Linear-1 10

Change of basis

\begin{matrix} Let V vector space \\ Let B, C bases of V \\ Then \exists! A : \forall v \in V : A [v]_{B} = [v]_{C} \\ Matrix A is denoted as [I]_{C}^{B} \\ [I]_{C}^{B} [v]_{B} = [v]_{C} \\ How do we find this matrix? \\ B = {v_{1}, \dots, v_{n}} \\ C = {w_{1}, \dots, w_{n}} \\ [I]_{C}^{B} = (\begin{matrix} ⋮ & ⋮ \\ [v_{1}]_{C} & \dots & [v_{n}]_{C} \\ ⋮ & ⋮ \end{matrix}) \end{matrix}

Example

\begin{matrix} Let V = R_{1} [x] \\ B = {2, x + 1} \\ C = {x, 2 x + 4} \\ [2]_{C} = (\begin{matrix} - 1 \\ \frac{1}{2} \end{matrix}) \\ [x + 1]_{C} = (\begin{matrix} \frac{1}{2} \\ \frac{1}{4} \end{matrix}) \\ ⟹ [I]_{C}^{B} = (\begin{matrix} - 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{4} \end{matrix}) \\ ⟹ \forall p (x) \in V : (\begin{matrix} - 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{4} \end{matrix}) [p (x)]_{B} = [p (x)]_{C} \\ Let p (x) = 3 x + 1 \\ ⟹ [p (x)]_{B} = (\begin{matrix} - 1 \\ 3 \end{matrix}) \\ ⟹ [p (x)]_{C} = (\begin{matrix} - 1 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{4} \end{matrix}) \cdot (\begin{matrix} - 1 \\ 3 \end{matrix}) = (\begin{matrix} \frac{5}{2} \\ \frac{1}{4} \end{matrix}) \end{matrix}

Exercise

\begin{matrix} Let V = R_{2} [x] \\ B = {1 + 2 x, 3 + 5 x, v} is a basis of V \\ [x^{2} + 1]_{B} = (\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix}) \\ Find v \\ Solution: \\ x^{2} + 1 = 2 + 4 x - 3 - 5 x + v \\ ⟹ v = x^{2} + x + 2 \end{matrix}

Properties

\begin{matrix} 1. [I]_{B}^{B} = I \\ 2. [I]_{B}^{C} [I]_{C}^{D} = [I]_{B}^{D} \\ 3. [I]_{B}^{C} is invertible, ([I]_{B}^{C})^{- 1} = [I]_{C}^{B} \end{matrix}

Exercise

\begin{matrix} Let V = F^{n} \\ B = {v_{1}, \dots, v_{n}} basis of V \\ S is a standard basis of V \\ Find [I]_{S}^{B} \\ Solution: \\ \forall v \in V : [v]_{S} = v \\ ⟹ [I]_{S}^{B} == (\begin{matrix} ⋮ & ⋮ \\ [v_{1}]_{S} & \dots & [v_{n}]_{S} \\ ⋮ & ⋮ \end{matrix}) = (\begin{matrix} ⋮ & ⋮ \\ v_{1} & \dots & v_{n} \\ ⋮ & ⋮ \end{matrix}) \\ ⟹ [I]_{C}^{B} = [I]_{C}^{S} \cdot [I]_{S}^{B} = ([I]_{S}^{C})^{- 1} [I]_{S}^{B} = \\ = {(\begin{matrix} ⋮ & ⋮ \\ w_{1} & \dots & w_{n} \\ ⋮ & ⋮ \end{matrix})}^{- 1} \cdot (\begin{matrix} ⋮ & ⋮ \\ v_{1} & \dots & v_{n} \\ ⋮ & ⋮ \end{matrix}) \end{matrix}

