3A. Vector Space of Linear Maps

Standing and new assumptions:

$F$ denotes either $C, R$ . Please note, sometimes we will just use $F, C, R$
$U, V, W$ all denote vector spaces over $F$

Definition of Linear map

A linear map or linear transform is a function between two vector spaces $T : V \to W$ that preserves addition and homogeneity:

Addition: $T (v + w) = T (v) + T (w)$ for $v, w \in V$
Homogeneity: $T (λ v) = λ T (v)$ for $λ \in F, v \in V$

Homogeneity is a lot like preserving scaler multiplication. For linear maps we could use either notation, usual function notation $T (v)$ or just $T v$

Through back from group theory:

Notation

We denote the set of linear maps from $V$ to $W$ as $L (V, W)$
The set of linear maps from $V$ to $V$ is denoted as $L (V)$

Examples:

Prove that the following are linear maps:

Zero
1. Zero is the function that is defined by taking every element to 0. I.E: $0 \in L (V, W) such that 0 v = v$
2. To prove this is a linear map lets prove it preserves addition and homogeneity:
  1. Addition
    1. Let $v_{1}, v_{2} \in V$ . Then: $0_{f} (v_{1} + v_{2}) = 0_{W} = 0_{W} + 0_{W} = 0_{f} (v_{1}) + 0_{f} (v_{2})$
    2. To help out I labeled which set 0 is in or if its the function. This will be assumed from now on.
  2. Homogeneity:
    1. Let $v \in V, λ \in F$ $λ (0 v) = λ 0_{T} = 0_{T} = 0 (λ v)$
Identity operator
1. By definition, the identity operator is a function $1 \in L (V)$ such that $1 v = v$
  1. Addition $1 (v_{1} + v_{2}) = v_{1} + v_{2} = 1 v_{1} + 1 v_{2}$
  2. Homogeneity $λ 1 v = λ v = 1 λ v$
Integration
1. Integration as a function in $T \in L (P (R), R)$ defined as $Tp = \int_{0}^{1} p$
2. is a function because when you add you just move the int to each part, and if u multiply by a scaler you can take it out of the integration.
1. For this last example, lets prove its a linear map.
  1. Let $(x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) \in R^{3}$
    1. Addition:
      1. $T ((x_{1}, y_{1}, z_{1}) + (x_{2}, y_{2}, z_{2})) = T (x_{1} + x_{2}, y_{1} + y_{2}, z_{1} + z_{2})$ $= (2 (x_{1} + x_{2}) - (y_{1} + y_{2}) + 3 (z_{1} + z_{2}), 7 (x_{1} + x_{2}) + 5 (y_{1} + y_{2}) - 6 (z_{1} + z_{2}))$
      2. From the other side: $T (x_{1}, y_{1}, z_{1}) + T (x_{2}, y_{2}, z_{2}) = (2 (x_{1}) - (y_{1}) + 3 (z_{1}), 7 (x_{1}) + 5 (y_{1}) - 6 (z_{1})) + = (2 (x_{2}) - (y_{2}) + 3 (z_{2}), 7 (x_{2}) + 5 (y_{2}) - 6 (z_{2}))$ $= (2 (x_{1} + x_{2}) - (y_{1} + y_{2}) + 3 (z_{1} + z_{2}), 7 (x_{1} + x_{2}) + 5 (y_{1} + y_{2}) - 6 (z_{1} + z_{2}))$
    2. Homogeneity:
      1. I will skip this proof for now.
1. This is a linear map as the following properties hold:
  1. Addition
    1. Let $p, q \in P (R)$ . Then notice:
      1. $(T (p + q)) (x) = x^{2} (p (x) + q (x)) = x^{2} p (x) + x^{2} q (x) = Tp (x) + Tq (x)$
  2. Homogenity:
    1. Let $λ \in R$ and $p \in P (R)$ .
    2. Then: $(T λ p) (x) = x^{2} λ p (x) = λ x^{2} p (x) = λ Tp (x)$
1. This is a linear map due to the following proof:
  1. Addition:
    1. Let $a, b \in F^{\infty}$ . Then: $a = (a_{1}, a_{2}, a_{3}, .....) and b = (b_{1}, b_{2}, b_{3}, ...)$
    2. Then $T (a + b) = T ((a_{1}, a_{2}, a_{3}, ....) + (b_{1}, b_{2}, b_{3}, .....))$ $= T (a_{1} + b_{1}, a_{2} + b_{2}, a_{3} + b_{3}, ....) = (a_{2} + b_{2}, a_{3} + b_{3}, ....)$ $= (a_{2}, a_{3}, .....) + (b_{2}, b_{3}, ...) = T a + T b$
  2. Homogeneity:
    1. Let $λ \in F$ and $(a_{1}, a_{2}, ...) \in F^{\infty}$
    2. Then, notice: $T (λ (a_{1}, a_{2}, ...)) = T ((λ a_{1}, λ a_{2}, ...)) = (λ a_{2}, λ a_{3}, ...) = λ (a_{2}, a_{3}, ....) = λ T (a_{1}, a_{2}, ....)$

