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.

• Are you familiar with the Gram-Schmidt orthonormalization? May 18 '20 at 23:29
• Here is a MathJax tutorial tutorial. May 19 '20 at 18:23
• May 19 '20 at 18:49
• Also, an abstract duplicate of Find a basis for $W^\perp$ for $W=\{(x_1,x_2,x_3)\in\mathbb{R}^3:x_1-x_2-x_3=0\}$. May 19 '20 at 18:55
• @Orchid_2.718281828 Oh, I've been just learning the system I forgot, thank you I will
– Kaan
May 19 '20 at 22:46

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.

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 $$w_1=-\frac12 v_1-\frac32 v_2$$, $$w_2=x-w_1$$.

• 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. May 19 '20 at 19:49
• We can also solve it via matrix. Since $W$ is a row-space of the coefficient matrix $A\in M_{2\times 4},\Omega=W^\perp$ May 19 '20 at 20:43