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I had difficulties understanding this question. Could you give me some advice how to approach this question? I couldn't create the relationship between the given features.

Let $x =\begin{bmatrix}1&2&0&-2\end{bmatrix}$ and $W=\operatorname{span}\left\{\underbrace{\begin{bmatrix}1\\1\\1\\1\end{bmatrix}}_{v_1},\underbrace{\begin{bmatrix}-1\\-1\\-1\\1\end{bmatrix}}_{v_2}\right\}\leqslant\Bbb R^4$. Find vectors $w_1\in W$ and $w_2\in W^\perp$ with $x = w_1 + w_2.$

What can I use to solve this question? Thank you very much.

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You can just write down the following system: $AX=0$ $$A=\begin{bmatrix}1&1&1&1\\-1&-1&-1&1\end{bmatrix}\cdot\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=0$$ The solution space to this system is $\Omega\leqslant\Bbb R^4$ and $\Omega=W^\perp$. Solve the system, find a basis and express $x$ in terms of it. Obviously $x_4=0$. $x_1+x_2=-x_3$ $x_1=t, x_2=s$

Your solution is of the following form: $$\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}t\\s\\-t-s\\0\end{bmatrix}=t\cdot\begin{bmatrix}1\\0\\-1\\0\end{bmatrix}+s\cdot\begin{bmatrix}0\\1\\-1\\0\end{bmatrix}$$ $$\implies\Omega=W^\perp=\operatorname{span}\left\{\begin{bmatrix}1\\0\\-1\\0\end{bmatrix},\begin{bmatrix}0\\1\\-1\\0\end{bmatrix}\right\}$$

Now we have: $$W\oplus\Omega=\Bbb R^4\iff\Bbb R^4=\operatorname{span}\left\{\begin{bmatrix}1\\1\\1\\1\end{bmatrix},\begin{bmatrix}-1\\-1\\-1\\1\end{bmatrix},\begin{bmatrix}1\\0\\-1\\0\end{bmatrix},\begin{bmatrix}0\\1\\-1\\0\end{bmatrix}\right\}$$

I hope it is now straight-forward.

This is the most efficient method. Just bear in mind the solution space $\Omega=W^\tau$, where $W$ is the row-space.

This is a concrete-explanation of @PedroTamaroff's answer.

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As caffeinemachine didn't implement the answer yet, I'd stole the idea and implement both approaches.
The first is more simple.
Let's $v_1'=\frac{v_1}{|v_1|}$ and $v_2'=v_2+av_1'$ for such $a$, that $v_1'.v_2'=0$: $$v_1'.v_2'=0$$ $$v_1'.(v_2+av_1')=0$$ $$v_1'.v_2+a=0$$ $$v_1'.v_2=-a,$$ so $v_2'=v_2-(v_1'.v_2)v_1'$ and $v_2''=\frac{v_2'}{|v_2'|}$.
Further, include $x$ into the set: $$\begin{cases} x'=x-bv_1'-cv_2''\\ x'.v_1'=0\\ x'.v_2''=0 \end{cases}$$ so $x'$ will be $w_2$ and $bv_1'+cv_2''$ will be $w_1$. $$\begin{cases} x'=x-bv_1'-cv_2''\\ (x-bv_1'-cv_2'').v_1'=0\\ (x-bv_1'-cv_2'').v_2''=0 \end{cases}$$ $$\begin{cases} x'=x-bv_1'-cv_2''\\ x.v_1'-bv_1'.v_1'-cv_2''.v_1'=0\\ x.v_2''-bv_1'.v_2''-cv_2''.v_2''=0 \end{cases}$$ $$\begin{cases} x'=x-bv_1'-cv_2''\\ x.v_1'-b\cdot 1-c\cdot 0=0\\ x.v_2''-b\cdot 0-c\cdot 1=0 \end{cases}$$ so $x'=x-(x.v_1')v_1'-(x.v_2'')v_2''$ and we're done.

The second, straightforward approach:
Let's say $x=w_1+w_2,\, w_1=av_1+bv_2,\, v_1.w_2=0,\, v_2.w_2=0$ so we have the system of linear equations and solve them for $a$, $b$: $a=-\frac12$, $b=-\frac32$, so $w_1=-\frac12 v_1-\frac32 v_2$, $w_2=x-w_1$.

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    $\begingroup$ It's not needed to find a $W^\perp$ basis here therefore it's much more simple than to guess $2$ more linearly independent vectors to extend $W$ to $\mathbb{R}^4$, IMHO. $\endgroup$ May 19 '20 at 19:49
  • $\begingroup$ We can also solve it via matrix. Since $W$ is a row-space of the coefficient matrix $A\in M_{2\times 4},\Omega=W^\perp$ $\endgroup$
    – Invisible
    May 19 '20 at 20:43

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