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Consider inner product space $\mathbb{R}^4$ with respect to the dot product.

Find an orthonormal basis for the subspace $W$ of $\mathbb{R}^4$ generated by:

$\left\lbrace \begin{bmatrix} 4\\2\\6\\-2 \end{bmatrix}, \begin{bmatrix} 1\\-1\\3\\-1 \end{bmatrix}, \begin{bmatrix} 1\\2\\0\\0 \end{bmatrix}, \begin{bmatrix} 1\\5\\-3\\1 \end{bmatrix} \right\rbrace $

Ok so I tried using Gram-Schmidt, however, these vectors are linearly dependent. I've read somewhere that Gram-Schmidt can only be applied to a set of linearly independent vectors..is this true? I end up getting two zero vectors in my set of orthogonal vectors! Is this against the rules? Please help!

Please be very elaborate as I have searched everywhere for help!

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  • $\begingroup$ You have to remove vectors so the remaining set is linearly independent $\endgroup$
    – Triatticus
    Commented Dec 2, 2017 at 6:43
  • $\begingroup$ what do you mean remove vectors $\endgroup$
    – TopGoober
    Commented Dec 2, 2017 at 6:44
  • $\begingroup$ For instance the first three are linearly independent, so remove the last vector and the subspace is the one spanned by the remaining vectors $\endgroup$
    – Triatticus
    Commented Dec 2, 2017 at 6:44
  • $\begingroup$ so you are saying use gram schmidt on the first three vectors and thats my answer?...the set of 3 orthogonal vectors i get at the end? after I normalize them? $\endgroup$
    – TopGoober
    Commented Dec 2, 2017 at 6:47
  • $\begingroup$ Yes that is exactly what I'm saying $\endgroup$
    – Triatticus
    Commented Dec 2, 2017 at 6:47

1 Answer 1

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Gram-Schmidt can be used to reduce down to a linearly independent basis! Normally, the way it works is you find a vector orthogonal to the vectors so far, then normalise it. This orthogonal vector will be $0$ if and only if the vector was dependent on the previous ones. So, if you get a zero, ignore that vector and move on to the next one.

It works pretty simply, if you understand Gram-Schmidt geometrically. What Gram-Schmidt does is, at stage $j$, project $v_j$ onto the span of $v_1, \ldots v_{j-1}$, and take the orthogonal complement of this projection. When this returns a zero, this means that the projection onto the span coincides with $v_j$ in the first place, which can only happen when $v_j$ belongs in the span to begin with. Thus, removing $v_j$ from the span will not reduce it.

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