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$\left( \begin{bmatrix}\frac{2}{\sqrt{10}}\\\frac{2}{\sqrt{10}}\\1\\1\end{bmatrix}, \begin{bmatrix}\frac{1}{\sqrt{3}}\\0\\-\frac{1}{\sqrt{3}}\\-\frac{1}{\sqrt{3}}\end{bmatrix}, \begin{bmatrix}0\\\frac{1}{\sqrt{3}}\\-\frac{1}{\sqrt{3}}\\-\frac{1}{\sqrt{3}}\end{bmatrix}, \begin{bmatrix}0\\0\\\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}\end{bmatrix} \right)$

So what I initially did is check the dot product of all the possible pairs of matrices? So I found v3 and v4, v1 and v4 and v2 and v4 are orthogonal? Is this correct? If so, where do I go from here?

(I labeled the vectors v1-v4 respectively)

For something to be orthonormal the dot product between every vector in S is zero AND the norm(magnitude) of each vector in S is one, correct? So this wouldn't be orthonormal because all the pairs of vectors aren't orthogonal?

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  • $\begingroup$ What means rad3 etc? $\endgroup$ May 4, 2016 at 12:31
  • $\begingroup$ radical symbol I am so sorry didn't know syntax for mathjax $\endgroup$
    – Lil
    May 4, 2016 at 12:32
  • $\begingroup$ You simply have to check the dot product, or even easier, put $A$ equal to the matrix having $v_1,\dots ,v_4$ in the columns, then if $AA^{T}=Id$ the vectors are orthonormal. (Why?) Also do you mean $\sqrt{10}$ instead of $rad10$? The radical of an integer also exists, but I doubt that is what you want here. $\endgroup$ May 4, 2016 at 12:34
  • $\begingroup$ the dot product of all 4 vectors at once? and yes I meant the square root. $\endgroup$
    – Lil
    May 4, 2016 at 12:36
  • $\begingroup$ To display $\sqrt2$ write $\sqrt2$. More generally, for other formatting questions, see meta.math.stackexchange.com/questions/5020/… $\endgroup$
    – David K
    May 4, 2016 at 12:42

1 Answer 1

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Orthogonals pairs are {v1,v4}, {v3,v4} only

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  • $\begingroup$ Also $v_2$ and $v_4$, which OP already discovered, but I agree there are no others. $\endgroup$
    – David K
    May 4, 2016 at 12:45
  • $\begingroup$ Yes. I missunderstand original "rad" form of formulas $\endgroup$ May 4, 2016 at 12:54

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