Thm: Let T : V → W be a linear transformation. Then T (0) = 0 Pf: T (0) = T (0 + 0) = T (0) + T (0) Thm: Let A be an m × n matrix. Then the function

\begin{matrix}T : R^{n} → R^{m} \\ T (x) = Ax \end{matrix}

is a linear transformation. Thm: If T : R^{n} → R^{m} is a linear transformation, then T (x) = Ax where

A = [T (e_{1})...T (e_{n})]

Ex: If T : R^{2} → R^{2} and

T ( \begin{matrix}1 \\ 0 \end{matrix} ) = ( \begin{matrix}2 \\ 3 \end{matrix} ), T ( \begin{matrix}0 \\ 1 \end{matrix} ) = ( \begin{matrix}1 \\ 4 \end{matrix} ), then

T ( \begin{matrix}x \\ y \end{matrix} ) = xT ( \begin{matrix}1 \\ 0 \end{matrix} )+yT ( \begin{matrix}0 \\ 1 \end{matrix} ) = x ( \begin{matrix}2 \\ 3 \end{matrix} )+y ( \begin{matrix}1 \\ 4 \end{matrix} ) = ( \begin{matrix}2 & 1 \\ 3 & 4 \end{matrix} ) ( \begin{matrix}x \\ y \end{matrix} )

Change of basis:

Suppose S = \{( \begin{matrix}1 \\ 0 \end{matrix} ) , ( \begin{matrix}0 \\ 1 \end{matrix} )\}

Suppose B = \{( \begin{matrix}2 \\ 3 \end{matrix} ) , ( \begin{matrix}1 \\ 4 \end{matrix} )\}

Defn: ( \begin{matrix}x \\ y \end{matrix} )_{B} = xb_{1} + yb_{2}

Thus ( \begin{matrix}1 \\ 0 \end{matrix} )_{B} = ( \begin{matrix}2 \\ 3 \end{matrix} )_{S} and ( \begin{matrix}0 \\ 1 \end{matrix} )_{B} = ( \begin{matrix}1 \\ 4 \end{matrix} )_{S}

( \begin{matrix}x \\ y \end{matrix} )_{B} = x ( \begin{matrix}2 \\ 3 \end{matrix} )_{S} + y ( \begin{matrix}1 \\ 4 \end{matrix} )_{S} = ( \begin{matrix}2x + y \\ 3x + 4y \end{matrix} )_{S}

( \begin{matrix}2 & 1 \\ 3 & 4 \end{matrix} ) ( \begin{matrix}x \\ y \end{matrix} )_{B} = ( \begin{matrix}u \\ v \end{matrix} )_{S}

Suppose C = \{( \begin{matrix}5 \\ 2 \end{matrix} ) , ( \begin{matrix}2 \\ 1 \end{matrix} )\}

( \begin{matrix}5 & 2 \\ 2 & 1 \end{matrix} ) ( \begin{matrix}w \\ z \end{matrix} )_{C} = ( \begin{matrix}u \\ v \end{matrix} )_{S}

( \begin{matrix}1 & −2 \\ −2 & 5 \end{matrix} ) ( \begin{matrix}u \\ v \end{matrix} )_{S} = ( \begin{matrix}w \\ z \end{matrix} )_{C}

( \begin{matrix}1 & −2 \\ −2 & 5 \end{matrix} ) ( \begin{matrix}2 & 1 \\ 3 & 4 \end{matrix} ) ( \begin{matrix}x \\ y \end{matrix} )_{B} = ( \begin{matrix}1 & −2 \\ −2 & 5 \end{matrix} ) ( \begin{matrix}u \\ v \end{matrix} )_{S} = ( \begin{matrix}w \\ z \end{matrix} )_{C}

( \begin{matrix}−4 & −7 \\ 11 & 18 \end{matrix} ) ( \begin{matrix}x \\ y \end{matrix} )_{B} = ( \begin{matrix}w \\ z \end{matrix} )_{C}

