If $V$ and $W$ are two different vector spaces, a \textit{linear map} $L$ from $V$ to $W$ is a function $L:V\to W$ such that
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Problem Ⅱ: Linear Transformations (You are currently browsing this post)
Proposition 2.a.1. Let $V, W, X$ be vector spaces over a field $\mathbb{R}$, and let $L: V \to W$ and $H: W \to X$ be linear maps. Then the composition $H \circ L: V \to X$ defined by $(H \circ L)(v) = H(L(v))$ for all $v \in V$ is linear.
Proof. By hypothesis, $L: V \to W$ is linear, hence for all $v_{1}, v_{2} \in V$ and all scalars $s \in \mathbb{R}$, the following identities hold:
\[L(v_{1} + v_{2}) = L(v_{1}) + L(v_{2}) \quad\text{and}\quad L(s \cdot v_{1}) = s \cdot L(v_{1}).\]Similarly, since $H: W \to X$ is linear, for all $w_{1}, w_{2} \in W$ and all $t \in \mathbb{R}$, one has:
\[H(w_{1} + w_{2}) = H(w_{1}) + H(w_{2}) \quad\text{and}\quad H(t \cdot w_{1}) = t \cdot H(w_{1}).\]To establish linearity of the composition $H \circ L: V \to X$, one must verify that for all $v_{1}, v_{2} \in V$ and all $r \in \mathbb{R}$, the following equalities hold:
\[(H \circ L)(v_{1} + v_{2}) = (H \circ L)(v_{1}) + (H \circ L)(v_{2}) \quad\text{and}\quad\] \[(H \circ L)(r \cdot v_{1}) = r \cdot (H \circ L)(v_{1}).\]Consider arbitrary vectors $v_{1}, v_{2} \in V$. Since $H \circ L$ is defined by composition, one has
\[(H \circ L)(v_{1} + v_{2}) = H(L(v_{1} + v_{2})).\]By linearity of $L$,
\[L(v_{1} + v_{2}) = L(v_{1}) + L(v_{2}),\]thus
\[(H \circ L)(v_{1} + v_{2}) = H(L(v_{1}) + L(v_{2})).\]Since $H$ is linear, it preserves addition, so
\[H(L(v_{1}) + L(v_{2})) = H(L(v_{1})) + H(L(v_{2})).\]Substitute back the definition of composition:
\[H(L(v_{1})) + H(L(v_{2})) = (H \circ L)(v_{1}) + (H \circ L)(v_{2}).\]Hence we have shown that for all $v_{1}, v_{2} \in V$,
\[(H \circ L)(v_{1} + v_{2}) = (H \circ L)(v_{1}) + (H \circ L)(v_{2}).\]Next, consider an arbitrary vector $v \in V$ and an arbitrary scalar $r \in \mathbb{R}$. By the definition of composition,
\[(H \circ L)(r \cdot v) = H(L(r \cdot v)).\]Since $L$ is linear, it respects scalar multiplication, yielding
\[L(r \cdot v) = r \cdot L(v).\]Therefore,
\[(H \circ L)(r \cdot v) = H(r \cdot L(v)).\]Because $H$ is linear, it also respects scalar multiplication, implying
\[H(r \cdot L(v)) = r \cdot H(L(v)).\]Rewriting in terms of composition, this becomes
\[r \cdot H(L(v)) = r \cdot (H \circ L)(v).\]Hence, for all $r \in \mathbb{R}$ and $v \in V$,
\[(H \circ L)(r \cdot v) = r \cdot (H \circ L)(v).\]Both of the defining properties of linear maps, namely the preservation of vector addition and scalar multiplication, have been verified for the map $H \circ L$. Consequently, $H \circ L$ is linear. $\textbf{Q.E.D.}$
Commentary
The essential elements of the proof rest on the axioms of linearity, which are stable under composition. Thus, the space of linear maps between vector spaces is closed under composition.
Proposition 2.b.1. Let $V$ and $W$ be vector spaces over the field $\mathbb{R}$, and let $L: V \to W$ be a linear map. Then $L(\vec{0}_V) = \vec{0}_W$, where $\vec{0}_V$ and $\vec{0}_W$ denote the additive identities in $V$ and $W$, respectively.
