Calculating an orthonormal basis of $\{(x,y,z) \mid 2x + y -z = 0\}$

Question

We have the real euclidean vector space $\mathbb{R}^3$ with the standard inner product and the standard basis $B = (e_1, e_2, e_3)$.

$W \subset \mathbb{R}^3$ is the subspace which is defined by: $$W = \{(x,y,z) \in \mathbb{R}^3 \mid 2x+y-z=0 \}$$

a.) Calculate an orthonormal basis of $W$.

I'm not sure how to solve this question. Do I have to use the Gram-Schmidt method?

How do I write out the matrix? Is this the matrix? $$ \begin{pmatrix} 2&1&-1 \end{pmatrix} $$

And then with the Gram-Schmidt algorithm: $$ w_1 = \frac{v_1}{\| v_1\|} = \frac{1}{\sqrt{5}}\begin{pmatrix} 2\\1\\-1\end{pmatrix} $$

Is my idea correct? If not, can you tell me where I went wrong?

Thank you!

Answer 1

Hint

$W=\{(x,y,z)\in\mathbb{R}^3: 2x+y-z=0\}=\{(x,y,z)\in\mathbb{R}^3: z=2x+y\}=\{(x,y,z)\in\mathbb{R}^3: (x,y,z)=(x,y,2x+y)\}=\{(x,y,z)\in\mathbb{R}^3: (x,y,z)=(x,0,2x)+(0,y,y)\}=\{(x,y,z)\in\mathbb{R}^3: (x,y,z)=x(1,0,2)+y(0,1,1)\}=\langle(1,0,2),(0,1,1)\rangle$

Answer 2

First, you should find a basis of $W$ (Hint: $W$ has dimension $2$). Then you can apply Gram-Schmidt to this basis.

To find vectors of $W$, you want to look for solutions of $2x+y-z=0$. The vector $v=\begin{pmatrix} 2\\1\\-1 \end{pmatrix}$ you normalized is in fact the orthogonal complement of $W$: the condition $2x+y-z$ shows that $v$ is orthogonal to every vector of $W$.

If you know any vector $w\neq 0$ of $W$ (by guessing? It's not that hard), you can use the crossproduct $\times$ to find a third vector, which is already orthogonal to both $v$ and $w$, so you only have to normalize $w$ and $v\times w$ to obtain a orthonormal basis. (Note that this only works in $\mathbb R^3$)

Answer 3

Start with any basis for $W$, such as $$\left\{\pmatrix{0 \\ 1 \\ 1}, \pmatrix{1 \\ 0 \\ 2}\right\}$$

and apply Gram-Schmidt to obtain the orthonormal basis $$\left\{\frac1{\sqrt2}\pmatrix{0 \\ 1 \\ 1}, \frac1{\sqrt3}\pmatrix{1 \\ -1 \\ 1}\right\}$$