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Wyatt's Notes — University

Linear Transformations

6.1 Definition

A linear transformation (or linear map) T:V→WT : V \to W between vector spaces VV and WW over FF Is a function satisfying:

  1. T(u+v)=T(u)+T(v)T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) for all u,v∈V\mathbf{u}, \mathbf{v} \in V
  2. T(αv)=αT(v)T(\alpha \mathbf{v}) = \alpha T(\mathbf{v}) for all α∈F\alpha \in F, v∈V\mathbf{v} \in V

Equivalently, T(αu+βv)=αT(u)+βT(v)T(\alpha\mathbf{u} + \beta\mathbf{v}) = \alpha T(\mathbf{u}) + \beta T(\mathbf{v}) for all α,β∈F\alpha, \beta \in F and u,v∈V\mathbf{u}, \mathbf{v} \in V.

The set of all linear transformations from VV to WW is denoted L(V,W)\mathcal{L}(V, W).

Proposition 6.1. For any linear transformation TT:

  1. T(0)=0T(\mathbf{0}) = \mathbf{0}.
  2. T(−v)=−T(v)T(-\mathbf{v}) = -T(\mathbf{v}).
  3. T(∑i=1kαivi)=∑i=1kαiT(vi)T\left(\sum_{i=1}^k \alpha_i \mathbf{v}_i\right) = \sum_{i=1}^k \alpha_i T(\mathbf{v}_i).

Proof. T(0)=T(0⋅0)=0⋅T(0)=0T(\mathbf{0}) = T(0 \cdot \mathbf{0}) = 0 \cdot T(\mathbf{0}) = \mathbf{0}. T(−v)=T((−1)v)=(−1)T(v)=−T(v)T(-\mathbf{v}) = T((-1)\mathbf{v}) = (-1)T(\mathbf{v}) = -T(\mathbf{v}). Property (3) follows by Induction. ■\blacksquare

2D Rotation of the Unit Circle

6.2 Matrix Representation

If VV and WW are finite-dimensional with bases BV={v1,…,vn}\mathcal{B}_V = \{\mathbf{v}_1, \ldots, \mathbf{v}_n\} and BW={w1,…,wm}\mathcal{B}_W = \{\mathbf{w}_1, \ldots, \mathbf{w}_m\}Then every T∈L(V,W)T \in \mathcal{L}(V, W) is Uniquely represented by a matrix [T]BVBW∈Mm×n(F)[T]_{\mathcal{B}_V}^{\mathcal{B}_W} \in \mathcal{M}_{m \times n}(F) Where the jj-th column is the coordinate vector of T(vj)T(\mathbf{v}_j) with respect to BW\mathcal{B}_W.

6.3 Kernel and Image

The kernel (null space) and image (range) of TT are:

ker⁡(T)={v∈V:T(v)=0}\ker(T) = \{\mathbf{v} \in V : T(\mathbf{v}) = \mathbf{0}\} im(T)={T(v):v∈V}\mathrm{im}(T) = \{T(\mathbf{v}) : \mathbf{v} \in V\}

Proposition 6.2. ker⁡(T)\ker(T) is a subspace of VV and im(T)\mathrm{im}(T) is a subspace of WW.

6.4 Rank-Nullity Theorem for Linear Maps

Theorem 6.3 (Rank-Nullity). For T∈L(V,W)T \in \mathcal{L}(V, W) with VV finite-dimensional:

dim⁡(ker⁡(T))+dim⁡(im(T))=dim⁡(V)\dim(\ker(T)) + \dim(\mathrm{im}(T)) = \dim(V)

Proof. Let {u1,…,uk}\{\mathbf{u}_1, \ldots, \mathbf{u}_k\} be a basis for ker⁡(T)\ker(T)Where k=dim⁡(ker⁡(T))k = \dim(\ker(T)). Extend to a basis {u1,…,uk,uk+1,…,un}\{\mathbf{u}_1, \ldots, \mathbf{u}_k, \mathbf{u}_{k+1}, \ldots, \mathbf{u}_n\} of VV Where n=dim⁡(V)n = \dim(V).

We claim {T(uk+1),…,T(un)}\{T(\mathbf{u}_{k+1}), \ldots, T(\mathbf{u}_n)\} is a basis for im(T)\mathrm{im}(T).

Spanning: For any w∈im(T)\mathbf{w} \in \mathrm{im}(T)Write w=T(v)\mathbf{w} = T(\mathbf{v}) for some v=∑i=1nαiui∈V\mathbf{v} = \sum_{i=1}^n \alpha_i \mathbf{u}_i \in V. Then

w=T(∑i=1nαiui)=∑i=1nαiT(ui)=∑i=k+1nαiT(ui)\mathbf{w} = T\left(\sum_{i=1}^n \alpha_i \mathbf{u}_i\right) = \sum_{i=1}^n \alpha_i T(\mathbf{u}_i) = \sum_{i=k+1}^n \alpha_i T(\mathbf{u}_i)

Since T(ui)=0T(\mathbf{u}_i) = \mathbf{0} for i≤ki \leq k.

