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Determine whether the given set of vectors are orthogonal? $$S=\{ (1,0,-1),(0,3,-6),(0,2,-4) \}.$$ I just know that orthogonality of vectors in a vector space on case of symmetric bilinear form (also known as scalar product) is defined as follows:

$$B(e_i,e_j)=0$$ for $i\ne j$ where $${e_1,e_2,...,e_n}$$ is basis for V.

I guess the definition for $B$ here is $$B(x,y)=x_1y_1+x_2y_2+x_3y_3$$ for $x=(x_1,x_2,x_3)$ & $y=(y_1,y_2,y_3)$

Then should I check $$B(e_i,e_j)=0$$ for $i\ne j$ for $i,j=1,2,3$ ? And if condition is satisfied for all cases then should I say the given set is orthogonal?

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    $\begingroup$ Yes, you should check this with $e_1 = (1,0,-1), e_2 = (0,3,-6),e_3 = (0,2,-4)$. $\endgroup$ – Kenny Wong May 2 '17 at 11:26
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    $\begingroup$ So I concluded the set is not orthogonal. Since for each case I am getting $$B\ne 0$$ Am i right? $\endgroup$ – Kavita May 2 '17 at 11:34
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    $\begingroup$ Yes, that seems correct. $\endgroup$ – gue May 2 '17 at 11:41
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Let the columns of $3 \times 3$ matrix $\rm V$ be the given vectors, i.e.,

$$\rm V := \begin{bmatrix}1 & 0 & 0\\ 0 & 3 & 2\\ -1 & -6 & -4\end{bmatrix}$$

Computing the Gram matrix, we obtain

$$\rm V^{\top} V = \begin{bmatrix} 2 & 6 & 4\\ 6 & 45 & 30\\ 4 & 30 & 20\end{bmatrix}$$

What can we conclude, then?

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  • $\begingroup$ since $V^TV \not=I$; $I$ is the identity matrix. So the column vectors of $V$ are not orthogonal. $\endgroup$ – Hirakjyoti Das May 2 '17 at 12:09
  • $\begingroup$ That would be orthonormal. $\endgroup$ – Rodrigo de Azevedo May 2 '17 at 12:18
  • $\begingroup$ I believe "not orthonormal" would work for us. I think it will imply non-orthogonality. $\endgroup$ – Hirakjyoti Das May 2 '17 at 12:23
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For first question: you don't need to do that. Since $B$ is symmetric, you only need to check three pairs.

For the second question: yes, by definition. So just do it.


"I guess the definition for $B$ here is..."

Don't guess. Check your book or ask your professor what is the definition of $B$. (In general, a symmetric bilinear form is not necessarily the standard inner product.) Otherwise this problem is undoable.

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