Gram Schmidt procedure

Question

For the subspace on the left in the image I am to find an orthogonal basis for the subspace. The answer is to the right in the image, and when I follow the procedure I get those vectors, however I also get one vector to be (0, 0, -120/15, 0). Why is that not included in the basis?

Answer 1

In the third step, $\mathbb{S}_2=\mathrm{span}\{v_1,v_2\}$ also equals $\mathbb{S}_2=\mathrm{span}\{w_1,w_2\}$. The Gram-Schmidt process produces a vector $v_3=\mathrm{perp}_{\mathbb{S}_2}w_3\in \mathbb{S}_2^\perp$ with $\mathrm{span}\{v_1,v_2,v_3\}=\mathrm{span}\{w_1,w_2,w_3\}$. Since $v_3=0$, $$ \mathrm{span}\{w_1,w_2,w_3\}=\mathrm{span}\{v_1,v_2,v_3\}=\mathrm{span}\{v_1,v_2,0\}=\mathrm{span}\{v_1,v_2\}=\mathrm{span}\{w_1,w_2\}. $$ Therefore, $w_3\in\mathrm{span}\{w_1,w_2\}=\mathbb{S}_2$.

Answer 2

You made a mistake somewhere. The four original vectors span a $3$-dimensional space and therefore there is no way of obtaining a basis with $4$ elements. Exactly which error you made is something I cannot tell you without seeing your computations.