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$n>1$ Given is $$V = \left\{ \vec{x} \in \mathbb R^n : x_1+x_2 + ... + x_n = 0 \right\} $$ a) Find orthogonal basis of $V^{\perp} $
b) Find orthogonal projection $\vec{x} = [n,0,0,...,0]^T$ on subspace $V$

If it comes to a)
$$\dim V^{\perp} = n - \dim V = \dim V^{\perp} = n - n + 1 = 1$$ So $V^{\perp} = span$ one_vector_perpendicular_to_v

Put $[1,1,1,...,1,1]^T$ - it is perpendicular to $V$

Let's start Gram–Schmidt process - but we have $1$ vector so $u_1 = [1,1,1,...,1,1]^T$ = orthogonal basis of $V^{\perp}$

b) It seems to be very interesting and hard. I found basis of $V$:
$$[-1,1,0,0,...,0] = \vec{v_1}$$ $$[-1,0,1,0,...,0] = \vec{v_2}$$ $$[-1,0,0,1,...,0] = \vec{v_3}$$ $$...$$ $$[-1,0,0,0,...,1] = \vec{v_{n-1}}$$

Now I start Gram–Schmidt process $$u_1 = v_1 $$ $$u_2 = v_2 - \frac{1}{2} \cdot v_1$$ $$u_ 3 = v_3 - \frac{1}{4} \cdot v_2 + \frac{1}{8} \cdot v_1 $$ $$u_4 = v_4 - \frac{7}{16} \cdot v_3 + \frac{7}{64} \cdot v_2 - \frac{7}{64} \cdot v_1$$ I don't even know if I don't take mistake.
Moreover the calculations getting harder and harder and I still don't see any regular sequence in it. Can somebody help me with this task?

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  • $\begingroup$ Why do you need an orthogonal basis of $V$? One can find the projection if one follows the vector $(1\ 1\ \ldots\ 1)$ until the line meets $V$. $\endgroup$
    – A.Γ.
    Jan 3, 2019 at 14:15
  • $\begingroup$ I need orthogonal basis because I have that formula for orthogonal projection: $ P_z(x) = \sum_{j=1}^k <z_j,x>z_j $ where $z_1,...,z_k $ is orthogonal basis and I don't have any other idea how to do this task @A.Γ. $\endgroup$
    – user617243
    Jan 3, 2019 at 14:18
  • $\begingroup$ You can save yourself a lot of work by taking advantage of the fact that the orthogonal projection onto a subspace is what’s left after subtracting the orthogonal projection onto its complement. $\endgroup$
    – amd
    Jan 3, 2019 at 21:40

1 Answer 1

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Let us call $\vec u$ the projection of $\vec{x} = [n,0,0,...,0]^T$ on $V$.
$\vec x - \vec u$ is orthogonal to $V$, i.e. $\vec x - \vec u = a \vec v$ with $v = (1, \dots, 1)^T$ as you already showed

Therefore $\vec x = a \vec v + \vec u$ with $\vec u$ satisfying $\sum_i u_i = 0$

Then, $$ \sum_i (x_i - a) = 0 $$

And finally $a = 1$ and

$$\vec u = (n-1, -1, \dots, -1) ^T $$

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  • $\begingroup$ the answer should be $\vec u = (n-1, 1, \dots, 1) ^T $ or $ \vec u = (n-1, -1, \dots, -1) ^T$? $\endgroup$
    – user617243
    Jan 3, 2019 at 14:36
  • $\begingroup$ Great, it is a loooot of simpler way than my idea, thanks $\endgroup$
    – user617243
    Jan 3, 2019 at 14:39
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    $\begingroup$ @VirtualUser Corrected. You read my answer before I had time to check it! $\endgroup$
    – Damien
    Jan 3, 2019 at 14:39

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