So I have the following question:
$Let \\ u_1 = \begin{pmatrix} \frac{1}{3\sqrt{2}}\\\frac{1}{3\sqrt{2}}\\\frac{-4}{3\sqrt{2}}\end{pmatrix} u_2 = \begin{pmatrix} \frac{2}{3}\\\frac{2}{3}\\\frac{1}{3}\end{pmatrix} u_3 = \begin{pmatrix} \frac{1}{\sqrt{2}}\\\frac{-1}{\sqrt{2}}\\0\end{pmatrix}\\$

(a)Show that $\{u_1, u_2, u_3\}$ is an orthonormal basis for $\mathbb{R}^3$

I know that I simply find the magnitude of all the vectors and if they are 1 then they are of unit length and meet 1 of 2 criterion for being a orthonormal basis. The next thing is take the dot product and if it's 0 then it's orthogonal. The question i have is do I take 3 diferrent dot products like:
$u_1\cdot u_2, u_1\cdot u_3, u_2\cdot u_3$

And if all of them are 0 then the set if a orthonormal basis. Or am I supposed to take the dot product of all three at once (is that even possible), like:
$u_1\cdot u_2\cdot u_3$

  • $\begingroup$ What is the definition of an orthogonal set of vectors that you are using? $\endgroup$
    – wj32
    Commented Oct 11, 2012 at 23:39

1 Answer 1


Yes you have to take three dot products. Another way of checking would be computing (with $A= (u_1,u_2,u_3)$) $$A A^T$$ if this equals the identity matrix, you have an orthonormal base(because $A$ being an othogonal matrix is equal to $\{u_1,u_2,u_3\}$ being a ONB).

The dot product of all three at once is not defined, because the dot product is a map, that maps two vectors to a real number. So after performing the first dot product, the term you have left would be undefined.

  • $\begingroup$ ah okay that makes sense, thank you very much. $\endgroup$ Commented Oct 11, 2012 at 23:42

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