Linear independence: If ∑i=k+1nαiT(ui)=0\sum_{i=k+1}^n \alpha_i T(\mathbf{u}_i) = \mathbf{0}Then T(∑i=k+1nαiui)=0T\left(\sum_{i=k+1}^n \alpha_i \mathbf{u}_i\right) = \mathbf{0}So ∑i=k+1nαiui∈ker⁡(T)\sum_{i=k+1}^n \alpha_i \mathbf{u}_i \in \ker(T). Thus ∑i=k+1nαiui=∑j=1kβjuj\sum_{i=k+1}^n \alpha_i \mathbf{u}_i = \sum_{j=1}^k \beta_j \mathbf{u}_j For some βj\beta_j. By linear independence of the full basis, all coefficients are zero.

Therefore dim⁡(im(T))=n−k\dim(\mathrm{im}(T)) = n - kGiving dim⁡(ker⁡(T))+dim⁡(im(T))=n\dim(\ker(T)) + \dim(\mathrm{im}(T)) = n. ■\blacksquare

6.5 Isomorphisms

A linear transformation T:V→WT : V \to W is an isomorphism if it is bijective. We write V≅WV \cong W.

Theorem 6.4. TT is an isomorphism if and only if ker⁡(T)={0}\ker(T) = \{\mathbf{0}\} and im(T)=W\mathrm{im}(T) = W.

Corollary 6.5. If dim⁡(V)=dim⁡(W)<∞\dim(V) = \dim(W) \lt \inftyThen TT is injective if and only if TT is surjective.

Proof. If TT is injective, ker⁡(T)={0}\ker(T) = \{\mathbf{0}\}So dim⁡(im(T))=dim⁡(V)=dim⁡(W)\dim(\mathrm{im}(T)) = \dim(V) = \dim(W) Hence im(T)=W\mathrm{im}(T) = W (a subspace of full dimension equals the whole space). Conversely, If TT is surjective, dim⁡(im(T))=dim⁡(W)=dim⁡(V)\dim(\mathrm{im}(T)) = \dim(W) = \dim(V)So dim⁡(ker⁡(T))=0\dim(\ker(T)) = 0Giving ker⁡(T)={0}\ker(T) = \{\mathbf{0}\}. ■\blacksquare

6.6 Change of Basis

If PP is the change-of-basis matrix from basis B\mathcal{B} to basis B′\mathcal{B}'Then for a Linear transformation TT with matrix representations [T]B[T]_{\mathcal{B}} and [T]B′[T]_{\mathcal{B}'}:

[T]B′=P−1[T]BP[T]_{\mathcal{B}'} = P^{-1}[T]_{\mathcal{B}} P

This is the similarity transformation. Similar matrices represent the same linear transformation In different bases and share the same eigenvalues, determinant, and trace.

6.7 Worked Example: Matrix of a Transformation with Change of Basis

Problem. Let T:R2→R2T : \mathbb{R}^2 \to \mathbb{R}^2 be defined by T(x,y)=(2x+y,x+2y)T(x, y) = (2x + y, x + 2y). (a) Find [T]E[T]_{\mathcal{E}} where E\mathcal{E} is the standard basis. (b) Find [T]B[T]_{\mathcal{B}} where B={(1,1),(1,−1)}\mathcal{B} = \{(1, 1), (1, -1)\}. (c) Verify that [T]B=P−1[T]EP[T]_{\mathcal{B}} = P^{-1}[T]_{\mathcal{E}} P.

Solution

(a) T(1,0)=(2,1)T(1, 0) = (2, 1) and T(0,1)=(1,2)T(0, 1) = (1, 2)So

[T]E=(2112)[T]_{\mathcal{E}} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}

(b) Compute TT on the basis B\mathcal{B}:

T(1,1)=(3,3)=3(1,1)+0(1,−1)T(1, 1) = (3, 3) = 3(1, 1) + 0(1, -1)So coordinates are (30)B\begin{pmatrix} 3 \\ 0 \end{pmatrix}_{\mathcal{B}}.

T(1,−1)=(1,−1)=0(1,1)+1(1,−1)T(1, -1) = (1, -1) = 0(1, 1) + 1(1, -1)So coordinates are (01)B\begin{pmatrix} 0 \\ 1 \end{pmatrix}_{\mathcal{B}}.