Proof. By the definition of a linear map, $L$ satisfies the following properties for all vectors $v, v’ \in V$ and all scalars $s \in \mathbb{R}$:
\[L(v + v') = L(v) + L(v') \quad \text{and} \quad L(s \cdot v) = s \cdot L(v).\]In particular, consider the vector $\vec{0}_V$, the additive identity in $V$. By the property of additive identities in vector spaces, for any vector $v \in V$, we have:
\[v + \vec{0}_V = v.\]Applying the linear map $L$ to both sides of this equation yields: \(L(v + \vec{0}_V) = L(v).\) Using the additivity of $L$, the left-hand side can be expressed as:
\[L(v + \vec{0}_V) = L(v) + L(\vec{0}_V).\]Thus, we have:
\[L(v) + L(\vec{0}_V) = L(v).\]To isolate $L(\vec{0}_V)$, subtract $L(v)$ from both sides:
\[L(v) + L(\vec{0}_V) - L(v) = L(v) - L(v).\]Simplifying both sides, we obtain:
\[L(\vec{0}_V) = \vec{0}_W,\]where $\vec{0}_W$ is the additive identity in $W$, since the only element in $W$ that satisfies $w + \vec{0}_W = w$ for all $w \in W$ is $\vec{0}_W$ itself.
Alternatively, one may employ the homogeneity property of linear maps. Consider the scalar multiplication by zero, which yields:
\[L(0 \cdot v) = 0 \cdot L(v).\]Since $0 \cdot v = \vec{0}_V$ for any $v \in V$, and $0 \cdot L(v) = \vec{0}_W$ in $W$, it follows that:
\[L(\vec{0}_V) = \vec{0}_W.\]$\textbf{Q.E.D.}$
Proposition 2.c.1. Let $V$ and $W$ be vector spaces over the field $\mathbb{R}$, and let $L: V \to W$ be a linear transformation. Then $L$ is injective (one-to-one) if and only if the kernel of $L$, denoted by $\ker(L)$, consists solely of the zero vector in $V$, that is, $\ker(L) = { \vec{0}_V }$.
Proof. We proceed by establishing both implications of the equivalence:
Necessity: Assume that $L: V \to W$ is injective. We aim to show that $\ker(L) = { \vec{0}_V }$.
By definition, the kernel of $L$ is the set:
\[\ker(L) = \{ v \in V \vert L(v) = \vec{0}_W \}.\]To demonstrate that $\ker(L) = { \vec{0}_V }$, we must show two inclusions:
\[\ker(L) \subseteq \{ \vec{0}_V \} \quad \text{and} \quad \{ \vec{0}_V \} \subseteq \ker(L).\]The second inclusion is trivial since $L(\vec{0}_V) = \vec{0}_W$ by the linearity of $L$ (as established in Proposition 2.b.1).
For the first inclusion, suppose $v \in \ker(L)$. Then, by definition:
\[L(v) = \vec{0}_W.\]Given that $L$ is injective, the only solution to $L(v) = \vec{0}_W$ is $v = \vec{0}_V$. To see this, consider the injectivity of $L$, which implies that if $L(v) = L(v’)$, then $v = v’$. Specifically, taking $v’ = \vec{0}_V$, we have:
\[L(v) = \vec{0}_W = L(\vec{0}_V) \implies v = \vec{0}_V.\]Thus, $v$ must be the zero vector in $V$, and therefore:
\[\ker(L) \subseteq \{ \vec{0}_V \}.\]Combining both inclusions, we conclude:
\[\ker(L) = \{ \vec{0}_V \}.\]This establishes the necessity part of the proposition.
Sufficiency: Now, assume that $\ker(L) = { \vec{0}_V }$. We aim to demonstrate that $L$ is injective.
To prove that $L$ is injective, we must show that for any $v_1, v_2 \in V$, if $L(v_1) = L(v_2)$, then $v_1 = v_2$.
