11.18 Linear Maps

Linear Maps

Definition

Linear function

\(V\) and \(W\) are \(\mathbb{F}\)-vector spaces (not necessarily finite-dimensional).

A function \(T: V \to W\) is said to be linear if:

1. \(T(v+w)=T(v)+T(w),\,\forall v,w\in V\) (Additivity)

2. \(T(\lambda v) = \lambda T(v), \, \forall v \in V, \, \forall \lambda \in \mathbb{F}\) (Homogeneity)

Example

1. \(T: \mathbb{R}^2 \to \mathbb{R}, \, T(x, y) = xy.\)

   Not Linear

   \(T(v + w) = T\big((x, y) + (m, z)\big) = T\big((x + m, y + z)\big) = (x + m)(y + z)\\=xy+xz+my+mz\)

   \(T(v) + T(w) = T(x, y) + T(m, z) = xy + mz\)

   Clearly, \(T(v + w) \neq T(v) + T(w).\)

   \(T(\lambda(x,y))=T(\lambda x,\lambda y)=(\lambda x)(\lambda y)=\lambda^2xy\neq\lambda xy=\lambda T(x,y)\)

2. \(T: \mathbb{C}^2 \to \mathbb{C}^2, \, T(x, y) = (x + 1, y - 1);\)

   Not Linear

   \(T(\lambda x, \lambda y) = (\lambda x + 1, \lambda y - 1) \neq \lambda(x + 1, y - 1) = \lambda T(x, y).\)

   \(T(x + m, y + z) = (x + m + 1, y + z - 1) \neq (x + 1, y - 1) + (m + 1, z - 1) = T(x, y) + T(m, z).\)

3. \(T: \mathbb{Q}^2 \to \mathbb{Q}^2, \, T(x, y) = (2x - y, x + y);\)

   Linear

   \(T((x, y) + (m, z)) = T(x + m, y + z) = (2(x + m) - (y + z), (x + m) + (y + z)) = (2x - y + 2m - z, x + y + m + z).\)

   \(T(x, y) + T(m, z) = (2x - y, x + y) + (2m - z, m + z) = (2x - y + 2m - z, x + y + m + z).\)

   \(T(\lambda(x, y)) = T(\lambda x, \lambda y) = (2\lambda x - \lambda y, \lambda x + \lambda y) = \lambda(2x - y, x + y) = \lambda T(x, y).\)

Example

1. \(T: \mathbb{F}[x] \to \mathbb{F}[x], \, T: p(x) \mapsto p(x+1)\)

   \(T(p(x) + q(x)) = T((p+q)(x)) = (p+q)(x+1) = p(x+1) + q(x+1) = T(p(x)) + T(q(x)).\)

   \(T(\lambda p(x)) = T((\lambda p)(x)) = (\lambda p)(x+1) = \lambda p(x+1) = \lambda T(p(x)).\)

   Thus, \(T\) is linear.

2. \(S: \mathbb{F}[x] \to \mathbb{F}[x], \, S: p(x) \mapsto (x+1)p(x)\)

   \(p(x) + q(x) \mapsto (x+1)(p(x) + q(x)) = (x+1)p(x) + (x+1)q(x).\)

   \(\lambda p(x) \mapsto (x+1)(\lambda p(x)) = \lambda((x+1)p(x)).\)

   Thus, \(S\) is linear.

3. \(\text{tr}:M_{n\times n}\left(\mathbb{F})\to\mathbb{F},\,\text{tr}(A\right)=\sum_{i=1}^{n}a_{ii}\)

   \(\text{tr}(A + B) = \sum_{i=1}^n (A + B)_{ii} = \sum_{i=1}^n (A_{ii} + B_{ii}) = \sum_{i=1}^n A_{ii} + \sum_{i=1}^n B_{ii} = \text{tr}(A) + \text{tr}(B).\) ✔

   \(\text{tr}(\lambda A) = \sum_{i=1}^n (\lambda A)_{ii} = \sum_{i=1}^n \lambda \cdot A_{ii} = \lambda \sum_{i=1}^n A_{ii} = \lambda \cdot \text{tr}(A).\) ✔

4. \(t: M_{m \times n}(\mathbb{F}) \to M_{n \times m}(\mathbb{F}), \, \text{"the transpose"}\)

