Use the standard Euclidean inner product on $\mathrm R^4$.
Let $W$ be the subspace of $\mathrm R^4$ spanned by
$u_1 = (1, 1, 1, 1),$
$u_2 = (2, 4, 1, 5),$
$u_3 = (1, -5, 4, -8).$
Find an orthogonal basis for $W.$
I don't understand this topic at all. I know finding basis, what inner product is, norm, etc (basic parts).
Please try to explain it very very simply for me. Thank you!
The "Gram Schmidt" process is to be used..
The Gram-Schmidt process is a very useful method to convert a set of linearly independent vectors into a set of orthogonal (or even orthonormal) vectors, in this case we want to find an orthogonal basis $\{v_i\}$ in terms of the basis $\{u_i\}$. It is an inductive process, so first let's define: $$ v_1 := u_1 = (1,1,1,1). $$ Then, by Gram-Schmidt: $$ \begin{align} v_2 :&= u_2 - \frac{\langle u_1 , u_2\rangle}{\langle u_1 , u_1\rangle}u_1\\ &= u_2 - \frac{2+4+1+5}{4}u_1 = (2,4,1,5) - 3(1,1,1,1)\\ &= (-1,1,-2,2). \end{align} $$ and finally $$ \begin{align} v_3 &= u_3 - \frac{\langle v_2 , u_3\rangle}{\langle v_2 , v_2\rangle}v_2 - \frac{\langle v_1 , u_3\rangle}{\langle v_1 , v_1\rangle}v_1\\ &= u_3 + \frac{30}{10}v_2 + \frac{8}{4}v_1 \\ &= (1,-5,4,-8) + 3(-1,1,-2,2) + 2(1,1,1,1) \\ &= (0,0,0,0) \end{align} $$ Now,why are we getting zero? what does this mean!? Basically, getting zero in the last result means your original vectors aren't linearly independent (we should've checked this before doing Gram-Schmidt, but I didn't think about it)! Some basic calculations show that $u_3 -7u_1 + 3u_2 =0$.
Still the previous results work, since $v_1$ and $v_2$ are orthogonal and span $W$ (why?).