Linear-1 9

1a

\begin{matrix} T : R^{3} \to R^{3} \\ T ((\begin{matrix} x \\ y \\ z \end{matrix})) = (\begin{matrix} x \\ x + y \\ x + y + z \end{matrix}) \\ Determine whether T is a linear transformation \\ If yes, find basis and dim of I m (T) and k e r (T) \\ Solution: \\ Let u, v \in R^{3}, α \in R \\ T (u) = (\begin{matrix} u_{1} \\ u_{1} + u_{2} \\ u_{1} + u_{2} + u_{3} \end{matrix}), α T (v) = α (\begin{matrix} v_{1} \\ v_{1} + v_{2} \\ v_{1} + v_{2} + v_{3} \end{matrix}) \\ T (u) + α T (v) = T (u + α v) = (\begin{matrix} u_{1} + α v_{1} \\ (u_{1} + u_{2}) + α (v_{1} + v_{2}) \\ (u_{1} + u_{2} + u_{3}) + α (v_{1} + v_{2} + v_{3}) \end{matrix}) \\ ⟹ T is a linear transformation \\ I m (T) = {u \in R^{3} | \exists v \in R^{3} : T (v) = u} = {(\begin{matrix} x \\ x + y \\ x + y + z \end{matrix}) | x, y, z \in R} = \\ = {x (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) + y (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}) + z (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) | x, y, z \in R} = s p ({(\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})}) \\ {(\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})} is a linear independence \\ ⟹ {(\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})} is a basis of I m (T) and d i m (I m (T)) = 3 \\ d i m (I m (T)) + d i m (k e r (T)) = d i m (R^{3}) = 3 ⟹ d i m (k e r (T)) = 0 \\ ⟹ \emptyset is a basis of k e r (T) and d i m (k e r (T)) = 0 \end{matrix}

1b

\begin{matrix} T : R_{2} [x] \to R_{2} \\ T (P) = (\begin{matrix} P (0) \\ P^{'} (0) \end{matrix}) \\ Determine whether T is a linear transformation \\ If yes, find basis and dim of I m (T) and k e r (T) \\ Solution: \\ P \in R_{2} [x] ⟹ P (x) = a + b x + c x^{2} \\ P^{'} (x) = b + 2 c x \\ ⟹ P (0) = a, P^{'} (0) = b \\ Let p_{1}, p_{2} \in R_{2} [x], α \in R \\ T (p_{1} + α p_{2}) = (\begin{matrix} (p_{1} + α p_{2}) (0) \\ (p_{1} + α p_{2})^{'} (0) \end{matrix}) = (\begin{matrix} p_{1} (0) + α p_{2} (0) \\ p_{1}^{'} (0) + α p_{2}^{'} (0) \end{matrix}) = (\begin{matrix} a_{1} + α a_{2} \\ b_{1} + α b_{2} \end{matrix}) = T (p_{1}) + α T (p_{2}) \\ ⟹ T is a linear transformation \\ I m (T) = {u \in R^{2} | \exists P \in R_{2} [x] : T (P) = u} = {(\begin{matrix} a \\ b \end{matrix}) | a, b \in R} = R^{2} \\ = s p ({(\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix})}) \\ {(\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix})} is a linear independence \\ ⟹ {(\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix})} is a basis of I m (T) and d i m (I m (T)) = 2 \\ k e r (T) = {v \in R_{2} [x] | T (v) = 0} = {a + b x + c x^{2} | {\begin{matrix} a = 0 \\ b = 0 \\ a, b, c \in R \end{matrix}} = {c x^{2} | c \in R} \\ = s p ({x^{2}}) \\ {x^{2}} is a linear independence \\ ⟹ {x^{2}} is a basis of k e r (T) and d i m (k e r (T)) = 1 \end{matrix}

