Completing System of Vectors to an Orthogonal Basis I want to complete ((1, -2, 2, -3), (2, -3, 2, 4)) to an orthgonal basis.  I honestly don't really know how to do this.  This a practice problem to a test, were we haven't done problems related to this.  I know that an orthogonal basis has that for basis $(v_1, ... , v_n)$, $v_i\perp v_j$ for $i\ne j$.  So I'm assuming I'm working in $R^4$, as the question only specifies "Euclideam Spcae".  In terms of actually solving the problem, I'm strugging, thanks in advance!
 A: We set:
$$\begin{gathered}
  {e_1}{\text{ = (}}1, - 2,2, - 3) \hfill \\
  {e_2}{\text{ = }}(2, - 3,2,4) \hfill \\
  A = ({a_1},{a_2},{a_3},{a_4}) \hfill \\ 
\end{gathered}$$
and solve:
$$\begin{gathered}
  {e_1} \cdot A = 0 \hfill \\
  {e_2} \cdot A = 0 \hfill \\ 
\end{gathered}$$
This gives:
$$\begin{gathered}
  {a_1} = {\text{2 }}{a_3} - {\text{17 }}{a_{\text{4}}} \hfill \\
  {a_2} = {\text{2 }}{a_3} - {\text{10 }}{a_{\text{4}}} \hfill \\
  A = ({\text{2 }}{a_3} - {\text{17 }}{a_{\text{4}}},{\text{2 }}{a_3} - {\text{10 }}{a_{\text{4}}},{a_3},{a_4}) \hfill \\ 
\end{gathered}$$
We may set:
$$\begin{gathered}
  {a_3} = 1 \hfill \\
  {a_4} = 0 \hfill \\
  {e_3}{\text{ = }}(2,2,1,0) \hfill \\ 
\end{gathered}$$
Now with $$A = ({a_1},{a_2},{a_3},{a_4})$$ we solve:
$$\begin{gathered}
  {e_1} \cdot A = 0 \hfill \\
  {e_2} \cdot A = 0 \hfill \\
  {e_3} \cdot A = 0 \hfill \\ 
\end{gathered} $$
This gives:
$$A = {a_4}( - 5,2,6,1)$$
Set ${a_4}=1$ and
$${e_4} = ( - 5,2,6,1)$$
then 
$$\{ {e_1},{e_2},{e_3},{e_4}\}$$
is a set of linear independent and orthogonal vectors.
We can now normalize:
$${f_1} = \frac{{{e_1}}}{{\sqrt {{e_1}.{e_1}} }},{f_2} = \frac{{{e_2}}}{{\sqrt {{e_2}.{e_2}} }},{f_3} = \frac{{{e_3}}}{{\sqrt {{e_3}.{e_3}} }},{f_4} = \frac{{{e_4}}}{{\sqrt {{e_4}.{e_4}} }}$$
so$$\{ {f_1},{f_2},{f_3},{f_4}\}$$
is an orthonormal base.
Gram-Schmidt process is used, in case a complete base is given.
But here we only have to solve two linear systems of equations, because
${e_1}$ and ${e_2}$ are orthogonal.
A: here is what you can do. you already have two vectors $u = (1, -2, 2, -3)^T, v = (2,- 2, 3, 4)^T$ that are orthogonal. we will first find four orthogonal vectors; making them of unit length is easier.
what we will do is pick a vector $a$ and find the projection on to the space spanned by $u, v$ and subtract it. we can just start with $a = (1,0,0,0).$  so let $$a = ku + lv + w $$ with $l,k$ are to be chosen so that $u^Tw = v^Tw  = 0.$  that is $$k = u^Ta/u^Tu, l = v^Ta/v^Tv \to k = \frac 1{14}, l = \frac 1{33},\\ w =(1,0,0,0)^T - \frac 1{14}((1, -2, 2, -3)^T + \frac1{33}(2,- 2, 3, 4)^T$$ 
once you have $w,$ or a nice multiple of it,  we have $u, v, w$ orthogonal. now you can use another vector hopefully $(0,1,0,0)$ and do the same trick to find the fourth vector. 
A: You (still) want Gram-Schmidt: link
The idea roughly is to start with a vector and remove the part in its direction from the other vectors. Then continue in the same fashion with the next vector.
And you need have four linear independent vectors in total, that means at least two more.
You can keep those two, as they are orthogonal already. Let them name $v$ and $w$. Again you can add $e_1 = (1,0,0,0)$ and $e_2=(0,1,0,0)$ to have an initial set which is a base (not tested, but very likely, check the determinant of the four vectors to be non-zero, if in doubt).
But now:
$$
e_1' = 
e_1 
- \left(e_1 \cdot \frac{v}{\lVert v \rVert}\right) 
\frac{v}{\lVert v \rVert}  
- \left(e_1 \cdot \frac{w}{\lVert w \rVert}\right) 
\frac{w}{\lVert w \rVert}  
$$
$$
e_2' = 
e_2 
- \left(e_2 \cdot \frac{v}{\lVert v \rVert}\right) 
\frac{v}{\lVert v \rVert}  
- \left(e_2 \cdot \frac{w}{\lVert w \rVert}\right) 
\frac{w}{\lVert w \rVert}  
- \left(e_2 \cdot \frac{e_1'}{\lVert e_1' \rVert}\right) 
\frac{e_1'}{\lVert e_1' \rVert}  
$$
where $a \cdot b = a_1 b_1+a_2 b_2+a_3 b_3+a_4 b_4$ and $\lVert r \rVert = \sqrt{r \cdot r} = \sqrt{r_1^2 + r_2^2 + r_3^2 + r_4^2}$.
I get 
\begin{align}
e_1' &= (0.823232, 0.292929, -0.232323, -0.075758) \\
e_2' &= (0, 0.40082, 0.48671, 0.057260)
\end{align}
and $\mbox{det}(v,w,e_1',e_2') = 14 \ne 0$, so this is a base of $\mathbb{R}^4$. 
