Pregunta:
Sea $\mathbb{F}$ sea $\mathbb{Z}_7$ . Sea $A=\begin{bmatrix} 2 & 5\\ 5 & 3 \end{bmatrix} \in M_{2\times 2}( F)$ .
Sea $W=\left\{B\in M_{2\times 2}( F)\Bigl|( AB)^{t} =AB\right\}$ sea un subespacio sobre $\mathbb{F}$ .
Encuentre una base y una dimensión para $W$ .
Mi intento:
$Solution.$
$\text{By the given information we have the following: }$ \begin{gather*} AB=\begin{bmatrix} 2 & 5\\ 5 & 3 \end{bmatrix} \cdotp \begin{bmatrix} a & b\\ c & d \end{bmatrix} = \begin{bmatrix} 2a+5c & 2b+5d\\ 5a+3c & 5b+3d \end{bmatrix} \\ \fin \begin{gather*} AB^{t} =\begin{bmatrix} 2a+5c & 5a+3c\\ 2b+5d & 5b+3d \end{bmatrix} \\ \fin $\text{By W's condition we get:}$
\begin{gather*} AB^{t} =AB\Longrightarrow \begin{bmatrix} 2a+5c & 5a+3c\\ 2b+5d & 5b+3d \end{bmatrix} = \begin{bmatrix} 2a+5c & 2b+5d\\ 5a+3c & 5b+3d \end{bmatrix} \\ \\ \Flecha larga 2b+5d = 5a+3c \\ \Flecha larga d=5a+3c-2b}{5} =a+3c-2b}{5} =a+3c+overbrace{7c}^{0} -2b+overbrace{7b}^{0}{5} =a+2c+b \fin $\text{So $ AB $ is depends on 3 free-parameters:}$ \begin{equation*} \begin{aligned} AB=\begin{bmatrix} 2a+5c & 2b+5d\\ 5a+3c & 5b+3d \end{bmatrix} & & = & & \begin{bmatrix} 2a+5c & 5a+3c\\ 5a+3c & 5b+3( a+2c+b) \end{bmatrix} \\ & & & & \begin{bmatrix} 2a+5c & 5a+3c\\ 5a+3c & 5b+3a+6c+3b \end{bmatrix} \\ & & & & \begin{bmatrix} 2a+5c & 5a+3c\\ 5a+3c & 8b+3a+6c \end{bmatrix} \\ & & & & \begin{bmatrix} 2a+5c & 5a+3c\\ 5a+3c & b+3a+6c \end{bmatrix} \end{aligned} \end{equation*} $\text{We take all the paramaters out of the marices, so we can get the basis vectors.}$
$\text{We get the following:}$
\begin{equation*} \begin{bmatrix} 2a+5c & 5a+3c\\ 5a+3c & b+3a+6c \end{bmatrix} =a\cdotp \begin{bmatrix} 2 & 5\\ 5 & 3 \end{bmatrix} +b\cdotp \begin{bmatrix} 0 & 0\\ 1 & 0 \end{bmatrix} +c\cdotp \begin{bmatrix} 5 & 3\\ 3 & 6 \end{bmatrix} \end{equation*} $\text{Which isomorphic to the following vectors: }$ \begin{equation*} ( 2,5,5,3) ,( 0,0,1,0) ,( 5,3,3,6) \end{equation*} $\text{respectively.}$
$\text{Now, we shall check whether those vectors are linear independent. }$
\begin{gather*} \begin{bmatrix} 2 & 5 & 5 & 3\\ 0 & 0 & 1 & 0\\ 5 & 3 & 3 & 6 \end{bmatrix} \xflecha derecha[ \begin{array}{l} \mathcal{L}_{3} +\mathcal{L}_{1}\rightarrow \mathcal{L}_{3}\\ \end{array}]{}\begin{bmatrix} 2 & 5 & 5 & 3\\ 0 & 0 & 1 & 0\\ 0 & 1 & 1 & 2 \end{bmatrix}\xrightarrow[\mathcal{L}_{1} -5\mathcal{L}_{3}\rightarrow \mathcal{L}_{1}]{} \begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 1 & 1 & 2 \end{bmatrix} \\ \\ \xflecha derecha[ \begin{array}{l} \frac{\mathcal{L}_{1}}{2}\rightarrow \mathcal{L}_{1}\\ \mathcal{L}_{3} -\mathcal{L}_{2}\rightarrow \mathcal{L}_{3} \end{array}]{}\mathcal{\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 2 \end{bmatrix}}\xrightarrow[\mathcal{L}_{3}\leftrightarrow \mathcal{L}_{2}]{}\mathcal{ \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 0 \end{bmatrix}} \end{gather*} $\text{Therefore, the vectors are linear independent, so they are basis of $ \Estilo de visualización W $, }$ $\text{and since we have 3 linear independent vectors, we conclude that:}$ \begin{equation*} \dim W=3 \end{equation*}
Pensamientos: ¿Es correcto lo que he escrito? ¿O quizás me he perdido algo? No puedo ver si estoy bien o mal, porque no he resuelto cuestiones de encontrar una base para matrices subespacios. Estaré encantado de recibir ayuda. Gracias.