Example

\begin{matrix} B = {(\begin{matrix} 1 \\ 2 \\ 4 \end{matrix}), (\begin{matrix} 8 \\ 6 \\ 2 \end{matrix}), (\begin{matrix} - 1 \\ - 2 \\ 4 \end{matrix})} \\ C = {(\begin{matrix} 6 \\ - 1 \\ 4 \end{matrix}), (\begin{matrix} 2 \\ 0 \\ 5 \end{matrix}), (\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix})} \\ [I]_{C}^{B} = {(\begin{matrix} 6 & 2 & 1 \\ - 1 & 0 & 0 \\ 4 & 5 & - 1 \end{matrix})}^{- 1} \cdot (\begin{matrix} 1 & 8 & - 1 \\ 2 & 6 & - 2 \\ 4 & 2 & 4 \end{matrix}) \end{matrix}

Exercise

\begin{matrix} A = (\begin{matrix} 1 & 0 & 1 \\ 0 & 3 & 0 \\ 0 & 3 & 1 \end{matrix}) \\ B = {(\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 4 \\ 6 \end{matrix})} is a basis of R^{3} \\ Find C = {w_{1}, w_{2}, w_{3}} basis of R^{3} \\ Such that A = [I]_{B}^{C} \\ Solution: \\ [w_{1}]_{B} = C_{1} ([I]_{B}^{C}) = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) ⟹ w_{1} = (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) \\ [w_{2}]_{B} = C_{2} ([I]_{B}^{C}) = (\begin{matrix} 0 \\ 3 \\ 3 \end{matrix}) ⟹ w_{2} = 3 (\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}) + 3 (\begin{matrix} 0 \\ 4 \\ 6 \end{matrix}) = (\begin{matrix} 3 \\ 18 \\ 18 \end{matrix}) \\ [w_{3}]_{B} = C_{3} ([I]_{B}^{C}) = (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) ⟹ w_{3} = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + (\begin{matrix} 0 \\ 4 \\ 6 \end{matrix}) = (\begin{matrix} 1 \\ 4 \\ 6 \end{matrix}) \\ Another approach \\ A = [I]_{C}^{B} = [I]_{B}^{S} \cdot [I]_{S}^{C} \\ ⟹ [I]_{S}^{C} = [I]_{S}^{B} \cdot A = (\begin{matrix} 1 & 1 & 0 \\ 0 & 2 & 4 \\ 1 & 0 & 6 \end{matrix}) \cdot (\begin{matrix} 1 & 0 & 1 \\ 0 & 3 & 0 \\ 0 & 3 & 1 \end{matrix}) = (\begin{matrix} 1 & 3 & 1 \\ 0 & 18 & 4 \\ 1 & 18 & 6 \end{matrix}) = (\begin{matrix} w_{1} & w_{2} & w_{3} \end{matrix}) \\ Find C = {w_{1}, w_{2}, w_{3}} basis of R^{3} \\ Such that A = [I]_{C}^{B} \\ Solution: \\ A = [I]_{C}^{B} \\ A^{- 1} = [I]_{B}^{C} = [I]_{B}^{S} \cdot [I]_{S}^{C} \\ ⟹ (\begin{matrix} w_{1} & w_{2} & w_{3} \end{matrix}) = [I]_{S}^{C} = [I]_{S}^{B} \cdot A^{- 1} \end{matrix}

Exercise

\begin{matrix} [I]_{C}^{B} = (\begin{matrix} 0 & 0 & 2 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{matrix}) \\ Find connection between B and C \\ Solution: \\ B = {v_{1}, v_{2}, v_{3}} \\ C = {w_{1}, w_{2}, w_{3}} \\ [v_{1}]_{C} = C_{1} ([I]_{C}^{B}) = (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) ⟹ v_{1} = w_{2} \\ [v_{2}]_{C} = C_{2} ([I]_{C}^{B}) = (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) ⟹ v_{2} = w_{3} \\ [v_{3}]_{C} = C_{3} ([I]_{C}^{B}) = (\begin{matrix} 2 \\ 0 \\ 1 \end{matrix}) ⟹ v_{3} = 2 w_{1} + w_{3} \end{matrix}