1c

\begin{matrix} S = {(\begin{matrix} a & b \\ 0 & c \end{matrix}) \in R^{2 \times 2}} \\ B = (\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}) \\ T : S \to R^{2 \times 2} \\ T (A) = A B \\ Determine whether T is a linear transformation \\ If yes, find basis and dim of I m (T) and k e r (T) \\ Solution: \\ T (A) = A B = (\begin{matrix} a & b \\ 0 & c \end{matrix}) (\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}) = (\begin{matrix} a + b & a + b \\ c & c \end{matrix}) \\ Let A_{1}, A_{2} \in S, α \in R \\ T (A_{1} + α A_{2}) = (A_{1} + α A_{2}) B = A_{1} B + α A_{2} B = T (A_{1}) + α T (A_{2}) \\ ⟹ T is a linear transformation \\ I m (T) = {C \in R^{2 \times 2} | \exists A \in S : T (A) = C} = {(\begin{matrix} a + b & a + b \\ c & c \end{matrix}) | a, b, c \in R} = \\ = {(a + b) (\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}) + c (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}) | a, b, c \in R} = s p ({(\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix})}) \\ {(\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix})} is a linear independence \\ ⟹ {(\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix})} is a basis of I m (T) and d i m (I m (T)) = 2 \\ k e r (T) = {A \in S | T (A) = 0} = {(\begin{matrix} a & b \\ 0 & c \end{matrix}) | {\begin{matrix} a + b = 0 \\ c = 0 \\ a, b, c \in R \end{matrix}} = {(\begin{matrix} a & - a \\ 0 & 0 \end{matrix}) | a \in R} = \\ {a (\begin{matrix} 1 & - 1 \\ 0 & 0 \end{matrix}) | a \in R} = s p ({(\begin{matrix} 1 & - 1 \\ 0 & 0 \end{matrix})}) \\ {(\begin{matrix} 1 & - 1 \\ 0 & 0 \end{matrix})} is a linear independence \\ ⟹ {(\begin{matrix} 1 & - 1 \\ 0 & 0 \end{matrix})} is a basis of k e r (T) and d i m (k e r (T)) = 1 \end{matrix}

2a

\begin{matrix} Prove or disprove: \exists a non-invertible linear transformation T : R^{3} \to R^{3} such that: \\ T = T^{3} \neq T^{2} \\ Proof: \\ A = (\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{matrix}) \\ Let T (v) = A v \\ T^{2} (v) = T (T (v)) = A A v = A^{2} v \\ T^{3} (v) = T (T (T (v))) = A A A v = A^{3} v \\ A^{2} = (\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) \neq A ⟹ T \neq T^{2} \\ A^{3} = (\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{matrix}) = A ⟹ T = T^{3} \end{matrix}

2b

\begin{matrix} Prove or disprove: \exists different non-invertible linear transformations S, T : R^{3} \to R^{3} \\ such that: S T - T S = 0 \\ Proof: \\ Let A = (\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{matrix}), B = (\begin{matrix} 0 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{matrix}) \\ Let T (v) = A v, S (v) = B v \\ A \neq B ⟹ T \neq S \\ A B = B A = (\begin{matrix} 0 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{matrix}) \\ ⟹ \forall v \in R^{3} : (S T - T S) (v) = B A v - A B v = (B A - A B) v = 0 \\ ⟹ S T - T S = 0 \end{matrix}

3

\begin{matrix} Let W = R^{2 \times 2}, V = R_{2} [x] \\ Determine whether exists a linear transformation T : V \to W \end{matrix}

3a

\begin{matrix} Such that: T (x^{2} + 1) = (\begin{matrix} 0 & 2 \\ 0 & - 5 \end{matrix}), T (x - 1) = (\begin{matrix} 1 & 0 \\ 0 & - 2 \end{matrix}), T (x + 1) = (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) \\ Solution: \\ Let p (x) \in V \\ p (x) = a + b x + c x^{2} = (\begin{matrix} a \\ b \\ c \end{matrix}) \\ x^{2} + 1 = (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}), x - 1 = (\begin{matrix} - 1 \\ 1 \\ 0 \end{matrix}), x + 1 = (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) \\ {(\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} - 1 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix})} is a linear independence of size 3 \\ ⟹ {x^{2} + 1, x - 1, x + 1} is a linear independence of size 3 \\ d i m (R_{2} [x]) = 3 ⟹ {x^{2} + 1, x - 1, x + 1} is a basis of R_{2} [x] \\ ⟹ By the "definition theorem" there exists such unique linear transformation T \end{matrix}

3b

\begin{matrix} Such that: T (2 x) = (\begin{matrix} 1 & 2 \\ - 3 & - 5 \end{matrix}), T (x - 1) = (\begin{matrix} 1 & 0 \\ 0 & - 2 \end{matrix}), T (x + 1) = (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) \\ Solution: \\ (x - 1) + (x + 1) = 2 x \\ T (x - 1) + T (x + 1) = (\begin{matrix} 2 & 2 \\ 3 & 2 \end{matrix}) \neq T (2 x) \\ ⟹ T is not a linear tranformation \end{matrix}