Consider arbitrary vectors $v_1, v_2 \in V$ such that:
\[L(v_1) = L(v_2).\]Subtracting $L(v_2)$ from both sides yields:
\[L(v_1) - L(v_2) = \vec{0}_W.\]Utilizing the linearity of $L$, this can be rewritten as:
\[L(v_1 - v_2) = \vec{0}_W.\]By the definition of the kernel, this implies:
\[v_1 - v_2 \in \ker(L).\]Given that $\ker(L) = { \vec{0}_V }$, it follows that:
\[v_1 - v_2 = \vec{0}_V.\]Thus, we conclude:
\[v_1 = v_2.\]This establishes that $L$ is injective, as required. $\textbf{Q.E.D.}$
Proposition 2.d.1 Let $V$ and $W$ be vector spaces over the field $\mathbb{R}$, and let $L: V \to W$ be a linear transformation. Then the kernel of $L$, denoted by $\ker(L)$, defined as
\[\ker(L) = \{ v \in V \vert L(v) = \vec{0}_W \},\]is a subspace of $V$.
Proof. To establish that $\ker(L)$ is a subspace of $V$, we must verify that $\ker(L)$ satisfies the three axioms defining a subspace within a vector space. Specifically, we need to confirm that:
We proceed by verifying each of these properties in turn.
Non-emptiness of $\ker(L)$: By definition, a subspace must contain the zero vector of the ambient vector space. Consider the additive identity $\vec{0}_V \in V$. Applying the linear transformation $L$ to $\vec{0}_V$, we obtain:
\[L(\vec{0}_V) = \vec{0}_W,\]as established in Proposition 2.b.1. Therefore, $\vec{0}_V \in \ker(L)$, which implies that $\ker(L)$ is non-empty.
Closure under Vector Addition: Let $u, v \in \ker(L)$. By the definition of the kernel, this means:
\[L(u) = \vec{0}_W \quad \text{and} \quad L(v) = \vec{0}_W.\]We must show that $u + v \in \ker(L)$, i.e., $L(u + v) = \vec{0}_W$.
Applying the linearity of $L$, we have:
\[L(u + v) = L(u) + L(v).\]Substituting the known values from the kernel, this becomes:
\[L(u + v) = \vec{0}_W + \vec{0}_W = \vec{0}_W.\]Thus, $u + v \in \ker(L)$, establishing closure under vector addition.
Closure under Scalar Multiplication: Let $v \in \ker(L)$ and let $c \in \mathbb{R}$ be an arbitrary scalar. We must demonstrate that $c \cdot v \in \ker(L)$, i.e., $L(c \cdot v) = \vec{0}_W$.
Applying the linearity of $L$ with respect to scalar multiplication, we obtain:
\[L(c \cdot v) = c \cdot L(v).\]Since $v \in \ker(L)$, it follows that $L(v) = \vec{0}_W$. Substituting this into the equation above yields:
\[L(c \cdot v) = c \cdot \vec{0}_W = \vec{0}_W.\]Therefore, $c \cdot v \in \ker(L)$, establishing closure under scalar multiplication. $\textbf{Q.E.D.}$
Proposition 2.e.1. Let $V$ and $W$ be vector spaces over the field $\mathbb{R}$, and let $L: V \to W$ be a linear map. Then the image of $L$, defined by \(\text{Im}(L) = \{ L(v) \mid v \in V \},\) is a subspace of $W$.
Proof. To establish that $\text{Im}(L)$ is a subspace of $W$, it is necessary to verify that it satisfies the three axioms defining a subspace. Specifically, we must demonstrate that:
Containment of the Zero Vector: By the linearity of $L$, for any vector space $V$, the map $L$ satisfies
\[L(\vec{0}_V) = \vec{0}_W.\]Here, $\vec{0}_V$ denotes the additive identity in $V$, and $\vec{0}_W$ denotes the additive identity in $W$. Since $\vec{0}_V \in V$, it follows directly from the definition of the image that
\[\vec{0}_W = L(\vec{0}_V) \in \text{Im}(L).\]Thus, the zero vector of $W$ is contained within $\text{Im}(L)$.
Closure Under Vector Addition: Let $u, v \in \text{Im}(L)$. By the definition of the image, there exist vectors $u’, v’ \in V$ such that
\[u = L(u') \quad \text{and} \quad v = L(v').\]Consider the sum $u + v$ in $W$. Applying the linearity of $L$, we have
\[u + v = L(u') + L(v') = L(u' + v').\]Since $u’ + v’ \in V$ (as $V$ is a vector space and thus closed under addition), it follows that
\[u + v = L(u' + v') \in \text{Im}(L).\]Therefore, $\text{Im}(L)$ is closed under vector addition.
