Let $f$ : ${R}^4\times{{R}^4}$ $\longrightarrow$ ${R}$ be a bilinear map
such that $f(($x$,$y$,$z$,$t$), ($a$,$b$,$c$,$d$))$ = $a$$x$ + $b$$y$ + $c$$z$
$1$. If $M$ = {($x$,$y$,$z$,$t$) | $x$ + $y$ - $z$ = $0$ and $t$ = $x$ + $y$ } is a subspace of ${R}^4$ then
what is the dimension of $M^\top$ + $M$ and $M^\top$ $\cap$ $M$
$2$ . If $T$ = { ($1$,$1$,$1$,$1$) , ($1$,$1$,$1$,$0$) , ($1$,$1$,$0$,$0$) , ( $1$ ,$0$ , $0$ , $0$ ) } and $S$ = {$e_{1}$ , $e_{2}$ , $e_{3}$ , $e_{4}$ } are basis of ${R}^4$
then find the matrix $A$ and matrix $B$ such that the matrix representation $M($f$,$T$,$S$)$ = $A$ $M($f$,$T$,$S$)$ $B^\top$ where $B$ is regular matrix
Edit : For the first part I found that
$M^\top$ = { ($1$ , $1$ , $-1$ , $x$ ) | $x$ $\in$ $R$ }
so { ($1$ , $1$ , $-1$ , $0$) , ($0$ , $0$ , $0$ , $1$)} is a basis of $M^\top$
and the set { ($1$ , $0$ , $1$ , $1$) , ($0$ , $1$ , $1$ , $1$) is a basis of $M$
and $M^\top$ $\cap$ $M$ is empty set but I know that empty set is not subspace what I can say about it's dimension $?$
For the second part I don't know how I can start any help please,
In general the intersection $M\cap M^{\bot}$ is $\{0\}$, not the empty set since for $v\in M\cap M^{\bot}$ you have $||v||^2=\langle v,v \rangle = 0 \Rightarrow v=0$ by assumption and the other inclusion is trivial. Thus the intersection has dimension $0$