Exam 2023 (C)

1

\begin{matrix} Let L = {x | A x = b, b \neq 0} \\ Let H = {x | A x = 0} \\ Let v_{0} \in L \\ Prove: H + v_{0} = L \\ Proof: \\ Let v \in H + v_{0} \\ ⟹ \exists h \in H : v = h + v_{0} \\ A v = A (h + v_{0}) = A h + A v_{0} = 0 + b = b ⟹ v \in L ⟹ H + v_{0} \subseteq L \\ Let v \in L \\ A (v - v_{0}) = A v - A v_{0} = b - b = 0 ⟹ (v - v_{0}) \in H \\ ⟹ (v - v_{0}) + v_{0} \in H + v_{0} ⟹ v \in H + v_{0} ⟹ L \subseteq H + v_{0} \\ ⟹ H + v_{0} = L \end{matrix}

2a

\begin{matrix} U = {p (x) \in R_{2} [x] | \begin{matrix} p^{'} (1) = 0 \\ p (1) = 0 \end{matrix}} \\ W = s p ({1 + x + x_{2}, x + x^{2}}) \\ Find basis and dimension of U + W, U \cap W \\ Solution: \\ Let p (x) \in R_{2} [x] \\ p (x) = a + b x + c x^{2} ⟹ p^{'} (x) = b + 2 c x \\ p^{'} (1) = b + 2 c \\ p (1) = a + b + c \\ {\begin{matrix} b + 2 c = 0 \\ a + b + c = 0 \end{matrix} ⟹ {\begin{matrix} b = - 2 c \\ a = c \end{matrix} \\ ⟹ U = s p ({1 - 2 x + x^{2}}) \\ U + W = s p (U \cup W) = s p ({1 + x + x^{2}, x + x^{2}, 1 - 2 x + x^{2}}) \\ (\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 1 & 1 & - 2 & 0 \\ 1 & 1 & 1 & 0 \end{array}) \to (\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & - 3 & 0 \\ 0 & 1 & 0 & 0 \end{array}) \to (\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & - 3 & 0 \\ 0 & 0 & 3 & 0 \end{array}) \\ ⟹ {1 + x + x^{2}, x + x^{2}, 1 - 2 x + x^{2}} is a linear independence and is a basis of U + W \\ ⟹ d i m (U + W) = 3 \\ \underset{3}{\underset{⏟}{d i m (U + W)}} = \underset{1}{\underset{⏟}{d i m (U)}} + \underset{2}{\underset{⏟}{d i m (W)}} - d i m (U \cap W) \\ ⟹ d i m (U \cap W) = 0 ⟹ {0} is a basis of U \cap W \end{matrix}

2b

\begin{matrix} Let A, B \in F^{n \times n} \\ Prove: d i m (N (A B)) \leq d i m (N (A)) + d i m (N (B)) \\ Proof: \\ Let v \in N (B) \\ B v = 0 ⟹ A B v = 0 ⟹ v \in N (B) ⟹ N (B) \subseteq N (A B) \\ Let B_{k} = {v_{1}, v_{2}, \dots, v_{k}} be a basis of N (B), d i m (N (B)) = k \\ Let B_{k + t} = {v_{1}, v_{2}, \dots, v_{k}, u_{1}, u_{2}, \dots, u_{t}} be a basis of N (A B), d i m (N (A B)) = k + t \\ \forall i \in [1, t] : A B u_{i} = 0 ⟹ B u_{i} \in N (A) \\ ⟹ {B u_{1}, B u_{2}, \dots, B u_{t}} \subseteq N (A) \\ Let \sum_{i = 1}^{t} α_{i} B u_{i} = 0 \\ ⟹ \sum_{i = 1}^{t} B α_{i} u_{i} = B \sum_{i = 1}^{t} α_{i} u_{i} = 0 \\ ⟹ \sum_{i = 1}^{t} α_{i} u_{i} \in N (B) ⟹ \sum_{i = 1}^{t} α_{i} u_{i} = \sum_{i = 1}^{k} β_{i} v_{i} \\ ⟹ \underset{Linear combination of B_{k + t}}{\underset{⏟}{\sum_{i = 1}^{t} α_{i} u_{i} - \sum_{i = 1}^{k} β_{i} v_{i}}} = 0 \\ B_{k + t} is a linear independence ⟹ α_{1} = α_{2} = \dots = α_{t} = β_{1} = β_{2} = \dots = β_{k} = 0 \\ ⟹ {B u_{1}, B u_{2}, \dots, B u_{t}} is a linear independence \\ ⟹ d i m (N (A)) \geq t ⟹ d i m (N (B)) + d i m (N (A)) \geq k + t = d i m (N (A B)) \\ ⟹ d i m (N (A B)) \leq d i m (N (A)) + d i m (N (B)) \end{matrix}

