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On p. 291 of Artin's Algebra, Artin writes that the rows of the character table of $S_3$ are of length 6. But in my mind the fact that the length of a vector $v=(a, b, c, d, e, f)\in{\mathbb{C}^6}$ is $\sqrt{a\bar{a}+b\bar{b}+c\bar{c}+d\bar{d}+e\bar{e}+f\bar{f}}$ seems to imply that the length of each of the row vectors is $\sqrt{6}$. Am I missing something?

Also, Artin writes that the columns are all orthogonal to each other. But the second and third columns are both $(1, 1, -1)$. Correct me if I'm wrong, but don't we have $(1, 1, -1)\cdot{(1, 1, -1)}=3\neq{0}$? And doesn't this mean those two columns are not orthogonal?

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By the way, the character table of $S_3$ can be found here.

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    The character table you link to obeys both column and row orthogonality. Not sure about the one in the book. – 2017-01-03
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    You're right, it's because the one Artin has on p. 291 has a separate column for each element, not for each conjugacy class. But he still claims that they are orthogonal. – 2017-01-03
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    They're identical for elements in the same conjugacy class, so can't be orthogonal, right? – 2017-01-03
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    I guess that's probably what you're saying too, in which case, I agree. I don't understand why you'd write a column for each element since any two elements in the same conjugacy class have the same character. I don't have the book, so I can't read and try and interpret what he's doing. – 2017-01-03

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By "length $6$" I think he means there are $6$ components (i.e. $6$ columns in the table, one for each element of $S_3$). Sorry, he definitely means square length is $6$, as you pointed out. If you look ahead to Theorem 10.4.6(a) (the Main Theorem) on page 300, the result is that the irreducible characters are orthonormal with respect to the inner product (10.4.3) $$\langle \chi, \chi' \rangle = \frac{1}{|G|} \sum_g \overline{\chi(g)} \chi'(g),$$ so the "square length" of a character is $|G|$.

Regarding column orthogonality, he of course means distinct columns are orthogonal. Since characters are constant on conjugacy classes, sometimes the character table is compressed (see (10.4.12) on page 302) to have one column per conjugacy class.

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    On p. 291 he also calls both of these "astonishing facts," so I find it doubtful he meant that there are 6 columns in the table, since that's (pretty clearly) what you will get if you evaluate the character functions on each of the six elements of $S_3$. As for orthogonality, I take your point that it works if we only have one column per conjugacy class. But the fact remains that the table on p.291 has a distinct column for each element of $S_3$, not for each conjugacy class . . . – 2017-01-03
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    @Rasputin Sorry, see my edit. – 2017-01-03
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    No worries, I'm still a little unclear though as to how the square length is 6, because taking $\chi_A$ for example (to use Artin's notation), we have $\langle{\chi_A, \chi_A}\rangle=\frac{1}{\mid{G}\mid}(\chi_A(1)\overline{\chi_A(1)}+\dots+(\chi_A(x^2y)\overline{\chi_A(x^2y)})=\frac{1}{6}(2^2+(-1)^2+(-1)^2+0^2+0^2+0^2)=1\neq{6}$. – 2017-01-03
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    But I guess you're saying that the "square length" is only the stuff being summed, i.e. is not multiplied by $\frac{1}{\mid{G}\mid}$? – 2017-01-03
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    @Rasputin Yes, I am using "length" in the context of your post (the usual inner product *without* the $\frac{1}{|G|}$). – 2017-01-03