# Creating an orthonormal basis with Gram schmidt procedure error.

I have a question which says the following:

Let $$V$$ be the span of $$v_{1}=(0,1,2)$$, $$v_{2}=(-1,0,1)$$ and $$v_{3}=(-1,1,3)$$.

Construct an orthonormal basis $$B'$$ for $$V$$ (usual dot product).

I know how to do the question. That's easy. I need to apply the gram schmidt procedure thrice and I have my answer. My problem is the 3rd iteration.

$$U_{1}=V_{1}=(0,1,2)$$

Orthonormalizing, we obtain $$V_{1}'=(0,\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}})$$

Then $$U_2=V_2-\frac{V_2 \cdot U_1}{||U_1||^2}U_1$$

Doing so gives: $$U_2=(-1,\frac{-2}{5},\frac{1}{5})$$

Orthonormalizing, we get $$V_{2}'=(0,\frac{-2}{\sqrt{30}},\frac{1}{\sqrt{30}})$$

Finally, $$U_3=V_3-\frac{V_3 \cdot U_1}{||U_1||^2}U_1 - \frac{V_3 \cdot U_2}{||U_2||^2}U_2$$

However this time, if I plug in everything, I get $$U_3=(0,0,0)$$. While that is indeed orthogonal to both $${v_1}$$ and $${v_2}$$ there is no way I can orthonormalize $$U_3$$ since I get division by $$0$$.

Is the question just badly asked? Because there is simply no way to obtain an orthonormal basis.

Hint: what is the dimension of $$V$$? This should tell you how big a basis should be.
• Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not? – Future Math person Dec 12 '18 at 6:12
• This doesn’t mean that the dimension is exactly $3$, only that it is at most $3$. For example, $span\{(1,0),(2,0)\}$ does not have dimension $2$. – platty Dec 12 '18 at 6:14
• Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less. – Larsson Dec 12 '18 at 6:15