3a

\begin{matrix} Let B = {1, 1 + x, 1 + x^{2}} be a basis of R_{2} [x] \\ Let C = {(\begin{matrix} 0 & 0 \\ 0 & - 1 \end{matrix}), (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 2 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix})} be a basis of R^{2 \times 2} \\ Let T : R_{2} [x] \to R^{2 \times 2} be a linear transformation \\ Let [T]_{C}^{B} = (\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 1 & 1 & 5 \\ 1 & - 1 & - 1 \end{matrix}) \\ Find basis and dimension of I m (T), k e r (T) \\ Solution: \\ (\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 1 & 1 & 5 \\ 1 & - 1 & - 1 \end{matrix}) \to (\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 1 & 3 \\ 0 & - 1 & - 3 \end{matrix}) \to (\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) \\ [k e r (T)]_{B} = N ([T]_{C}^{B}) = s p {(\begin{matrix} - 2 \\ - 3 \\ 1 \end{matrix})} ⟹ {\begin{matrix} k e r (T) = s p {- 4 - 3 x + x^{2}} \\ d i m (k e r (T)) = 1 \end{matrix} \\ [I m (T)]_{C} = C ([T]_{C}^{B}) = s p {(\begin{matrix} 1 \\ 0 \\ 1 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 1 \\ - 1 \end{matrix})} ⟹ {\begin{matrix} I m (T) = s p {(\begin{matrix} 0 & 2 \\ 1 & 0 \end{matrix}), (\begin{matrix} 1 & 2 \\ - 1 & - 1 \end{matrix})} \\ d i m (I m (T)) = 2 \end{matrix} \end{matrix}

3b

\begin{matrix} Let V, W be finitely generated vector spaces \\ Let T, S : V \to W be linear transformations, S is invertible \\ T S^{- 1} is injective \\ Is T neccessarily invertible? \\ Solution: \\ T S^{- 1} : W \to W is injective \\ d i m (W) = d i m (W) ⟹ T S^{- 1} is surjective \\ ⟹ T is surjective \\ S is invertible ⟹ V ≅ W ⟹ d i m (V) = d i m (W) \\ ⟹ T is injective ⟹ T is bijective \\ ⟹ T is invertible \end{matrix}

4a

\begin{matrix} Let A, B \in F^{n \times n} \\ Let r a n k (A) < \frac{n}{2}, r a n k (B) < \frac{n}{2} \\ Prove or disprove: A B = 0 \\ Disproof: \\ A = B = (\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) ⟹ A B = (\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) \neq 0 \\ r a n k (A) = r a n k (B) = 1 < \frac{3}{2} \end{matrix}

