# Let $W_{1},W_{2}$ be sub-spaces of $\mathbb{R}^{4}$, find a subspace $W_{3}$ s.t $W_3\subset W_{2}$ and $W_{1}\oplus W_{3}=W_{1}+W_{2}$

Let $W_{1},W_{2}$ be linear sub-spaces of $\mathbb{R}^{4}$.

$W_{1}=\text{sp}\{(1,2,3,4),(3,4,5,6),(7,8,9,10)\}$

$W_{2}=\text{sp }\{(x,y,z,w)| \ x+y=0\}$

Find a linear subspace of $\ \mathbb{R}^{4} \ ; W_{3}$ , such that: $W_{3}\subset W_{2} \ \text {*and*} \ W_{1}\oplus W_{3}=W_{1}+W_{2}$

My attempt:

I applied Gaussian elimination on the vectors of $W_1$, such that:

$$W_{1}=\left\{ \left(\begin{array}{c} x\\ y\\ z\\ w \end{array}\right)=a\left(\begin{array}{c} 1\\ 0\\ -2\\ -1 \end{array}\right)+b\left(\begin{array}{c} 0\\ 1\\ -2\\ -1 \end{array}\right)|a,b\in\mathbb{R}\right\}$$

At that point I got stuck. I'm not sure how to continue.

• Your Gaussian elimination seems wrong. How can $(1,2,3,4)$ belong to $W_{1}$ if your Gaussian elimination is correct ? – Pjonin Aug 8 '18 at 14:09
• checked it in wolfam alpha, it's correct – Jneven Aug 8 '18 at 14:13
• It is « obviously » as you say, not correct. Give me your $a$ and your $b$ such as $(1,2,3,4)$ belongs to your $W_{1}$. – Pjonin Aug 8 '18 at 14:53
• @Jneven Pjonin is right there is some problem in the basis you have found for $W_1$. Check again it is trivial also by inspection that (1,1,1,1) and (0,1,2,3) work fine. – gimusi Aug 8 '18 at 15:46

HINT

We should write as a span (check again you derivation by RREF)

$$W_{1}=\left\{ \left(\begin{array}{c} x\\ y\\ z\\ w \end{array}\right)=s\left(\begin{array}{c} 1\\ 1\\ 1\\ 1 \end{array}\right)+t\left(\begin{array}{c} 0\\ 1\\ 2\\ 3 \end{array}\right)\right\}$$

and we can also easily find that

$$W_{2}=\left\{ \left(\begin{array}{c} x\\ y\\ z\\ w \end{array}\right)=r\left(\begin{array}{c} 1\\ -1\\ 0\\ 0 \end{array}\right)+s\left(\begin{array}{c} 0\\ 0\\ 1\\ 0 \end{array}\right)+t\left(\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right)\right\}$$

then check by RREF on the $5$ basis vectors that $W_{1}+W_{2}$ has dimension $4$ and finally select 2 basis vectors for $W_3$ from the basis vectors of $W_2$ such that $W_{1}\oplus W_{3}=\mathbb{R^4}$.

(I won't change the notation although it's not 100% correct)

Hint: Rewrite the basis of $W_1$ as $$W_{1}=\left\{ \left(\begin{array}{c} x\\ y\\ z\\ w \end{array}\right)=a\left(\begin{array}{c} 1\\ 0\\ -2\\ -1 \end{array}\right)+b\left(\begin{array}{c} 1\\ -1\\ 0\\ 0 \end{array}\right)\right\}$$

A basis of $W_2$ is $$W_{2}=\left\{ \left(\begin{array}{c} x\\ y\\ z\\ w \end{array}\right)=a\left(\begin{array}{c} 1\\ -1\\ 0\\ 0 \end{array}\right)+b\left(\begin{array}{c} 0\\ 0\\ 1\\ 0 \end{array}\right)+c\left(\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right)\right\}$$

can you take it from here?

• to see that I got this right (this what I was thinking needed to be done) - I looked for a vector $x$ that applies $x \in W_1 \cap W_2$, and then tried to transform this expression into a matrix. but my matrix was in size of $4 \times 5$ which got me completely confused. – Jneven Aug 8 '18 at 14:22

Hint : what is the dimension of $W_{1}$ ? As a consequence, what should be the dim of $W_{3}$ ? What must verify a vector in $W_{3}$?

• obviously dim $W_1 =2$ and dim $W_1 \leq 3$ – Jneven Aug 8 '18 at 14:14
• Be more modest. Seems not that obvious since you are giving a wrong answer. – Pjonin Aug 8 '18 at 17:05