3c

\begin{matrix} Such that: T (2 x) = (\begin{matrix} 2 & 2 \\ 3 & 2 \end{matrix}), T (x - 1) = (\begin{matrix} 1 & 0 \\ 0 & - 2 \end{matrix}), T (x + 1) = (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) \\ Solution: \\ (x - 1) + (x + 1) = 2 x \\ T (x - 1) + T (x + 1) = T (2 x) \\ x^{2} does not depend on x - 1, x + 1, 2 x \\ ⟹ T (x^{2}) does not depend on T (x - 1), T (x + 1), T (2 x) \\ Let T (x^{2}) = 0 \\ {x - 1, x + 1, x^{2}} is a basis of R_{2} [x] \\ ⟹ By the "definition theorem" there exists such linear transformation \end{matrix}

4

\begin{matrix} Let V, W be vector spaces over F \\ d i m (V) \leq d i m (W) \\ Prove: \exists a linear transformation T : V \to W that is injective \\ Proof: \\ Let B_{V} = {v_{1}, v_{2}, \dots, v_{n}} be a basis of V \\ Let B_{W} = {w_{1}, w_{2}, \dots, w_{m}} be a basis of W \\ n \leq m \\ Let \forall i \in [1, n] : T (v_{i}) = w_{i} \\ ⟹ By the "definition theorem": I m (T) = s p ({w_{1}, w_{2}, \dots, w_{n}}) \\ {w_{1}, w_{2}, \dots, w_{n}} \subseteq B_{W} ⟹ {w_{1}, w_{2}, \dots, w_{n}} is a linear independence \\ T (v) = T (\sum_{i = 1}^{n} α_{i} v_{i}) = \sum_{i = 1}^{n} α_{i} T (v_{i}) = \sum_{i = 1}^{n} α_{i} w_{i} - linear combination of {w_{1}, w_{2}, \dots, w_{n}} \\ T (v) = 0 ⟹ \sum_{i = 1}^{n} α_{i} w_{i} = 0 ⟹ α_{1} = α_{2} = \dots = α_{n} = 0 \\ ⟹ v = \sum_{i = 1}^{n} α_{i} v_{i} = 0 \\ ⟹ k e r (T) = {0} ⟹ T is injective \end{matrix}

5

\begin{matrix} Let V, W be vector spaces over F \\ d i m (V) \geq d i m (W) \\ Prove: \exists a linear transformation T : V \to W that is surjective \\ Proof: \\ Let B_{V} = {v_{1}, v_{2}, \dots, v_{n}} be a basis of V \\ Let B_{W} = {w_{1}, w_{2}, \dots, w_{m}} be a basis of W \\ n \geq m \\ Let \forall i \in [1, m] : T (v_{i}) = w_{i} \\ Let \forall i \in [m + 1, n] : T (v_{i}) = 0 \\ I m (T) = s p ({T (v_{1}), T (v_{2}), \dots, T (v_{n})}) = s p ({w_{1}, w_{2}, \dots, w_{m}, 0, \dots, 0}) \\ = s p ({w_{1}, w_{2}, \dots, w_{m}}) = W \\ I m (T) = W ⟹ T is surjective \end{matrix}

6

\begin{matrix} Let V, W, U be vector spaces over F \\ S, R : V \to W, T : W \to U are linear transformations \\ Prove: T \circ (S + R) = T \circ S + T \circ R \\ Proof: \\ Let v \in V \\ (T \circ (S + R)) (v) = T ((S + R) (v)) = T (S (v) + R (v)) \\ S (v), R (v) \in W, \forall w_{1}, w_{2} \in W : T (w_{1} + w_{2}) = T (w_{1}) + T (w_{2}) \\ ⟹ T (S (v) + R (v)) = T (S (v)) + T (R (v)) = (T \circ S) (v) + (T \circ R) (v) \\ ⟹ \forall v \in V : (T \circ (S + R)) (v) = (T \circ S) (v) + (T \circ R) (v) \\ ⟹ T \circ (S + R) = T \circ S + T \circ R \end{matrix}