Answer to Question #144525 in Linear Algebra for Sourav Mondal

Question

Let
W = ((X1, X2, X3) belongs to R³ : X2 + X3 = 0).
Show that W is a subspace of R³ . Find two
subspaces W1 and W2 of R3 such that
R³ = W (direct sum )W1 and R³ = W (direct sum) W2 but W1 (not equal to) W2.

Answer 1

Given

$$W=\{(x_1,x_2,x_3)\R^3:x_2+x_3=0\}$$

Let $a,b\in \R\&(x_1,x_2,x_3),(y_1,y_2,y_3)\in\R^3$

Now,

$$a(x_1,x_2,x_3)+b(y_1,y_2,y_3)=(ax_1+by_1,ax_2+by_2,ax_3+by_3)$$

Note that

$$ax_2+by_2+ax_3+by_3=a(x_2+x_3)+b(y_2+y_3)=0\\ \implies (ax_1+by_1,ax_2+by_2,ax_3+by_3)\in W$$

Hence, W is a subspace of $\R^3$ .

Let us define a subspace

$$W_1=\{(x_2,x_2,0)\}\\ W_2=\{(x_2,2x_2,0)\}$$

Clearly,

$$(\star)\,W_1\cap W_2=\{0\},W_1\cap W=\{0\},W_2\cap W=\{0\}$$

And,

$$\R^3=W\oplus W_1\\$$

As, for any $\vec v\in \R^3$ we can write $\vec v=\vec u+\vec w$ for some $\vec u\in W,\vec w\in W_1$ and hold $(\star)$ .

Similarly for

$$\R^3=W\oplus W_2$$