# Finding an orthogonal basis for a subspace of $\mathbb R^5$

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.

• Solve the system, which will give you the vectors you want. Use Gram-Schmidt to orthonormalize.
– cmk
Oct 17, 2019 at 20:23
• After you apply Gauss elimination, take the independent variables and gives them all value zero except to one. For the different choices of which of them you choose to be non-zero, you get a different vector from $V$. Those form a basis of $V$. Oct 17, 2019 at 20:25
• Does this mean that the basis of vectors I need is in terms of $x_n$? Oct 17, 2019 at 20:26
• No, you give concrete values to the independent variables: All of them zero, except one, to which you give some non-zero value (like $1$). Then you solve for the dependent variables. Oct 17, 2019 at 20:33
• I've edited the MathJax slightly, please check to ensure I haven't inadvertently changed the intention. Oct 18, 2019 at 4:13

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.