   \(A \mapsto A^t, \, (A^t)_{ij} = A_{ji}\) for \(1 \leq i \leq n, \, 1 \leq j \leq m.\)

   Example:

   \(\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \mapsto \begin{pmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{pmatrix}\)

Example

1. $T:\mathbb{F}^{n}\to\mathbb{F}^{m},\,T(x_1,x_2,\ldots,x_{n})=(a_1^1x_1+a_1^2x_2+\cdots+a_1^{n}x_{n},\,a_2^1x_1+\cdots+a_2^{n}x_{n},\,\ldots,\,a_{m}^1x_1+\cdots+a_{m}^{n}x_{n})\= \left(a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n, \, a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n, \, \ldots, \, a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n\right) $

   Thus we can use this to define the vector transformation (coordinate)

2. Given \(A \in M_{m \times n}(\mathbb{F})\), \(T_A: \mathbb{F}^n \to \mathbb{F}^m\) is defined by \(T_A(v) = A \cdot v \quad \big((A \cdot v^t)^t\big)\)

   \(v=\left(v_1,...,v_{n}\right)\)

   \(A\cdot v=\left(u_1,..,u_{m}\right)\)​

   \(T_A(v + w) = A \cdot (v + w) = A \cdot v + A \cdot w\)

   \(T_{A}(\lambda v)=A(\lambda v)=A(\lambda I\cdot v)=(\lambda I\cdot A)\cdot v=\lambda\cdot(Av)\)​

   Thus we can use this to define the vector transformation (matrix)

Some general properties:

1. \(T(0) = 0\)

   \([T(0)=T(0+0)=T(0)+T(0)\implies T(0)-T(0)=T(0)+T(0)-T(0)\implies0=T(0)]\)

   \(T(0)=T(0\cdot v)=0\cdot T(v)=0\)

   ​

2. \(T(-v) = -T(v)\)

   \([T(-v)=T((-1)\cdot v)=(-1)\cdot T(v)=-T(v)]\)

\(L(V,W)\) and \(L(V,V)\)

\(L(V,W)=\{T:V\to W,\text{ linear }\}=\text{Hom}(V,W)\quad(\text{homomorphism})\)

In particular, \(L(V,V)=\{T:V\to V,\text{ linear }\}=\text{End}(V)\quad(\text{endomorphism})\)

Definition

Given \(T,S\in L(V,W)\) and \(\lambda \in \mathbb{F}\), we define:

1. \(T+S\in L(V,W)\) by: \((T + S)(v) = T(v) + S(v)\)

2. \(\lambda T\in L(V,W)\) by: \((\lambda T)(v) = \lambda \cdot T(v)\)

Since the essence is function, thus these definition is just same with functions

Remark

1. In fact, \(T + S\) is linear, and also \(\lambda \cdot T\) is linear

   \(T + S\) is linear:

   \((T + S)(u + v) = T(u + v) + S(u + v) = T(u) + T(v) + S(u) + S(v) = (T(u) + S(u)) + (T(v) + S(v)) = (T + S)(u) + (T + S)(v)\)

   \((T + S)(\lambda u) = T(\lambda u) + S(\lambda u) = \lambda T(u) + \lambda S(u) = \lambda (T(u) + S(u)) = \lambda (T + S)(u)\)

   Thus, \((T + S)\) is linear.

2. Recall that \(\mathbb{R}^S\) is a vector space, the set of all functions from \(S\) to \(\mathbb{R}\) with \(+\) and \(\cdot\) as just defined.

Proposition

\(L(V, W)\) is an \(\mathbb{F}\)-vector space.

Proof (sketch):

1. \(W^V = \{ \text{Functions from } V \text{ to } W \}\) is an \(\mathbb{F}\)-vector space (as \(\mathbb{R}^S\) is).

