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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.

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Hint: what is the dimension of $V$? This should tell you how big a basis should be.

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  • $\begingroup$ Forgive my elementary question but is it not just $3$? There are $3$ vectors after all which span V is there not? $\endgroup$ – Future Math person Dec 12 '18 at 6:12
  • $\begingroup$ 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 $2$. $\endgroup$ – platty Dec 12 '18 at 6:14
  • $\begingroup$ Notice that $v_1+v_2=v_3$, so they are not linearly independent. Dimension is $2$ or less. $\endgroup$ – Larsson Dec 12 '18 at 6:15
  • $\begingroup$ Yes I was just about to comment that haha. I only have 2 basis vectors so the dimension would be 2. Thanks! $\endgroup$ – Future Math person Dec 12 '18 at 6:16

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