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Find a set of vector B such that $$\operatorname{span}(B)=\{(x,y,z,w)^T\mid 2x+y+w=0, y+2z=0\}$$ in $\mathbb{R}^4$.

This is what I have so far I don't know if it is correct.

$$ x\begin{pmatrix} a_1 \\ b_1 \\ c_1 \\ d_1 \end{pmatrix}+ y\begin{pmatrix} a_2 \\ b_2 \\ c_2 \\ d_2 \end{pmatrix}+ z\begin{pmatrix} a_3 \\ b_3 \\ c_3 \\ d_3 \end{pmatrix}+w \begin{pmatrix} a_4 \\ b_4 \\ c_4 \\ d_4 \end{pmatrix} $$ where $2x+y+w=0$ and $y+2z=0$. Then $w=-2x-y$, $y=2z$, $w=-2x+2z$. Then $$ x\begin{pmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{pmatrix}+ y\begin{pmatrix} α_1 \\ α_2 \\ α_3 \\ α_4 \end{pmatrix}+ z\begin{pmatrix} -2c_1 \\ -2c_2 \\ -2c_3 \\ -2c_4 \end{pmatrix} +w\begin{pmatrix} -2b-α_1 \\ -2b-α_2 \\ -2b-α_3 \\ -2b-α_4 \end{pmatrix}$$

Thus $$2b+α_1+-2b_1-α_1=0$$ and $$α_1-2c=0$$ where $$c_1=1/2α_1$$

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Not sure where you deduced $w=-2x-y$ and $y+2z=0$.

What you need to find is the solution set of $$ \begin{cases} 2x+y+w=0 \\ y+2z=0 \end{cases} $$ so $w=-2x-y$ and $y=-2z$; hence $w=-2x+2z$, so the vectors are of the form $$ \begin{pmatrix} a \\ -2b \\ b \\ -2a-2b \end{pmatrix} $$ where $a$ and $b$ are arbitrary. Since $$ \begin{pmatrix} a \\ -2b \\ b \\ -2a-2b \end{pmatrix} = a\begin{pmatrix} 1 \\ 0 \\ 0 \\ -2 \end{pmatrix} + b\begin{pmatrix} 0 \\ -2 \\ 1 \\ -2 \end{pmatrix} $$ you are done.

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  • $\begingroup$ Thanks, it helped me a lot. $\endgroup$ – marie Feb 15 '18 at 17:27
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The following constrains imply that the Span (B) is a two dimentional subspace of $\mathbb{R}^4$

$$ \begin{cases} 2x+y+w=0 \\ y+2z=0 \end{cases}$$

Thus it suffices to find two linearly independent vectors such as $$b_1 =\begin{pmatrix} 1 \\ -2 \\ 1 \\ 0 \end{pmatrix}$$

and

$$ b_2= \begin{pmatrix} -1 \\ 0 \\ 0 \\ 2 \end{pmatrix}$$

Which satisfy the constrains and let $$B= \text {{$b_1$, $b_2$}} $$

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