Closure Under Scalar Multiplication: Let $v \in \text{Im}(L)$ and let $c \in \mathbb{R}$ be an arbitrary scalar. By the definition of the image, there exists a vector $v’ \in V$ such that
\[v = L(v').\]Consider the scalar multiple $c \cdot v$ in $W$. Utilizing the linearity of $L$, we obtain
\[c \cdot v = c \cdot L(v') = L(c \cdot v').\]Since $c \cdot v’ \in V$ (as $V$ is a vector space and thus closed under scalar multiplication), it follows that
\[c \cdot v = L(c \cdot v') \in \text{Im}(L).\]Hence, $\text{Im}(L)$ is closed under scalar multiplication. $\textbf{Q.E.D.}$
Proposition 2.f.1. Let $V$ and $W$ be vector spaces over the field $\mathbb{R}$. Define the set $\mathcal{L}(V, W)$ to consist of all linear maps from $V$ to $W$. Equip $\mathcal{L}(V, W)$ with the operations of addition and scalar multiplication as defined below:
\[(F + G)(v) = F(v) + G(v) \quad \text{for all } F, G \in \mathcal{L}(V, W) \text{ and } v \in V,\] \[(c \cdot F)(v) = c \cdot F(v) \quad \text{for all } F \in \mathcal{L}(V, W), \, c \in \mathbb{R}, \text{ and } v \in V.\]Then $\mathcal{L}(V, W)$ endowed with these operations constitutes a vector space over $\mathbb{R}$.
Proof. To establish that $\mathcal{L}(V, W)$ constitutes a vector space under the defined operations of addition and scalar multiplication, it is requisite to verify that it adheres to all the fundamental axioms that characterize a vector space. Specifically, we will verify the following properties:
Closure under Addition: For all $F, G \in \mathcal{L}(V, W)$, the map $F + G$ defined by $(F + G)(v) = F(v) + G(v)$ for all $v \in V$ is also an element of $\mathcal{L}(V, W)$.
Closure under Scalar Multiplication: For all $F \in \mathcal{L}(V, W)$ and $c \in \mathbb{R}$, the map $c \cdot F$ defined by $(c \cdot F)(v) = c \cdot F(v)$ for all $v \in V$ is also an element of $\mathcal{L}(V, W)$.
Associativity of Addition: For all $F, G, H \in \mathcal{L}(V, W)$, $(F + G) + H = F + (G + H)$.
Commutativity of Addition: For all $F, G \in \mathcal{L}(V, W)$, $F + G = G + F$.
Existence of Additive Identity: There exists a linear map $\Theta \in \mathcal{L}(V, W)$ such that for all $F \in \mathcal{L}(V, W)$, $F + \Theta = F$.
Existence of Additive Inverses: For each $F \in \mathcal{L}(V, W)$, there exists a linear map $-F \in \mathcal{L}(V, W)$ such that $F + (-F) = \Theta$.
Distributivity of Scalar Multiplication with Respect to Vector Addition: For all $c \in \mathbb{R}$ and $F, G \in \mathcal{L}(V, W)$, $c \cdot (F + G) = c \cdot F + c \cdot G$.
Distributivity of Scalar Multiplication with Respect to Field Addition: For all $c, d \in \mathbb{R}$ and $F \in \mathcal{L}(V, W)$, $(c + d) \cdot F = c \cdot F + d \cdot F$.
Compatibility of Scalar Multiplication with Field Multiplication: For all $c, d \in \mathbb{R}$ and $F \in \mathcal{L}(V, W)$, $c \cdot (d \cdot F) = (c d) \cdot F$.
Identity Element of Scalar Multiplication: For the scalar $1 \in \mathbb{R}$ and for all $F \in \mathcal{L}(V, W)$, $1 \cdot F = F$.
Closure under Addition: Let $F, G \in \mathcal{L}(V, W)$. We must show that $F + G$ is a linear map from $V$ to $W$, i.e., $F + G \in \mathcal{L}(V, W)$.
