$u,v,w\in V\ni \|u\|=\|v\|=\|w\|=2,\langle u,v\rangle=0,\langle u,w\rangle=1,\langle v,w\rangle=-1$ $V$ be an inner product space, $u,v,w\in V\ni \|u\|=\|v\|=\|w\|=2,\langle u,v\rangle=0,\langle u,w\rangle=1,\langle v,w\rangle=-1$. Then which of the following is (are) true?


*

*$\|w+v-u\|=2\sqrt{2}$ 

*$\{{1\over 2}u,{1\over 2}v\}$ forms an orthonormal basis for a two dimensional subspace of $V$

*$w$ and $4u-w$ are orthogonal to each other. 

*$u,v,w$ are necessarily linearly independent.
I have calculated myself and got $1$ is true.
$2$ is also true as the set satisfies all the conditions for forming an  an orthonormal basis for a two dimensional subspace of $V$
$3$ is also true as $\langle w,4u-w\rangle=0$
$4$ is False  I think? Thanks for correcting 
 A: *
1. 


$$\begin{align} \|{\bf w}+{\bf v}-{\bf u}\|^2 &= \langle {\bf w}+{\bf v}-{\bf u},{\bf w}+{\bf v}-{\bf u} \rangle \\ &= \langle {\bf w},{\bf w}\rangle+\langle {\bf v},{\bf v}\rangle  +\langle {\bf u},{\bf u}\rangle + 2\left(\langle {\bf w},{\bf v}\rangle - \langle {\bf w},{\bf u}\rangle - \langle {\bf v},{\bf u}\rangle\right)\rangle \\ &= 4+4+4+2(-1-1-0) \\ &= 12-4 = 8.\end{align}$$ So $\|{\bf w}+{\bf v}-{\bf u}\| = 2\sqrt{2}$, check.


*$\langle \frac{1}{2}{\bf u},\frac{1}{2}{\bf v} \rangle = \frac{1}{4}\langle {\bf u},{\bf v}\rangle = 0$, and $\|\frac{1}{2}{\bf u}\| = \frac{1}{2}\|{\bf u}\| = 1$, same for $\bf v$, so it is true.

*$\langle{\bf w},4{\bf u}-{\bf w} \rangle = 4\langle {\bf w},{\bf u}\rangle - \langle{\bf w},{\bf w}\rangle = 4-4 = 0$, so it is true.

*Write the linear combination $a{\bf u}+b{\bf v}+c{\bf w} = 0$. Applying $\langle \cdot, {\bf u}\rangle, \langle \cdot, {\bf v}\rangle$ and $\langle \cdot, {\bf w}\rangle$, we obtain the system: $$\begin{cases} 4a+0b+c = 0 \\ 0a+4b-c = 0 \\ a-b+4c = 0\end{cases}$$ Clearly $a=b=c=0$ is a solution. And we have: $$\begin{vmatrix} 4 & 0 & 1 \\ 0 & 4 & -1 \\ 1 & -1 & 4\end{vmatrix} = 4\begin{vmatrix}4 & -1 \\ -1 & 4\end{vmatrix} + \begin{vmatrix} 0 & 1 \\ 4 & -1\end{vmatrix} = 4\cdot 17 - 4 \neq 0,$$ so this solution is unique. Then the vectors are linearly independent.
(+1 for effort)
