Finding an orthonormal basis of the subspace

Question

Find an orthonormal basis of the subspace: $$V = {[x, y, z, w]^T:x+y+z+w=0}$$ of $\mathbb{R}^4$

First I found a $4\times4$ determinant to verify whether they are non-singular or not. $$\begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{vmatrix}=1\ne0$$

Then to orthogonalize the basis I applied Gram-Schmidt process, $$v_1=x_1=(1,1,1,1)$$

Is my basis $(1,1,1,1)$ correct?

Answer 1

Your proposed basis vector is not in your subspace $$x+y+z+w=0$$ because $$1+1+1+1=4 \ne 0$$

you need to find some linearly independent vectors in the subspace to form a basis and then apply Gram-Schmidt method to find an orthonormal basis.

For example $$(1,-1,0,0), (0,1,-1,0), (0,0,1,-1)$$ are linearly independent vectors in your subspace.

Can you apply Gram-Schmidt to that set to find an orthonormal basis?

Answer 2

Note that ($1,1,1,1)$ is not a basis vector and that we need

• to find a basis for the subspace (the dimension is three then we need 3 basis vectors),

• apply GS process and finally normalize them.

Note that we can easily find by inspection

• $v_1=(1,0,-1,0)$
• $v_2=(0,1,0,-1)$

which are independent and orthogonal, then we need only a third vector to complete a basis. What about $v_3=(0,1,-1,0)$?

From here we need only to apply GS process to vector $v_3$ since $v_1$ and $v_3$ are given for free.

Answer 3

How about:

$$\begin{bmatrix}1 \\ -1 \\ 0 \\ 0\end{bmatrix},\begin{bmatrix}1 \\ 0 \\ -1 \\ 0\end{bmatrix},\begin{bmatrix}1 \\ 0 \\ 0 \\ -1\end{bmatrix}$$