Note:

( \begin{matrix}−4 & −7 \\ 11 & 18 \end{matrix} ) ( \begin{matrix}1 \\ 0 \end{matrix} )_{B} = ( \begin{matrix}−4 \\ 11 \end{matrix} )_{C} and ( \begin{matrix}−4 & −7 \\ 11 & 18 \end{matrix} ) ( \begin{matrix}0 \\ 1 \end{matrix} )_{B} = ( \begin{matrix}−7 \\ 18 \end{matrix} )_{C}

Basis for T_{(1,0)}(S^{1}) (polar coordinates):

ϕ : S^{1} − {e^{iπ}} → (−π, π), ϕ(e^{iθ}) = θ

Note: ϕ(e^{i0}) = 0,

The standard basis for T_{(1,0)}(S^{1}) w.r.t. (U, ϕ) = {D_{α}} where

α : (−ϵ, ϵ) → M, α(t) = ϕ^{−1}(t) = e^{it} = (1, t)_{p} = (cost, sint)_{E} for some ϵ > 0.

G((1, 0)) = {f^{smooth} : U ⊂ S^{1} → R}

D_{α} : G((1, 0)) → R

D_{α}(g) = \frac{d(g∘α)}{dt}|_{t=0} = \frac{d(g(ϕ^{−1}(t)))}{dt}|_{t=0} = \frac{d(g(cos(t),sin(t)))}{dt}|_{t=0}

If v ∈ T_{p}(M), then v = cD_{α} where c = v([π_{1} ∘ ϕ]) = v([ϕ])

Basis for T_{(1,0)}(S^{1}) (projection):

ψ_{yp} : {θ | − π/2 < θ < π/2} → (−1, 1), ϕ(x, y) = y

The standard basis for T_{(1,0)}(S^{1}) w.r.t. (U′, ϕ) = {D_{β}} where

β : (−ϵ, ϵ) → M, β(t) = ψ_{yp}^{−1}(t) = (x, t) = (\sqrt{1 − t^{2}}, t) for some ϵ > 0.

D_{β}(g) = \frac{d(g∘α)}{dt}|_{t=0} = \frac{d(g(ψ_{yp}^{−1}(t)))}{dt}|_{t=0} = ^{\frac{d(g(\sqrt{1−t^{2}},t))}{dt}}|_{t=0}

If v ∈ T_{p}(M), then v = cD_{β} where c = v([π_{1} ∘ ψ_{yp}]) = v([ψ_{yp}])

D_{α}, D_{β} ∈ T_{p}(M),

D_{α} ∈ T_{p}(M) implies D_{α} = cD_{β} where c = D_{α}([ψ_{yp}])

D_{β} ∈ T_{p}(M) implies D_{β} = cD_{α} where c = D_{β}([π_{1} ∘ϕ]) = D_{β}([ϕ])

c = D_{β}(ϕ) = ^{\frac{d(ϕ(\sqrt{1−t^{2}},t))}{dt}}|_{t=0} = \frac{d(sin^{−1}(t))}{dt}|_{t=0} = \frac{1}{\sqrt{1−t^{2}}} |_{t=0} = 1 Thus if Ax_{polar} = x_{proj}, then A =

Suppose Bx_{B} = x_{S} and Cx_{C} = x_{D}

If T x_{B} = y_{C}, then T B^{−1}Bx_{B} = C^{−1}Cy_{C}

CT B^{−1}(Bx_{B}) = Cy_{C}

CT B^{−1}x_{S} = y_{D}

Suppose C = {3D_{β}}

Then [3]x_{C} = x_{proj}

Suppose the derivative of f : S^{1} → S^{1} with respect to ψ_{yp} is Df

Then Dfv_{proj} = w_{proj}, then Df[1/3][3]v_{proj} = w_{proj}.

Hence (1/3)Dfv_{C} = w_{proj}.