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Just was wondering if someone can confirm if my solution is correct or not.

using these three vectors:

$[-3,-1,1,0,0]$,

$[1,-1,0,1,0]$,

$[1,-1,0,0,1]$.

Find the orthogonal basis by using the Gram Schmidt process

my answer:

$v_1 = [-3,-1,1,0,0]$,

$v_2 = [\frac{5}{11},\frac{-13}{11},\frac{2}{11},1,0]$,

$v_3 = [\frac{5}{29},\frac{-13}{29},\frac{2}{29},\frac{-18}{29},1]$.

I just want to know if those 3 vectors are correct. I'm questioning it because I have those fractions. If I'm wrong, I don't want an answer but a tip in the right direction would be appreciated. Thanks

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  • $\begingroup$ You will almost always have these fractions, mostly even square roots. To check orthogonality real fast, calculate the inner product of any 2 distinct vectors. It should always give $0$. $\endgroup$
    – user370967
    Apr 3, 2017 at 7:45
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    $\begingroup$ @Math_QED I don't understand what you mean by your first sentence: the OP does not look for unit-normed vectors. On the contrary, one could say him/her that vectors $v_2$ and $v_3$ can be multiplied by $11$ and $29$ resp. in order to give a simpler answer. $\endgroup$
    – Jean Marie
    Apr 3, 2017 at 8:10
  • $\begingroup$ Ah I'm sorry I thought he was looking for an orthonormal basis, not an orthogonal one. You are absolutely correct. $\endgroup$
    – user370967
    Apr 3, 2017 at 8:15

1 Answer 1

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I check that the orthogonality is correct, but your new vector $v_2$ cannot be generated by the original basis, so something may be wrong. Maybe you should tell us your process and we can point out where you should correct! ( $\textbf{Wrong!}$)

$\textbf{Edit:}$ The orthogonality is correct, and $v_1,v_2,v_3$ are all generated by the original basis and linearly independent (one can easily check this seeing that the last three columns of the original basis is like identity), so the basis is really an orthogonal basis.

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  • $\begingroup$ No: $v_2=\tfrac{2}{11}[−3,−1,1,0,0]+[1,−1,0,1,0]$. $\endgroup$
    – Jean Marie
    Apr 3, 2017 at 8:02
  • $\begingroup$ Oh yeh... I made bad mistake. $\endgroup$
    – User
    Apr 3, 2017 at 8:04

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