Linear-1 2

1a

\begin{matrix} A = (\begin{matrix} 3 & 4 & 5 \\ 5 & 4 & 8 \\ 9 & 2 & 9 \end{matrix}), B = (\begin{matrix} 9 & 2 & 9 \\ 5 & 4 & 8 \\ 3 & 4 & 5 \end{matrix}) \end{matrix}

1b

\begin{matrix} A B = (\begin{matrix} 3 & 4 & 5 \\ 5 & 4 & 8 \\ 9 & 2 & 9 \end{matrix}) \times (\begin{matrix} 9 & 2 & 9 \\ 5 & 4 & 8 \\ 3 & 4 & 5 \end{matrix}) = (\begin{matrix} 62 & 42 & 84 \\ 89 & 58 & 117 \\ 118 & 62 & 142 \end{matrix}) \\ B A = (\begin{matrix} 9 & 2 & 9 \\ 5 & 4 & 8 \\ 3 & 4 & 5 \end{matrix}) \times (\begin{matrix} 3 & 4 & 5 \\ 5 & 4 & 8 \\ 9 & 2 & 9 \end{matrix}) = (\begin{matrix} 118 & 62 & 142 \\ 107 & 52 & 129 \\ 74 & 38 & 92 \end{matrix}) \end{matrix}

Given: A, B \in F^{3 \times 4}; C \in F^{4 \times 5}

2a

\begin{matrix} Can a change in A_{21} affect ((A + B) \cdot C)_{23} ? \\ If so, give an example \\ D = A + B \\ ((A + B) \cdot C)_{23} = (D \cdot C)_{23} = \sum_{k = 1}^{4} D_{2 k} \cdot C_{k 3} = \\ = D_{21} \cdot C_{13} + D_{22} \cdot C_{23} + \dots \\ D_{21} = A_{21} + B_{21} ⟹ A_{21} can affect ((A + B) \cdot C)_{23} \\ Example: \\ A = (\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}), B = (\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}), C = (\begin{matrix} 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}) \\ (A + B) \cdot C = (\begin{matrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}) \\ A^{'} = (\begin{matrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) ⟹ (A^{'} + B) \cdot C = (\begin{matrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}) \end{matrix}

2b

\begin{matrix} Can a change in C_{21} affect ((A + B) \cdot C)_{23} ? \\ If so, give an example \\ D = A + B \\ ((A + B) \cdot C)_{23} = (D \cdot C)_{23} = \sum_{k = 1}^{4} D_{2 k} \cdot C_{k 3} = \\ = D_{21} \cdot C_{13} + D_{22} \cdot C_{23} + D_{23} \cdot C_{33} + D_{24} \cdot C_{43} \\ ⟹ C_{21} cannot affect ((A + B) \cdot C)_{23} \end{matrix}

Given: A \in F^{2 \times 3}; B \in F^{3 \times 4}; C \in F^{4 \times 5}

3a

\begin{matrix} Can a change in B_{22} affect (A B C)_{13} ? \\ If so, give an example \\ A B_{i j} = \sum_{k = 1}^{3} A_{i k} B_{k j} \\ A B C_{13} = \sum_{m = 1}^{4} A B_{1 m} C_{m 3} = \sum_{m = 1}^{4} (\sum_{k = 1}^{3} A_{1 k} B_{k m}) C_{m 3} = \\ = \dots + A_{12} B_{22} C_{23} + \dots \\ ⟹ B_{22} can affect (A B C)_{13} \\ Example: \\ A = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}), B = (\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}), C = (\begin{matrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}) \\ A B C = (\begin{matrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}) \\ B^{'} = (\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}), A B^{'} = (\begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) ⟹ A B^{'} C = (\begin{matrix} 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}) \end{matrix}