Additivity: For all $v_1, v_2 \in V$,
\[(F + G)(v_1 + v_2) = F(v_1 + v_2) + G(v_1 + v_2).\]Since $F$ and $G$ are linear maps,
\[F(v_1 + v_2) = F(v_1) + F(v_2),\] \[G(v_1 + v_2) = G(v_1) + G(v_2).\]Thus,
\[(F + G)(v_1 + v_2) = [F(v_1) + F(v_2)] + [G(v_1) + G(v_2)]\] \[= [F(v_1) + G(v_1)] + [F(v_2) + G(v_2)] = (F + G)(v_1) + (F + G)(v_2).\]Homogeneity: For all $c \in \mathbb{R}$ and $v \in V$,
\[(F + G)(c \cdot v) = F(c \cdot v) + G(c \cdot v).\]Again, since $F$ and $G$ are linear maps,
\[F(c \cdot v) = c \cdot F(v),\] \[G(c \cdot v) = c \cdot G(v).\]Thus,
\[(F + G)(c \cdot v) = c \cdot F(v) + c \cdot G(v) = c \cdot [F(v) + G(v)] = c \cdot (F + G)(v).\]Therefore, $F + G$ preserves both vector addition and scalar multiplication, and hence $F + G$ is linear. Consequently, $F + G \in \mathcal{L}(V, W)$, establishing closure under addition.
Closure under Scalar Multiplication: Let $F \in \mathcal{L}(V, W)$ and $c \in \mathbb{R}$. We must show that $c \cdot F$ is a linear map from $V$ to $W$, i.e., $c \cdot F \in \mathcal{L}(V, W)$.
Additivity: For all $v_1, v_2 \in V$,
\[(c \cdot F)(v_1 + v_2) = c \cdot F(v_1 + v_2).\]Since $F$ is linear,
\[F(v_1 + v_2) = F(v_1) + F(v_2).\]Thus,
\[(c \cdot F)(v_1 + v_2) = c \cdot [F(v_1) + F(v_2)] = c \cdot F(v_1) + c \cdot F(v_2) = (c \cdot F)(v_1) + (c \cdot F)(v_2).\]Homogeneity: For all $d \in \mathbb{R}$ and $v \in V$,
\[(c \cdot F)(d \cdot v) = c \cdot F(d \cdot v).\]Since $F$ is linear,
\[F(d \cdot v) = d \cdot F(v).\]Thus,
\[(c \cdot F)(d \cdot v) = c \cdot [d \cdot F(v)] = (c d) \cdot F(v) = d \cdot (c \cdot F)(v).\]Therefore, $c \cdot F$ preserves both vector addition and scalar multiplication, and hence $c \cdot F$ is linear. Consequently, $c \cdot F \in \mathcal{L}(V, W)$, establishing closure under scalar multiplication.
Associativity of Addition: Let $F, G, H \in \mathcal{L}(V, W)$. We must show that $(F + G) + H = F + (G + H)$.
For all $v \in V$,
\[[(F + G) + H](v) = (F + G)(v) + H(v) = [F(v) + G(v)] + H(v) =\] \[F(v) + [G(v) + H(v)] = F(v) + (G + H)(v) = [F + (G + H)](v).\]Since this holds for all $v \in V$, we conclude that $(F + G) + H = F + (G + H)$.
Commutativity of Addition: Let $F, G \in \mathcal{L}(V, W)$. We must show that $F + G = G + F$.
For all $v \in V$,
\[(F + G)(v) = F(v) + G(v) = G(v) + F(v) = (G + F)(v).\]Since this holds for all $v \in V$, we conclude that $F + G = G + F$.
Existence of Additive Identity: We must exhibit an element $\Theta \in \mathcal{L}(V, W)$ such that for all $F \in \mathcal{L}(V, W)$, $F + \Theta = F$.
Define $\Theta: V \to W$ by $\Theta(v) = \vec{0}_W$ for all $v \in V$, where $\vec{0}_W$ is the zero vector in $W$.
Linearity of $\Theta$: For all $v_1, v_2 \in V$ and $c \in \mathbb{R}$,
\[\Theta(v_1 + v_2) = \vec{0}_W = \vec{0}_W + \vec{0}_W = \Theta(v_1) + \Theta(v_2),\] \[\Theta(c \cdot v) = \vec{0}_W = c \cdot \vec{0}_W = c \cdot \Theta(v).\]Thus, $\Theta$ is linear, and hence $\Theta \in \mathcal{L}(V, W)$.
