Question

Considere os subconjuntos V e W de M2,2 (conjunto das matrizes quadraticas
de ordem 2).

Answer 1

Resposta:

Explicação passo-a-passo:

A)

 $V=\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] =\left[\begin{array}{ccc}a&b\\b&a\\\end{array}\right]=a\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]+b\left[\begin{array}{ccc}0&1\\1&0\\\end{array}\right]\\\\B_V=({\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right],\left[\begin{array}{ccc}0&1\\1&0\\\end{array}\right]})$

Base de V

 

$W=\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] =\left[\begin{array}{ccc}a&b\\a&b\\\end{array}\right]=a\left[\begin{array}{ccc}1&0\\1&0\\\end{array}\right]+b\left[\begin{array}{ccc}0&1\\0&1\\\end{array}\right]\\\\B_W=({\left[\begin{array}{ccc}1&0\\1&0\\\end{array}\right],\left[\begin{array}{ccc}0&1\\0&1\\\end{array}\right]})$

Base de W

Fácil ver que São subespaços de $M_2_x_2$

$B_M_2_x_2=(\left[\begin{array}{ccc}1&0\\0&0\\\end{array}\right],\left[\begin{array}{ccc}0&1\\0&0\\\end{array}\right],\left[\begin{array}{ccc}0&0\\1&0\\\end{array}\right],\left[\begin{array}{ccc}0&0\\0&1\\\end{array}\right])$

B) V∩W Para que haja esta intersecção precisamos verificar se existem escalares: x,y,z,w tais que:

$x\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]+y\left[\begin{array}{ccc}0&1\\1&0\\\end{array}\right]=z\left[\begin{array}{ccc}1&0\\1&0\\\end{array}\right]+w\left[\begin{array}{ccc}0&1\\0&1\\\end{array}\right]\\\\ \left[\begin{array}{ccc}x&y\\y&x\\\end{array}\right]=\left[\begin{array}{ccc}z&w\\z&w\\\end{array}\right] \\\\ \left \{ {{x=z=y} \atop {x=w=y}} \right.$

V∩W Será formado pelos vetores da forma:

$\left[\begin{array}{ccc}x&x\\x&x\\\end{array}\right]=x\left[\begin{array}{ccc}1&1\\1&1\\\end{array}\right]$

V∩W $=(\left[\begin{array}{ccc}1&1\\1&1\\\end{array}\right])$  

É subespaço De M2x2 (Fácil verificar)

Seja v,u ∈ V∩W

i) Vetor Nulo está em  V∩W

ii) Soma u+v estará em V∩W

iii) Tome um escalar s∈ R, sv∈V∩W