3b

\begin{matrix} Can a change in B_{13} affect (A B C)_{22} ? \\ If so, give an example \\ A B_{i j} = \sum_{k = 1}^{3} A_{i k} B_{k j} \\ A B C_{22} = \sum_{m = 1}^{4} A B_{2 m} C_{m 2} = \sum_{m = 1}^{4} (\sum_{k = 1}^{3} A_{2 k} B_{k m}) C_{m 2} = \\ = \dots + A_{21} B_{13} C_{32} + \dots \\ ⟹ B_{13} can affect (A B C)_{22} \\ Example: \\ A = (\begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \end{matrix}), B = (\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}), C = (\begin{matrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}) \\ A B C = (\begin{matrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}) \\ B^{'} = (\begin{matrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}), A B^{'} = (\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{matrix}) ⟹ A B^{'} C = (\begin{matrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \end{matrix}) \end{matrix}

4a

\begin{matrix} Given: A, B \in F^{n \times n} \\ Prove: t r (A + B) = t r (A) + t r (B) \\ t r (A + B) = \sum_{i = 1}^{n} (A + B)_{i i} = \sum_{i = 1}^{n} (A_{i i} + B_{i i}) = \\ = \sum_{i = 1}^{n} A_{i i} + \sum_{i = 1}^{n} B_{i i} = t r (A) + t r (B) \\ ⟹ t r (A + B) = t r (A) + t r (B) \end{matrix}

4b

\begin{matrix} Given: A \in F^{m \times n}, B \in F^{n \times m} \\ Prove: t r (A B) = t r (B A) \\ t r (A B) = \sum_{i = 1}^{m} (A B)_{i i} = \sum_{i = 1}^{m} (\sum_{k = 1}^{n} A_{i k} B_{k i}) = \sum_{k = 1}^{n} (\sum_{i = 1}^{m} A_{i k} B_{k i}) = \\ = \sum_{k = 1}^{n} (\sum_{i = 1}^{m} B_{k i} A_{i k}) = \sum_{k = 1}^{n} B A_{k k} = t r (B A) \\ ⟹ t r (A B) = t r (B A) \end{matrix}

5a

\begin{matrix} Given: α \in F; A \in F^{m \times n} \\ Prove: (α \cdot A)^{T} = α \cdot A^{T} \\ {\begin{matrix} (α \cdot A)_{j i}^{T} = (α \cdot A)_{i j} & = α \cdot A_{i j} \\ α \cdot A_{j i}^{T} & = α \cdot A_{i j} \end{matrix} \\ ⟹ (α \cdot A)_{j i}^{T} = α \cdot A_{j i}^{T} ⟹ (α \cdot A)^{T} = α \cdot A^{T} \end{matrix}

5b

\begin{matrix} Given: A \in F^{m \times n}; B \in F^{n \times p} \\ Prove: (A B)^{T} = B^{T} A^{T} \\ (A B)_{j i}^{T} = \sum_{k = 1}^{n} A_{i k} B_{k j} = \sum_{k = 1}^{n} A_{k i}^{T} B_{j k}^{T} = \sum_{k = 1}^{n} B_{j k}^{T} A_{k i}^{T} = (B^{T} A^{T})_{j i} \\ ⟹ (A B)_{j i}^{T} = (B^{T} A^{T})_{j i} ⟹ (A B)^{T} = B^{T} A^{T} \end{matrix}

5c

\begin{matrix} Given: A \in R^{n \times n} \\ Prove or disprove: A \cdot A^{T} = A^{T} \cdot A \\ Example: \\ A = (\begin{matrix} 1 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 1 \end{matrix}), A^{T} = (\begin{matrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 1 & 1 & 1 \end{matrix}) \\ A A^{T} = (\begin{matrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{matrix}), A^{T} A = (\begin{matrix} 3 & 0 & 3 \\ 0 & 0 & 0 \\ 3 & 0 & 3 \end{matrix}) \neq A A^{T} \\ ⟹ Disproved \end{matrix}

