Given
W = { ( x 1 , x 2 , x 3 ) R 3 : x 2 + x 3 = 0 } W=\{(x_1,x_2,x_3)\R^3:x_2+x_3=0\} W = {( x 1 , x 2 , x 3 ) R 3 : x 2 + x 3 = 0 } Let a , b ∈ R & ( x 1 , x 2 , x 3 ) , ( y 1 , y 2 , y 3 ) ∈ R 3 a,b\in \R\&(x_1,x_2,x_3),(y_1,y_2,y_3)\in\R^3 a , b ∈ R & ( x 1 , x 2 , x 3 ) , ( y 1 , y 2 , y 3 ) ∈ R 3
Now,
a ( x 1 , x 2 , x 3 ) + b ( y 1 , y 2 , y 3 ) = ( a x 1 + b y 1 , a x 2 + b y 2 , a x 3 + b y 3 ) a(x_1,x_2,x_3)+b(y_1,y_2,y_3)=(ax_1+by_1,ax_2+by_2,ax_3+by_3) a ( x 1 , x 2 , x 3 ) + b ( y 1 , y 2 , y 3 ) = ( a x 1 + b y 1 , a x 2 + b y 2 , a x 3 + b y 3 ) Note that
a x 2 + b y 2 + a x 3 + b y 3 = a ( x 2 + x 3 ) + b ( y 2 + y 3 ) = 0 ⟹ ( a x 1 + b y 1 , a x 2 + b y 2 , a x 3 + b y 3 ) ∈ W ax_2+by_2+ax_3+by_3=a(x_2+x_3)+b(y_2+y_3)=0\\
\implies (ax_1+by_1,ax_2+by_2,ax_3+by_3)\in W a x 2 + b y 2 + a x 3 + b y 3 = a ( x 2 + x 3 ) + b ( y 2 + y 3 ) = 0 ⟹ ( a x 1 + b y 1 , a x 2 + b y 2 , a x 3 + b y 3 ) ∈ W Hence, W is a subspace of R 3 \R^3 R 3 .
Let us define a subspace
W 1 = { ( x 2 , x 2 , 0 ) } W 2 = { ( x 2 , 2 x 2 , 0 ) } W_1=\{(x_2,x_2,0)\}\\
W_2=\{(x_2,2x_2,0)\} W 1 = {( x 2 , x 2 , 0 )} W 2 = {( x 2 , 2 x 2 , 0 )} Clearly,
( ⋆ ) W 1 ∩ W 2 = { 0 } , W 1 ∩ W = { 0 } , W 2 ∩ W = { 0 } (\star)\,W_1\cap W_2=\{0\},W_1\cap W=\{0\},W_2\cap W=\{0\} ( ⋆ ) W 1 ∩ W 2 = { 0 } , W 1 ∩ W = { 0 } , W 2 ∩ W = { 0 } And,
R 3 = W ⊕ W 1 \R^3=W\oplus W_1\\ R 3 = W ⊕ W 1 As, for any v ⃗ ∈ R 3 \vec v\in \R^3 v ∈ R 3 we can write v ⃗ = u ⃗ + w ⃗ \vec v=\vec u+\vec w v = u + w for some u ⃗ ∈ W , w ⃗ ∈ W 1 \vec u\in W,\vec w\in W_1 u ∈ W , w ∈ W 1 and hold ( ⋆ ) (\star) ( ⋆ ) .
Similarly for
R 3 = W ⊕ W 2 \R^3=W\oplus W_2 R 3 = W ⊕ W 2
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