Exercise

\begin{matrix} U, V, W vector spaces over F \\ T : U \to V linear transformation \\ S : V \to W linear transformation \\ Let \hat{S} : I m (T) \to W, \forall v \in I m (T) : \hat{S} (v) = S (v) \\ 1. Prove: k e r (\hat{S}) = k e r (S) \cap I m (T) \\ I m (\hat{S}) = I m (S T) \\ 2. d i m (k e r (S T)) \leq d i m (k e r (S)) + d i m (k e r (T)) \\ 3. d i m (k e r (S T)) = d i m (k e r (S)) + d i m (k e r (T)) ⟺ k e r (S) \subseteq I m (T) \end{matrix}

\begin{matrix} Proof for 1. \\ Let v \in k e r (\hat{S}) ⟹ \hat{S} (v) = 0 = S (v) ⟹ v \in k e r (S) \\ k e r (\hat{S}) \subseteq I m (T) ⟹ v \in I m (T) \\ ⟹ v \in k e r (S) \cap I m (T) ⟹ k e r (\hat{S}) \subseteq k e r (S) \cap I m (T) \\ Let v \in k e r (S) \cap I m (T) ⟹ v \in k e r (S) \land v \in I m (T) \\ ⟹ S (v) = 0 \land \exists \hat{S} (v) = S (v) = 0 \\ ⟹ v \in k e r (\hat{S}) ⟹ k e r (S) \cap I m (T) \subseteq k e r (\hat{S}) \\ ⟹ k e r (\hat{S}) = k e r (S) \cap I m (T) \\ Let w \in I m (\hat{S}) \\ ⟹ \exists v \in I m (T) : \hat{S} (v) = w = S (v) \\ ⟹ \exists u \in U : T (u) = v ⟹ w = S (T (u)) = (S T) (u) \\ ⟹ w \in I m (S T) ⟹ I m (\hat{S}) \subseteq I m (S T) \\ Let w \in I m (S T) \\ ⟹ \exists u \in U : (S T) (u) = w = S (T (u)) \\ ⟹ \exists v \in V : v = T (u) ⟹ v \in I m (T) \land S (v) = w \\ ⟹ \hat{S} (v) = w ⟹ w \in I m) \hat{S} ⟹ I m (S T) \subseteq I m (\hat{S}) \\ ⟹ I m (\hat{S}) = I m (S T) \end{matrix}

\begin{matrix} Proof for 2. \\ 1. d i m (I m (T)) + d i m (k e r (T)) = d i m (U) \\ 2. d i m (I m (S T)) + d i m (k e r (S T)) = d i m (U) \\ 3. d i m (I m (\hat{S})) + d i m (k e r (\hat{S})) = d i m (I m (T)) \\ 3. ⟹ d i m (I m (S T)) + d i m (k e r (S) \cap I m (T)) = d i m (I m (T)) \\ (1. - 2.) ⟹ d i m (k e r (S T)) = d i m (I m (T)) + d i m (k e r (T)) - d i m (I m (S T)) = \\ = d i m (I m (S T)) + d i m (k e r (S) \cap I m (T)) + d i m (k e r (T)) - d i m (I m (S T)) = \\ = d i m (k e r (S) \cap I m (T)) + d i m (k e r (T)) \leq d i m (k e r (S)) + d i m (k e r (T)) \\ ⟹ d i m (k e r (S T)) \leq d i m (k e r (S)) + d i m (k e r (T)) \\ Proof for 3. \\ d i m (k e r (S T)) = d i m (k e r (S)) + d i m (k e r (T)) \\ ⟺ d i m (k e r (S) \cap I m (T)) + d i m (k e r (T)) = d i m (k e r (S)) + d i m (k e r (T)) \\ ⟺ d i m (k e r (S) \cap I m (T)) = d i m (k e r (S)) ⟺ k e r (S) \cap I m (T) = k e r (S) \\ ⟺ k e r (S) \subseteq I m (T) \\ ⟹ d i m (k e r (S T)) = d i m (k e r (S)) + d i m (k e r (T)) ⟺ k e r (S) \subseteq I m (T) \end{matrix}