Propiedades de la transposición de matrices

1. $\left( A ^ t \right)^t = A \ \ \ \ \ \ \ \ \ \ \forall A_{m\times n} $
2. $\left( A + B \right)^t = A^t + B^t \ \ \ \ \ \ \ \ \ \ \forall A,B \in M(m,n) $
3. $\left( A x \right)^t = x^t A^t \ \ \ \ \ \ \ \ \ \ \forall A_{m\times n} , x \in \mathbb{R}^n$
4. $\left( y ^ t A \right)^t = A^t y \ \ \ \ \ \ \ \ \ \ \forall A_{m\times n} , y \in \mathbb{R}^m$
5. $\left( A B \right)^t = B^t A^t \ \ \ \ \ \ \ \ \ \ \forall A_{m\times n} , B_{n\times k} $

Demostración. $$ $$

1. Sea $A= \left[ \begin{array}{c|c|c|c} a_1 & a_2 & \ldots & a_n \end{array} \right] $ $$ \Rightarrow A^t = \begin{pmatrix}{a_1}^t \\ {a_2}^t \\ \vdots \\ {a_n}^t\end{pmatrix} \Rightarrow ( A^t ) ^ t = \left[ \begin{array}{c|c|c|c} a_1 & a_2 & \ldots & a_n \end{array} \right] = A $$ $$ \therefore \ \ ( A^t ) ^ t = A \ \ \blacksquare $$
2. $ (A + B) ^ t = \left[ \begin{array}{c|c|c|c} a_1 + b_1& a_2+b_2 & \ldots & a_n+b_n \end{array} \right]^ t $ $$= \begin{pmatrix} (a_1 + b_1)^t \\ (a_2 + b_2)^t \\ \vdots \\ (a_n + b_n)^t \end{pmatrix} = \begin{pmatrix} {a_1}^t+ {b_1}^t \\ {a_2}^t+ {b_2}^t \\ \vdots \\ {a_n}^t+ {b_n}^t \end{pmatrix} = \begin{pmatrix} {a_1}^t \\ {a_2}^t \\ \vdots \\ {a_n}^t \end{pmatrix} + \begin{pmatrix} {b_1}^t \\ {b_2}^t \\ \vdots \\ {b_n}^t \end{pmatrix} = A^t + B^t $$ $$ \therefore \ \ (A+B)^t = A^t + B^t \ \ \blacksquare $$
3. $Ax = \begin{pmatrix}{r_1}^t x \\ {r_2}^t x \\ \vdots \\ {r_m}^t x\end{pmatrix} $ $$\Rightarrow \ (Ax)^t = \left[ {r_1}^t x \ \ \ {r_2}^t x \ \ \ldots \ \ {r_m}^t x \right] = \left[ {x}^t r_1 \ \ \ x^t r_2 \ \ \ldots \ \ x^t r_m \right] = x^t A^t$$ $$\therefore \ \ (Ax)^t = x^t A^t \ \ \blacksquare $$
4. $ y^t A = \left[ y^t a_1 \ \ \ y^t a_2 \ \ \ldots \ \ y^t a_n \right] $ $$ \Rightarrow \ (y^t A) ^t = \begin{pmatrix}y^t a_1 \\ y^t a_2 \\ \vdots \\ y^t a_n \end{pmatrix} = \begin{pmatrix} {a_1}^t y \\ {a_2}^t y \\ \vdots \\ {a_n}^t y \end{pmatrix} = A^t y $$ $$\therefore \ \ (y^t A)^t = A^t y \ \ \blacksquare $$
5. $AB = \left[ \begin{array}{c|c|c|c} Ab_1 & Ab_2 & \ldots & Ab_k \end{array} \right]$ $$ \Rightarrow \ (AB)^t = \begin{pmatrix} (A b_1)^t \\ (A b_2)^t \\ \vdots \\ (A b_k)^t \end{pmatrix} = \begin{pmatrix} {b_1}^t A^t \\ {b_2}^t A^t \\ \vdots \\ {b_k}^t A^t \end{pmatrix} = B^t A^t$$ $$\therefore \ \ (AB)^t = B^t A^t \ \ \blacksquare $$