   Proof

   1. Commutativity

      \(\left(f+g\right)\left(v\right)=f\left(v\right)+g\left(v\right)=g\left(v\right)+f\left(v\right)=\left(g+f\right)\left(v\right)\)​ 2. Associativity

      \(\left(\left(f+g\right)+h\right)\left(v\right)=\left(f+g\right)\left(v\right)+h\left(v\right)=f\left(v\right)+g\left(v\right)+h\left(v\right)=f\left(v\right)+\left(g+h\right)\left(v\right)=\left(f+\left(g+h\right)\right)\left(v\right)\) 3. Additive identity

      \(\left(f+0\right)(v)=f(v)\)​ 4. Additive inverse

      \(\left(f+\left(-f\right))(v)\right.=0\left(v\right)=0\)​ 5. Scalar identity

      \(\left(1\cdot f\right)\left(x\right)=f\left(x\right)\)​ 6. \(\left(\lambda\cdot\left(f+g\right)\right)\left(v\right)=\left(\lambda f+\lambda g\right)\left(v\right)=\lambda f\left(v\right)+\lambda g\left(v\right)\)​ 7. \(\left(\left(\lambda+\mu\right)\cdot f\right)\left(v\right)=\left(\lambda f+\mu f\right)\left(v\right)=\lambda f\left(v\right)+\mu f\left(v\right)\)​ 8. \(\left(\left(\lambda\cdot\mu\right)\cdot f\right)\left(v\right)=\lambda\cdot\mu\cdot f\left(v\right)\)​

   Thus \(W^V = \{ \text{Functions from } V \text{ to } W \}\) is an \(\mathbb{F}\)-vector space

2. \(L(V, W)\) is a subset of \(W^V\), where the \(+\) and \(\cdot\) in \(L(V, W)\) are those from \(W^V\).

   1. \(0(v)=0\in L(V,W)\)​
   2. \(T(v)+S(v)=(T+S)(v)\in L\left(V,W\right)\)​
   3. \(\lambda\cdot T(v)=(\lambda T)(v)\in L\left(V,W\right)\)​

   Thus \(L(V, W)\) is a subspace of \(W^V\)

3. Thus \(L(V, W)\) is an \(\mathbb{F}\)-vector space.

Proposition

The composition of linear maps is linear.

Proof: Let \(T: V \to W\) and \(S: W \to Z\) be linear maps, so \(S \circ T: V \to Z\).

(a) \((S \circ T)(u + v) = S(T(u + v)) = S(T(u) + T(v)) = S(T(u)) + S(T(v)) = (S \circ T)(u) + (S \circ T)(v)\)

(b) \((S \circ T)(\lambda u) = S(T(\lambda u)) = S(\lambda T(u)) = \lambda S(T(u)) = \lambda (S \circ T)(u)\)

Thus, \(S \circ T\) is linear.

Corollary

\(L(V, V)\) is an \(\mathbb{F}\)-algebra (with unit).

i.e., \(L(V, V)\) satisfies \(\mathbb{I} \cdot T = T = T \cdot \mathbb{I}\).

Given \(T, S \in L(V, V)\), let \(S \cdot T\) be the composition \(S \circ T\) and call it "product."

This product is associative (in general not commutative). Moreover, it satisfies:

\(T \cdot (S + R) = T \circ (S + R) = T \circ S + T \circ R = T S + T R\)

\((T + S) \cdot R = (T + S) \circ R = T \circ R + S \circ R = T R + S R\)

Null space and range

Definition

Let \(T: V \to W\) be linear. Then:

1. The null space of \(T\) is \(\text{null } T = \{ v \in V : T(v) = 0 \}\).

2. The range of \(T\) is \(\text{range } T = \{ w \in W : \exists v \in V \text{ such that } T(v) = w \}\).

The null space emphasize on the domain and the range space emphasize on the range of linear maps

Remark

Both are subspaces

We prove these two by the criterion for subspaces.

1. Let \(v, w \in \text{null } T\). We need to show that \(v + w \in \text{null } T\).

   Now \(T(v+w)=T(v)+T(w)=0+0=0\implies v+w\in\text{null }T\).

   Let \(\lambda\cdot v\in\text{null }T\) if \(v\in\text{null }T\).

   Now \(T(\lambda\cdot v)=\lambda\cdot T(v)=\lambda\cdot0=0\).

2. Let \(w_1, w_2 \in \text{range } T\). We have that \(\exists v_1, v_2\) such that \(T(v_1) = w_1\) and \(T(v_2) = w_2\).

   Now \(w_1 + w_2 = T(v_1) + T(v_2) = T(v_1 + v_2) \implies w_1 + w_2 \in \text{range } T\).

   Similarly, \(\lambda \cdot w_1 = \lambda \cdot T(v_1) = T(\lambda \cdot v_1) \implies \lambda \cdot w_1 \in \text{range } T\).

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