4b

\begin{matrix} Let A, B \in F^{n \times n}, A B = 0 \\ Prove or disprove: r a n k (A) \leq \frac{n}{2} \lor r a n k (B) \leq \frac{n}{2} \\ Proof: \\ A B = 0 ⟹ r a n k (A B) = 0 ⟹ d i m (N (A B)) = n \\ \forall v \in F^{n} : A B v = 0 v = 0 \\ \forall v \in F^{n} : B v \in C (B) ⟹ C (B) \subseteq N (A) \\ ⟹ d i m (C (B)) \leq d i m (N (A)) \\ ⟹ r a n k (B) \leq n - r a n k (A) ⟹ r a n k (A) + r a n k (B) \leq n \\ Let r a n k (A) = k > \frac{n}{2} \\ ⟹ k + r a n k (B) \leq n ⟹ \frac{n}{2} + r a n k (B) \leq n ⟹ r a n k (B) \leq \frac{n}{2} \\ Let r a n k (B) = k > \frac{n}{2} \\ ⟹ k + r a n k (A) \leq n ⟹ \frac{n}{2} + r a n k (A) \leq n ⟹ r a n k (A) \leq \frac{n}{2} \end{matrix}

4c

\begin{matrix} Let A \in R^{3 \times 3} be anti-symmetric \\ Prove: A (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) = (\begin{matrix} a \\ b \\ c \end{matrix}) ⟹ a + b + c = 0 \\ Proof: \\ (\begin{matrix} 0 & a_{2} & a_{3} \\ - a_{2} & 0 & a_{5} \\ - a_{3} & - a_{5} & 0 \end{matrix}) (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) = (\begin{matrix} a_{2} + a_{3} \\ a_{5} - a_{2} \\ - a_{3} - a_{5} \end{matrix}) \\ {\begin{matrix} a_{2} + a_{3} = a \\ a_{5} - a_{2} = b \\ - a_{3} - a_{5} = c \end{matrix} ⟹ a + b + c = a_{2} + a_{3} + a_{5} - a_{2} - a_{3} - a_{5} = 0 \\ Note: this doesn’t work in Z_{2} \\ Because (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) \in Z_{2}^{3 \times 3} is anti-symmetric \end{matrix}

5

\begin{matrix} Let V, W be finitely generated vector spaces over F \\ Let T : V \to W be a linear transformation \\ Let S : W \to V be a linear transformation, T S T = T \end{matrix}

5a

\begin{matrix} Show that not neccesserily k e r (S) = {0} \\ Solution: \\ Let T = S = 0 \\ T S T = T = 0 \\ k e r (S) = W \neq {0} \end{matrix}

5b

\begin{matrix} Prove: k e r (S) \cap I m (T) = {0} \\ Proof: \\ Let w \in k e r (S) \cap I m (T) \\ ⟹ S (w) = 0, \exists v \in V : T (v) = w \\ S (T (v)) = S (w) = 0 ⟹ T (S (T (v))) = T (0) = 0 = T (v) = w \\ ⟹ k e r (S) \cap I m (T) = {0} \end{matrix}

5c

\begin{matrix} Prove: \forall T : V \to W : \exists S : W \to V : T S T = T \\ Proof: \\ Let B_{T} = {w_{1}, w_{2}, \dots, w_{k}} be a basis of I m (T) \\ ⟹ \forall i \in [1, k] : \exists v_{i} \in V : T (v_{i}) = w_{i} \\ I m (T) \subseteq W ⟹ \exists B = B_{T} \cup {w_{k + 1}, w_{k + 2}, \dots, w_{k + t}} basis of W \\ ⟹ By the defining theorem \exists S : W \to V, S (w_{i}) = {\begin{cases} v_{i} & i \leq k \\ 0 & i > k \end{cases} \\ Let v \in V \\ ⟹ \exists {α_{i}}_{i \in [1, n]} \subseteq F : v = \sum_{i = 1}^{n} α_{i} v_{i} \\ Let T (v) = w \in W \\ w \in I m (T) ⟹ \exists {β_{j}}_{j \in [1, k]} \subseteq F : w = \sum_{j = 1}^{k} β_{j} w_{j} \\ T S T (v) = T S (w) = T S (\sum_{i = 1}^{k} β_{j} w_{j}) = T (\sum_{i = 1}^{k} β_{j} S (w_{j})) = T (\sum_{i = 1}^{k} β_{j} v_{j}) = \sum_{j = 1}^{k} β_{j} T (v_{j}) = \\ = \sum_{i = 1}^{k} β_{j} w_{j} = w = T (v) \\ ⟹ T S T = T \end{matrix}