Additive Identity Property: For all $F \in \mathcal{L}(V, W)$ and $v \in V$,
\[(F + \Theta)(v) = F(v) + \Theta(v) = F(v) + \vec{0}_W = F(v).\]Therefore, $F + \Theta = F$ for all $F \in \mathcal{L}(V, W)$, establishing the existence of an additive identity in $\mathcal{L}(V, W)$.
Existence of Additive Inverses: For each $F \in \mathcal{L}(V, W)$, we must exhibit a linear map $-F \in \mathcal{L}(V, W)$ such that $F + (-F) = \Theta$, where $\Theta$ is the additive identity in $\mathcal{L}(V, W)$.
Define $-F: V \to W$ by $(-F)(v) = -F(v)$ for all $v \in V$.
Linearity of $-F$: For all $v_1, v_2 \in V$ and $c \in \mathbb{R}$,
\[(-F)(v_1 + v_2) = -F(v_1 + v_2) = -[F(v_1) + F(v_2)] =\] \[-F(v_1) - F(v_2) = (-F)(v_1) + (-F)(v_2),\] \[(-F)(c \cdot v) = -F(c \cdot v) = -[c \cdot F(v)] = c \cdot (-F(v)) = c \cdot (-F)(v).\]Thus, $-F$ is linear, and hence $-F \in \mathcal{L}(V, W)$.
Additive Inverse Property: For all $F \in \mathcal{L}(V, W)$ and $v \in V$,
\[(F + (-F))(v) = F(v) + (-F)(v) = F(v) - F(v) = \vec{0}_W = \Theta(v).\]Therefore, $F + (-F) = \Theta$, establishing the existence of additive inverses in $\mathcal{L}(V, W)$.
Distributivity of Scalar Multiplication with Respect to Vector Addition: Let $c \in \mathbb{R}$ and $F, G \in \mathcal{L}(V, W)$. We must show that
\[c \cdot (F + G) = c \cdot F + c \cdot G.\]For all $v \in V$,
\[[c \cdot (F + G)](v) = c \cdot (F + G)(v) = c \cdot [F(v) + G(v)] =\] \[c \cdot F(v) + c \cdot G(v) = [c \cdot F](v) + [c \cdot G](v) = [c \cdot F + c \cdot G](v).\]Since this holds for all $v \in V$, we conclude that $c \cdot (F + G) = c \cdot F + c \cdot G$.
Distributivity of Scalar Multiplication with Respect to Field Addition: Let $c, d \in \mathbb{R}$ and $F \in \mathcal{L}(V, W)$. We must show that
\[(c + d) \cdot F = c \cdot F + d \cdot F.\]For all $v \in V$,
\[[(c + d) \cdot F](v) = (c + d) \cdot F(v) = c \cdot F(v) + d \cdot F(v) =\] \[[c \cdot F](v) + [d \cdot F](v) = [c \cdot F + d \cdot F](v).\]Since this holds for all $v \in V$, we conclude that $(c + d) \cdot F = c \cdot F + d \cdot F$.
Compatibility of Scalar Multiplication with Field Multiplication: Let $c, d \in \mathbb{R}$ and $F \in \mathcal{L}(V, W)$. We must show that
\[c \cdot (d \cdot F) = (c d) \cdot F.\]For all $v \in V$,
\[[c \cdot (d \cdot F)](v) = c \cdot [d \cdot F(v)] = (c d) \cdot F(v) = [(c d) \cdot F](v).\]Since this holds for all $v \in V$, we conclude that $c \cdot (d \cdot F) = (c d) \cdot F$.
Identity Element of Scalar Multiplication: Let $F \in \mathcal{L}(V, W)$. We must show that
\[1 \cdot F = F.\]For all $v \in V$,
\[[1 \cdot F](v) = 1 \cdot F(v) = F(v).\]Thus, $1 \cdot F = F$. $\textbf{Q.E.D.}$
Commentray
In infinite-dimensional settings, while some properties may differ, the vector space structure of $\mathcal{L}(V, W)$ remains valid as long as operations are well-defined.