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Let $V=ℝ^3$ and the inner product on V is the dot product. Let

$v_1=(1,2,3)$

$v_2=(2,2,1)$

$v_3=(2,2,1)$

be given.

Find the orthogonal basis of $Span(v_1,v_2,v_3)$ using the Gram-Schmidt algorithm.

i began by saying the order of the elements as $v_1,v_2,v_3$

then, i let $w_1=v_1=(1,2,3)$

then i used the gram-schmidt formula and got

$w_2=(\frac{19}{14},\frac{5}{7},-\frac{13}{14})$

then got $w_3=(0,0,0)$

so would the answer of this be just be $\left\{\left(1,2,3\right),\left(\frac{19}{14},\frac{5}{7},-\frac{13}{14}\right),\left(0,0,0\right)\right\}$?

also, how do i check if it actually is Orthogonal?

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  • $\begingroup$ Toss the third vector out of your basis set. The space which is spanned by the three vectors is two dimensional. As you can see $v_{2}=v_{3}$. $\endgroup$ – Tucker Aug 7 '15 at 17:56
  • $\begingroup$ What you need to do from here is take an inner product $(1,2,3)\cdot(\frac{19}{14},\frac{5}{7},-\frac{13}{14})$. If the result of the inner product is zero then the vectors are orthogonal. $\endgroup$ – Tucker Aug 7 '15 at 17:58
  • $\begingroup$ When I did the arithmetic I did get that the inner product is zero and so it looks like you did it correctly. $\endgroup$ – Tucker Aug 7 '15 at 17:59
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The basis should just be {(1,2,3), $(\frac{19}{14}, \frac57, -\frac{13}{14})$}. It makes sense because the $span(v_1,v_2,v_3)$ also has dimension 2. You can tell that it's an orthogonal basis by checking if the dot product of any two vectors in the basis is equal to 0, and in this case it is.

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  • $\begingroup$ The process of finding it -- where did the OP go wrong ? Thats probaby what the OP needs. Not just the answer. $\endgroup$ – Shailesh Apr 15 '16 at 0:34
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The fact that you got the zero vector tells you that $\dim(V) < 3$. In this case this can be seen directly from the fact that $v_3=v_2$, which means that your vectors are not independent.

After applying the G-S, the non-zero vectors you get form an orthogonal basis. So you need to eliminate the zero vectors from your set to get the basis.

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