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In space $\mathbb R^3 $ f Find the orthogonal bases of the space $ V $ and $ V^{\perp}$ where $$V = \left\{ \vec{x} \in \mathbb R^3 : x_1 - 3x_2 + x_3 = 0 \right\} $$ On the begining, I know that may I ask you for basics of topic, but I truly have some doubts...

Ok, I found basis of $V$: $ [3,1,0]^T,[-1,0,1]^T$
and I have used Gram–Schmidt process to find orthogonal basis of $V$:
$[3,1,0]^T,[-1/10,3/10,1]^T$

Ok, now I should find basis of $ V^{\perp} $ and on the same algorithm find orthogonal basis of $V^{\perp}$ - but there is my question - if in basis of $V^{\perp}$ I have all vectors which are perpendicular to vectors from basis $V$, the basis of $ V^{\perp} $ will be just orthogonal basis of $V$? And orthogonal basis of $V^{\perp}$ will be just basis of $V$? Have I right or there is something other to do?

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  • $\begingroup$ I have asked this again with fixed symbols $\endgroup$ – VirtualUser Jan 3 at 12:07
  • $\begingroup$ You could simply edit the old question fixing those symbols (see the edit button). It is easier than retyping the whole thing as a new question. $\endgroup$ – A.Γ. Jan 3 at 12:25
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You have a mistake in your Gram-Schmidt, it should be:

$$\left\{\frac1{\sqrt{10}}\begin{bmatrix} 3 \\ 1 \\ 0\end{bmatrix}, \frac1{\sqrt{110}}\begin{bmatrix} -1 \\ 3 \\ 10\end{bmatrix}\right\}$$

To find the orthonormal basis for $V^\perp$, notice that $\dim V^\perp = \dim\mathbb{R}^3 - \dim V = 3-2 = 1$ so it suffices to find one vector perpendicular to $V$ and normalize it:

$$\frac1{\sqrt{11}}\begin{bmatrix} 1 \\ -3 \\ 1\end{bmatrix}$$

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  • $\begingroup$ There is no mistake. It says an orthogonal basis, not orthonormal. $\endgroup$ – Shubham Johri Jan 3 at 12:30
  • $\begingroup$ @ShubhamJohri OP said they used Gram-Schmidt. Gram-Schmidt yields the basis from my answer. $\endgroup$ – mechanodroid Jan 3 at 12:35
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    $\begingroup$ @VirtualUser Since you only want an orthogonal basis, normalization is not necessary, but it won't get you a wrong answer. An orthonormal basis is orthogonal too $\endgroup$ – Shubham Johri Jan 3 at 12:45
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    $\begingroup$ @VirtualUser That is correct. Take $(a,b,c)\in V^\perp$, then $\forall(x_1,x_2,x_3)\in V$, we have $(a,b,c)\cdot(x_1,x_2,x_3)=0$. Now take $(x_1,x_2,x_3)$ as any two vectors in $V$ to get $(a,b,c)=k(1,-3,1),k\in\Bbb R$. $\endgroup$ – Shubham Johri Jan 3 at 15:38
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    $\begingroup$ Alternatively, it is pretty easy to observe that $V$ is the plane with the equation $x-3y+z=0$, and any vector in $V^\perp$ will be parallel to the normal to the plane whose direction ratios are given by the coefficients of $x,y,z$ in the equation of the plane. $\endgroup$ – Shubham Johri Jan 3 at 15:42
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No, it's wrong. If you have an orthogonal basis $\mathcal{B} = \{v_1,\cdots, v_n\}$ of $V$, then for each $v_i$, it is perpendicular to other $v_j$'s but not to itself. So $v_i\notin V^\perp$ for all $i$. By the way, you can find an orthonormal basis of $V^\perp$ easily from the definition of $V$ as $$ V = \{(1,-3,1)\}^\perp. $$ (It's just $\{\frac{1}{\sqrt{11}}(1,-3,1)'\}$.)

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