6

\begin{matrix} Given: A \in R^{n \times n} \\ Prove: \exists B, C \in R^{n \times n}, B = B^{T}, C = - C^{T} : A = B + C \\ Three criteria must be met: \\ \begin{matrix} 1. & B_{i j} = B_{j i}; B = f (A) \\ 2. & C_{i j} = - C_{j i}; C = g (A) \\ 3. & A = B + C \end{matrix} \\ Where f (x), g (x) are some functions \\ Example: \\ B = \frac{A + A^{T}}{2}, C = \frac{A - A^{T}}{2} \\ B_{i j} = {(\frac{A + A^{T}}{2})}_{i j} = \frac{A_{i j} + A_{j i}}{2} = \frac{A_{j i} + A_{i j}}{2} = {(\frac{A + A^{T}}{2})}_{j i} = B_{j i} \\ ⟹ B = B^{T} \\ C_{i j} = {(\frac{A - A^{T}}{2})}_{i j} = \frac{A_{i j} - A_{j i}}{2} = \frac{- (A_{j i} - A_{i j})}{2} = {(\frac{- (A - A^{T})}{2})}_{j i} = - C_{j i} \\ ⟹ C = - C^{T} \\ B + C = \frac{A + A^{T}}{2} + \frac{A - A^{T}}{2} = \frac{2 A}{2} = A \\ ⟹ B = B^{T}, C = - C^{T}, A = B + C \end{matrix}

\begin{matrix} Matrices A, B are called "commuting" if A B = B A \\ Given: A, B \in R^{n \times n} \end{matrix}

7a

\begin{matrix} Prove or disprove: (A = - A^{T}, B = - B^{T}) ⟹ (A + B) = - (A + B)^{T} \\ A_{i j} = - A_{j i} \\ B_{i j} = - B_{j i} \\ (A + B)_{i j} = A_{i j} + B_{i j} = (- A_{j i}) + (- B_{j i}) = - (A_{j i} + B_{j i}) = - (A + B)_{j i} \\ ⟹ (A + B)_{i j} = - (A + B)_{j i} ⟹ A + B = - (A + B)^{T} \end{matrix}

7b

\begin{matrix} Prove or disprove: A B = - (A B)^{T} ⟹ A B = B A \\ Example: \\ A = (\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}), B = (\begin{matrix} - 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}) \\ A B = (\begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ - 1 & 0 & 0 \end{matrix}), (A B)^{T} = (\begin{matrix} 0 & 0 & - 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{matrix}) = - A B \\ B A = (\begin{matrix} 0 & 0 & - 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{matrix}) \neq A B ⟹ Disproved \end{matrix}

7c

\begin{matrix} Given: A = A^{T}, B = B^{T} \\ Prove or disprove: A B = (A B)^{T} ⟺ A B = B A \\ 1. A B = (A B)^{T} ⟹ A B = B A \\ A B = (A B)^{T} ⟹ (A B)_{i j} = (A B)_{i j}^{T} = (A B)_{j i} = \sum_{k = 1}^{n} A_{j k} B_{k i} = \\ = \sum_{k = 1}^{n} B_{k i}^{T} A_{j k}^{T} = \sum_{k = 1}^{n} B_{i k} A_{k j} = (B A)_{i j} \\ A B = (A B)^{T} ⟹ (A B)_{i j} = (B A)_{i j} ⟹ A B = B A \\ A B = (A B)^{T} ⟹ A B = B A \\ 2. A B = B A ⟹ A B = (A B)^{T} \\ A B = B A ⟹ (A B)_{i j} = (B A)_{i j} = \sum_{k = 1}^{n} B_{i k} A_{k j} = \sum_{k = 1}^{n} B_{i k}^{T} A_{k j}^{T} = \\ = \sum_{k = 1}^{n} A_{j k} B_{k i} = (A B)_{j i} \\ A B = B A ⟹ (A B)_{i j} = (A B)_{j i} ⟹ A B = (A B)^{T} \\ A B = B A ⟹ A B = (A B)^{T} \\ 1. and 2. ⟹ A B = (A B)^{T} ⟺ A B = B A \end{matrix}