\setminus q \quad \text{a)" \quad \ T \ \ "is linear [note how important it is that the second}$
$\text{coordinate of elements in" \ W \ "is 0]:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus T \left(\setminus \alpha r + \setminus \beta s\right) \setminus = \setminus \left(\setminus \alpha r + \setminus \beta s , 0\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus \left(\setminus \alpha r , 0\right) \setminus + \setminus \left(\setminus \beta s , 0\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus \setminus \alpha \left(r , 0\right) \setminus + \setminus \setminus \beta \left(s , 0\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus \setminus \alpha T \left(r\right) \setminus + \setminus \setminus \beta T \left(s\right) .$

$\setminus q \quad \setminus q \quad \text{Thus:" \qquad \qquad \qquad \qquad \qquad \ \ T \ \ "is linear.}$

$\setminus q \quad \text{b)" \quad \ T \ \ "is one-one [injective]:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{Suppose:} \setminus q \quad \setminus q \quad \setminus \quad \setminus T \left(r\right) \setminus = \setminus T \left(s\right) .$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus \quad \left(r , 0\right) \setminus = \setminus \left(s , 0\right) .$

$\therefore \setminus \quad \text{equating first components:} \setminus q \quad \setminus q \quad r \setminus = \setminus s .$

$\setminus q \quad \setminus q \quad \text{Thus:" \qquad \qquad \qquad \quad \ \ T \ \ "is one-one [injective].}$

$\setminus q \quad \text{c)" \quad \ T \ \ "is not invertible [is not (one-one and onto)]:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{Let:} \setminus q \quad \setminus q \quad \setminus \quad \setminus w \setminus = \setminus \left(0 , 1\right) .$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{Clearly:} \setminus q \quad \setminus q \quad \setminus \quad \setminus w \setminus \in {\mathbb{R}}^{2} \setminus = \setminus W .$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{Suppose:} \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus \exists \setminus r \setminus \in {\mathbb{R}}^{1} : q \quad T \left(r\right) \setminus = \setminus w .$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus \quad \left(r , 0\right) \setminus = \setminus \left(0 , 1\right) .$

$\therefore \setminus \quad \text{equating second components:} \setminus q \quad \setminus q \quad 0 \setminus = \setminus 1.$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus \quad \Rightarrow \Leftarrow .$

$\setminus q \quad \setminus q \quad \text{Thus:" \qquad \qquad \qquad \quad \ \ T \ \ "is not onto [surjective].}$

$\setminus q \quad \setminus q \quad \text{Thus:" \qquad \qquad \qquad \quad \ \ T \ \ "is not invertible .}$

$\setminus q \quad \text{d)" \quad "Having established (a), (b), and (c), the proof of the}$
$\text{falsity of (i) is complete.} \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \square$

$\setminus$

$\text{(ii) Proof that (ii) is True.}$

$\text{This is pretty much by definition.}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{Let:" \qquad \qquad \quad \ U \quad "be a unitary matrix} .$

$\text{Then by definition:" \qquad \quad \ U U^H \ = \ I;}$
$\setminus q \quad \setminus q \quad \text{where" \ \ U^H \ "is the Hermitian Transpose of} \setminus \setminus U .$

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

$\setminus$

$\setminus$

$\text{Quick proof of offered salvage of (i)}$
$\setminus q \quad \setminus q \quad \text{[some small steps I leave as exercises !!!]:}$

$\text{Suppose:}$
$\setminus q \quad \setminus \setminus T : V \rightarrow W , \setminus \quad T \setminus \setminus \text{is one-one, and} \setminus \dim \left(V\right) = \dim \left(W\right) .$

$\text{We want to show:" \qquad T \ \ "is invertible.}$

$\text{Since" \ \ T \ \ "is one-one:} \setminus q \quad \setminus q \quad k e r \left(T\right) \setminus = \setminus \setminus \left\{0 \setminus\right\} .$

$\text{As" \ \ T \ \ "is linear:} \setminus q \quad \setminus \quad \frac{V}{k e r \left(T\right)} \setminus \approx \setminus I m \left(T\right) \setminus \subseteq \setminus W .$

$\setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \frac{V}{\setminus \left\{0 \setminus\right\}} \setminus \approx \setminus I m \left(T\right) \setminus \subseteq \setminus W .$

$\setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad V \setminus \approx \setminus I m \left(T\right) \setminus \subseteq \setminus W .$

$\setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \dim \left(V\right) \setminus = \setminus \dim \left(I m \left(T\right)\right) .$

$\text{As given:" dim(V) = dim(W), \ "we have:}$

$\setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \dim \left(W\right) = \setminus \dim \left(I m \left(T\right)\right) .$

$\therefore \setminus I m \left(T\right) \setminus \setminus \text{is a subspace of" \ \ W, "and has same dimension as} \setminus W .$

$\setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus I m \left(T\right) \setminus = \setminus \setminus W .$

$\text{Thus:" \qquad T \ \ "is onto [surjective].}$

$\text{By the given here:" \qquad T \ \ "is one-one [injective].}$

$\text{Thus we have:}$

$\setminus q \quad \setminus q \quad \setminus q \quad T \setminus \setminus \text{is one-one and onto [injective and surjective].}$

$\text{Hence:" \ \ T \ \ "is bijective, and so invertible.} \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \square$

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