[T]B=(3001)[T]_{\mathcal{B}} = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}

(c) The change-of-basis matrix from E\mathcal{E} to B\mathcal{B} is

P=(111−1),P−1=(1/21/21/2−1/2)P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \quad P^{-1} = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}

P−1[T]EP=(1/21/21/2−1/2)(2112)(111−1)P^{-1}[T]_{\mathcal{E}} P = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}

=(3/23/21/2−1/2)(111−1)=(3001)=[T]B= \begin{pmatrix} 3/2 & 3/2 \\ 1/2 & -1/2 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix} = [T]_{\mathcal{B}}. ■\blacksquare

6.8 Dual Spaces

Definition. The dual space of a vector space VV over FFDenoted V∗V^*Is the space of all Linear functionals f:V→Ff : V \to F. Elements of V∗V^* are called covectors.

Proposition 6.6. If dim⁡(V)=n<∞\dim(V) = n \lt \inftyThen dim⁡(V∗)=n\dim(V^*) = n.

Proof. Let {e1,…,en}\{\mathbf{e}_1, \ldots, \mathbf{e}_n\} be a basis for VV. Define the dual basis {φ1,…,φn}⊆V∗\{\varphi_1, \ldots, \varphi_n\} \subseteq V^* by φi(ej)=δij\varphi_i(\mathbf{e}_j) = \delta_{ij} (the Kronecker Delta). Each φi\varphi_i is a well-defined linear functional since it is defined on a basis and Extended linearly. These are linearly independent: if ∑ciφi=0\sum c_i \varphi_i = 0Then applying to ej\mathbf{e}_j gives cj=0c_j = 0. They span V∗V^*: for any f∈V∗f \in V^*, f=∑i=1nf(ei)φif = \sum_{i=1}^n f(\mathbf{e}_i)\varphi_i. ■\blacksquare

Definition. The double dual of VV is V∗∗=(V∗)∗V^{**} = (V^*)^*.

Theorem 6.7. If VV is finite-dimensional, the map Φ:V→V∗∗\Phi : V \to V^{**} defined by Φ(v)(f)=f(v)\Phi(\mathbf{v})(f) = f(\mathbf{v}) is a natural isomorphism.

Intuition. The double dual “recovers” the original space. A vector v\mathbf{v} can be Identified with the functional on V∗V^* that evaluates each covector at v\mathbf{v}.

Example. For V=R3V = \mathbb{R}^3 with standard basis, the dual basis {φ1,φ2,φ3}\{\varphi_1, \varphi_2, \varphi_3\} Is given by φi(x1,x2,x3)=xi\varphi_i(x_1, x_2, x_3) = x_i. The functional f(x1,x2,x3)=3x1−2x2+x3f(x_1, x_2, x_3) = 3x_1 - 2x_2 + x_3 Corresponds to the covector 3φ1−2φ2+φ33\varphi_1 - 2\varphi_2 + \varphi_3 in V∗V^*.

Remark. In infinite dimensions, VV and V∗∗V^{**} need not be isomorphic. The double dual Isomorphism is a special feature of finite-dimensional spaces.

6.9 Annihilators

Definition. For a subset S⊆VS \subseteq VThe annihilator of SS is

S0={f∈V∗:f(s)=0 for all s∈S}S^0 = \{f \in V^* : f(s) = 0 \mathrm{~for~all~} s \in S\}

Proposition 6.8. S0S^0 is a subspace of V∗V^*And if WW is a subspace of VV with dim⁡(V)=n\dim(V) = nThen dim⁡(W0)=n−dim⁡(W)\dim(W^0) = n - \dim(W).

Proof. S0S^0 is the intersection of the kernels ker⁡(s)\ker(s) as ss ranges over SSWhere each ss Is viewed as an element of V∗∗V^{**} via Φ\Phi. Each ker⁡(s)\ker(s) is a subspace of V∗V^*And any Intersection of subspaces is a subspace.

For the dimension: let dim⁡(W)=k\dim(W) = k and extend a basis {w1,…,wk}\{\mathbf{w}_1, \ldots, \mathbf{w}_k\} Of WW to a basis {w1,…,wk,wk+1,…,wn}\{\mathbf{w}_1, \ldots, \mathbf{w}_k, \mathbf{w}_{k+1}, \ldots, \mathbf{w}_n\} of VV. Let {φ1,…,φn}\{\varphi_1, \ldots, \varphi_n\} be the dual basis. Then f∈W0f \in W^0 iff f(wi)=0f(\mathbf{w}_i) = 0 For i=1,…,ki = 1, \ldots, k. Writing f=∑cjφjf = \sum c_j \varphi_jWe need ci=0c_i = 0 for i=1,…,ki = 1, \ldots, k. So f=∑j=k+1ncjφjf = \sum_{j=k+1}^n c_j \varphi_jGiving dim⁡(W0)=n−k\dim(W^0) = n - k. ■\blacksquare