# Find the orthogonal bases of the space $V$ and $V^{\perp}$

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?

• I have asked this again with fixed symbols – VirtualUser Jan 3 at 12:07
• 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. – A.Γ. Jan 3 at 12:25

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:

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

• There is no mistake. It says an orthogonal basis, not orthonormal. – Shubham Johri Jan 3 at 12:30
• @ShubhamJohri OP said they used Gram-Schmidt. Gram-Schmidt yields the basis from my answer. – mechanodroid Jan 3 at 12:35
• @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 – Shubham Johri Jan 3 at 12:45
• @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 $(a,b,c)=k(1,-3,1),k\in\Bbb R$. – Shubham Johri Jan 3 at 15:38
• 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. – Shubham Johri Jan 3 at 15:42

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 $$\{\frac{1}{\sqrt{11}}(1,-3,1)'\}$$.)