Let $\textsf V$ be a vector subspace of $\mathbb R^5$ that is given by:
$$\begin{cases} x_1+x_2+x_3+x_4+x_5=0,\\ x_1-x_2+x_3-x_4+x_5=0. \end{cases}$$
Find an orthonormal basis of the vector subspace $\textsf V$.
I know how to use the Gram-Schmidt process, but I'm not sure what vectors to use it with.
Best Answer
Your system of equations can be converted into $$\left\{ \begin{align} x_1+x_3+x_5 = 0 \\ x_2+x_4 = 0 \end{align} \right.$$ so, $x_1=-x_3-x_5$ and $x_2=-x_4$. Hence, the vectors $(x_1,x_2,x_3,x_4,x_5)\in\mathbb R^5$ that satisfy the initial both equations also satisfy $$\begin{align} (x_1,x_2,x_3,x_4,x_5) &= (-x_3-x_5,-x_4,x_3,x_4,x_5) \\ &= x_3(-1,0,1,0,0)+x_4(0,-1,0,1,0)+x_5(-1,0,0,0,1) \end{align}$$ that is, all of them can be written as a linear combinations of the vectors $(-1,0,1,0,0)$, $(0,-1,0,1,0)$ and $(-1,0,0,0,1)$. Thus, $$\{(-1,0,1,0,0),(0,-1,0,1,0),(-1,0,0,0,1)\}$$ is a basis for $\textsf V$. Finally, apply the Gram-Schmidt process to this preceding set of vectors and we are done.