# Gram-Schmidt-Complex

I have the following complex vectors:

We need to apply Gram-Schmidt.

For $v_1$ I got the same vector as for $u_1$ that means $v_1=(0,0,0,0,1)$.
For $v_2' = u_2 \langle v_1,u_2 \rangle v1$
So I calculated $\langle v_1,u_2 \rangle = (0,0,0,0,1)\cdot(0,0,1,0,i) = i$
With that: $(0,0,1,0,i) - i\,(0,0,0,0,1) = (0,0,1,0,0)=v_2'$
Moreover for $v_2$ i also got $(0,0,1,0,0)$ with $\|v_2\|=\sqrt{0^2+0^2+1^2+0^2+0^2} = \sqrt1 = 1\,(0,0,1,0,0)=(0,0,1,0,0)$
For $v_3'=u_3-\langle v_1,u_3 \rangle v_1-\langle v_2,u_3 \rangle v_2$
For $\langle v_1,u_3 \rangle =(0,0,0,0,1)\cdot(2,0,i,1,1)=1$
For $\langle v_2,u_3 \rangle =(0,0,1,0,0)\cdot(2,0,i,1,1)=i$
With that $(2,0,i,1,1)-1\,(0,0,0,0,1)-i\,(0,0,1,0,0)=(2,0,i,0,0)-i\,(0,0,1,0,0)=(2,0,0,0,0)=v_3'$
There is definitly a mistake here. So it would be really helpful if someone could help me here..

• I would be really helpful if you showed more of your work. – amd May 22 '17 at 18:55
• Ok. I edited it – PhysX May 22 '17 at 19:20
• I’ve edited your question to make it more readable. Have a look at how my edit differs from what you had. For the future, please learn to use MathJax. You can find a tutorial here. You’re much more likely to attract answers when people don’t have to slog through nigh-unreadable questions. – amd May 22 '17 at 20:53
• Thank you so much man.. I am still a noob at this, so I will take a look – PhysX May 22 '17 at 20:55

$(2,0,i,1,1)-(0,0,0,0,1)\ne(2,0,i,0,0)$. You dropped the $1$ in the fourth slot.
• @PhysX They are. You just made a simple arithmetic error when computing $v_3'$. – amd May 22 '17 at 21:02
• I calculated the rest. For v'3=(2,0,0,1,0). For v3= $\sqrt5$ * (2,0,0,1,0). v'4 = (1/5, 2i, 0,-2/5, 0) and v4= $\sqrt24/5 \$ *(1/5, 2i, 0 ,-2/5,0) ^^ – PhysX May 22 '17 at 22:28