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.
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.