Linear Map Lemma

Suppose $v_{1}, v_{2}, ...., v_{n}$ is a basis for $V$ and $w_{1}, w_{2}, ..., w_{n} \in W$ . Then there exists a unique map $T \in L (V, W)$ such that $T v_{i} = w_{i}$ The range here is $span (w_{1}, w_{2}, ..., w_{n}) \subset W$ . Proof:

Existence
1. Consider the map $T (a_{1} v_{1} + a_{2} v_{2} + .... + a_{n} v_{n}) = a_{1} w_{1} + a_{2} w_{2} + ... + a_{n} w_{n}$ for each $a_{i} \in F$
2. for each $v_{k}$ you can take the constant to 1, and the rest to $0$ , showing that this is actually a function from $v_{k}$ to $w_{k}$ . This makes sense because you can see the constants are shared on both sides
3. Proof that T is a linear map
  1. Addition:
    1. Let $a, b \in V$ , then: $T (a + b) = T ((a_{i} + v_{i}) v_{i}) = (a_{i} + v_{i}) w_{i}$ $T (a) + T (b) = a_{i} w_{i} + b_{i} w_{i} = (a_{i} + b_{i}) w_{i}$
  2. Homogenity:
    1. Let $λ \in F$ and $a \in F$ , Then: $T (λa) = T (λ a_{i} v_{i}) = λ a_{i} w_{i} = λ (a_{i} w_{i}) = λ T (a_{i} v_{i}) = λ T (a)$
Uniqueness:
1. Lets say there is two functions where $T_{1} v_{i} = w_{i}$ and $T_{2} v_{i} = w_{i}$ . Then through the properties of additivity and homogeneity these functions must be the same as they take every linear combination to the same linear combination in the range.

Operations on $L (V, W)$

We can define addition and scaler multiplication on $L (V, W)$ as such: Let $S, T \in L (V, W)$ and $λ \in F$ . Then: $(S + T) v = S v + T v (λ T) v = λ (T v)$ for $v \in V$

Notice the lambda in the last one is also the same as $T (λ v)$ through homogeneity.

Proof:

Let $S, T \in L (V, W), α \in F, v, u \in V$
$(S + T)$
1. Additivity
  1. $(S + T) (v + u) = S (v + u) + T (v + u) = S v + S u + T v + T u = S v + T v + S u + T u$ $= (S + T) v + (S + T) u$
2. Homogenity
  1. $(S + T) (αv) = S αv + T αv = α S v + α T v = (α (S + T)) v$ #todo-week if needed.

$L (V, W)$ is a vector space

With the operations defined above, $L (V, W)$ is a vector space.

Additive identity was defined in an example above.

Multiplication in $L (V, W)$

Usually we do not multiply vectors because it does not make sense. For this case, however, a useful product exists

Definition of multiplication of linear maps

For $S \in L (V, W), T \in L (U, V)$ we define the product function $ST \in L (V, W)$ as: $(ST) (v) = S (T v)$ for $v \in V$

todo fix the definition above to show the proper domains. S and T are not in the same set of linear maps, they map different subspaces.

This is similar to composition of functions, think of $(S \circ T) (v)$ . Works only when $T$ maps to the domain of $S$ which makes sense as it needs to be an input.

todo Verify that $ST$ is indeed a linear map in $L (V, W)$

Algebraic properties of linear map products

Associativity $(T_{1} T_{2}) T_{3} = T_{1} (T_{2} T_{3})$ then the domains make sense, ( $T_{3}$ maps to domain of $T_{2}$ which maps to domain of $T_{1}$ )
Identity: $I T = T I = T$ for the identity map described before. First is identity operator on V and second is identity operator on W to keep domains consistant.
Distributive: $(S_{1} + S_{2}) T = S_{1} T + S_{2} T and S (T_{1} + T_{2}) = S T_{1} + S T_{2}$ for $T, T_{1}, T_{2} \in L (U, V)$ and $S, S_{1}, S_{2} \in L (V, W)$ (domains make sense)

Note that Commutativity is missing

todo do a rutine proof to prove the above properties. Idk how much this is needed but could learn smth.

Linear maps / transforms take identity to identity

For and $T \in L (V, W)$ , $T$ always maps $0$ to $0$ $T (0) = 0$

Straight from Abstract Algebra Home

Proof:

Notice $0 = T (0) - T (0) = T (0 + 0) - T (0) = T (0) + T (0) - T (0) = T (0)$

The above result shows us that, for some function $f \in L (R, R)$ , defined as: $f (x) = m x + b$ Is only a linear map if $f (0) = 0$ . This is only the case if $m (0) + b = 0$ , or $b = 0$

Thus linear functions from high school are not the same as linear maps.

Notes

Explorer

3A. Vector Space of Linear Maps