Question

W={(x1, x2, x3)∈R3: x2+x3=0}. Find two subspace W1,W2 of R such that R3= W⊕W1 and R3=W⊕W2 but W1≠W2

Answer 1

Let denote $V=\mathbb{R}^3$ be a real vector space.

Given that $W=\{(x_1,x_2,x_3) \in V|\: x_2+x_3=0\}$ subspace of $V$ .

Let, $W_1=\{(x_1,x_1,x_1)\in V : x_1\in\mathbb{R}\}$ and $W_2=\{(2x_1,-x_1,x_1)\in V: \: x_1\in \mathbb{R}\}$ ,clearly, $W,W_1,W_2$ all different as a set.

Claim 1: $W_1 ,W_2$ are subspace of $V$

Proof:

Let, $a,b,c,d \in \mathbb{R}$ and for every $u,v \in W_1 \: \&\: w,y \in W_2$ consider the linear combination as follows,

Case-I: clearly, $u=(u_1,u_1,u_1)\&v=(v_1,v_1,v_1)$ such that,

$$au+bv=(au_1+bv_1,au_1+bv_1,au_1+bv_1)$$

which implies,

$$au+bv\in W_1$$

Hence $W_1$ is vector subspace.

Case-II: In a similar manner as case-I , consider the

$$cw+dy\in W_2$$

Thus our claimed is proved.

Claim 2: $V=W \oplus W_1$

Proof:

Clearly, we see that

$$V=W+ W_1$$

Thus we will show that $W\cap W_1=\{0\}$ . Suppose for any $w\in W \cap W_1$ such that $w\neq0$ ,thus $w\in W \& w\in W_1$ which implies,

$$(x_1,x_2,-x_2)=(x_1,x_1,x_1)\\\implies (x_1,x_2,x_3)=(0,0,0)$$

Hence we arrived at contradiction that $w\neq 0$ .

Thus,

$$V=W\oplus W_1$$

We are done.

Claim 3: $V=W\oplus W_2$

Proof:

Observe that $W+W_2=\{(3x_1,x_2-x_1,x_3+x_1): x_1,x_2,x_3 \in \mathbb{R}\}\supset \mathbb{R}^3$ which implies

$$V=W+W_2 \hspace{1cm}(\because W+W_1\subset V)$$

Now, we will show that

$$W\cap W_2=\{0\}$$

Clearly, exactly applying the same arguments as in claim 2 we get,$W\cap W_2=\{0\}$ .

Thus,

$$V=W\oplus W_2$$

Hence, we are done.