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I'm currently in a linear algebra course and I am just looking for feedback or some sort of verification of what I'm doing is correct.

So I have to find all vectors that are orthogonal to $u=(1, -2, 2, 1)$.

Seeing as this vector is in $R^4$, we let the vector $v=(v_1, v_2, v_3, v_4)$.

We also know that a vector is orthogonal to another, when the dot product of $u$ and $v$, $$u\cdot v=0$$

$$u\cdot v=(1, -2, 2, 1)\cdot(v_1, v_2, v_3, v_4)=v_1-2v_2+2v_3+v_4=0$$

This implies that $v_1=2v_2-2v_3-v_4$

Which means every vector that is orthogonal to the vector $(1,-2,2,1)$ will be in the form $v = (t, 2t, -2t, t)$ or $v = t(1,2,-2,1)$, letting $t$ be any real number.

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  • $\begingroup$ how did you reach your last (wrong) conclusion? $\endgroup$ – Exodd Apr 11 '17 at 17:13
  • $\begingroup$ You should more focus on the parameters you have found. $\endgroup$ – Alessandro Blasetti Apr 11 '17 at 17:23
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There is a problem in your conclusion. Everything is correct until you say that a vector $\overrightarrow{v}=(v_1,v_2,v_3,v_4)$ is orthogonal to the vector $\overrightarrow{u}=(1,-2,2,1)$ implies $v_1=2v_2-2v_3-v_4$.

From that point, the use of the $t$ is a bit weird: notice that the only thing we know is that given values for $v_2,v_3,v_4$, the value of $v_1$ will be completely determined. That is, the variable $v_1$ DEPENDS on the values of $v_2,v_3,v_4$, and the latter might be seen as INDEPENDENT variables.

If we write $v_2=r,v_3=s,v_4=t$ (which is simply a change of variables, and although popular it is completely unnecessary) we would get:

\begin{align*} \overrightarrow{v}&=\begin{bmatrix}v_1\\ v_2\\ v_3\\ v_4\end{bmatrix}=\begin{bmatrix}2v_2-2v_3-v_4\\ v_2\\ v_3\\ v_4\end{bmatrix} &(\text{changing $v_1$ in terms of $v_2,v_3,v_4$)}\\ &=\begin{bmatrix}2r-2s-t\\r\\ s\\ t\end{bmatrix} &\text{(changing the variables)}\\ &=r\begin{bmatrix}2\\1\\0\\0\end{bmatrix} +s\begin{bmatrix}-2\\0\\1\\0\end{bmatrix}+t \begin{bmatrix}-1\\ 0\\ 0\\ 1\end{bmatrix} \end{align*}

Hence, you can describe all the vectors that orthogonal to $\overrightarrow{u}=(1,-2,2,1)$ in several (equivalent) ways:

  • Vectors of the form $\overrightarrow{v}=(2r-2s-t,r,s,t)$ where $r,s,t\in\mathbb{R}$.
  • All linear combinations of the vectors $(2,1,0,0),(-2,0,1,0),(-1,0,0,1)$.
  • Vectors in the subspace $W=\operatorname{span}\big((2,1,0,0),(-2,0,1,0),(-1,0,0,1)\big)$.
  • Vectors of the form $(x,y,z,w)$ solving the equation $x-2y+2z+w=0$.
  • Vectors contained in the hyperplane $x-2y+2z+w=0$.

Personally, I like the third and the fifth options.

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  • $\begingroup$ I'm a fan of option 3, but I don't quite understand the benefit of option 5. Would you mind explaining that to me? $\endgroup$ – Mitchell Faas Apr 11 '17 at 17:25
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    $\begingroup$ If the problem were in $\mathbb{R}^3$, you could use some software to see geometrically what is the set of vectors satisfying the condition. Usually, those programs can draw the plane given the equation of the plane, but they cannot if you just provide the generators. $\endgroup$ – Darío G Apr 11 '17 at 17:27
  • $\begingroup$ Ok, I see where you are getting at. I'm going to go work the problem some more. $\endgroup$ – pudge Apr 11 '17 at 17:52
  • $\begingroup$ I reworked and thought about the problem some more, I understand it now. Honestly, not sure where I went with my conclusion after looking at this. Thanks for the help! $\endgroup$ – pudge Apr 11 '17 at 18:11
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The last conclusion is wrong, $v_1 = 2v_2 - 2v_3 - v_4\space$implies that $V=\{(2a-2b-c,a,b,c);a,b,c\in \mathbb R \}$

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