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}

Take a chart (U, ϕ) at p where ϕ(p) = 0, The standard basis for T_{p}(M) w.r.t.(U, ϕ) = {v_{1}, ..., v_{m}}, where v_{i} = D_{α_{i}} and α_{i} : (−ϵ, ϵ) → M, α_{i}(t) = ϕ^{−1}(0, ..., t, ..., 0) for some ϵ > 0. If v ∈ T_{p}(M), then v = Σ_{i=1}^{m}a_{i}v_{i} where a_{i} = v([π_{i} ∘ ϕ]) 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}