Fix the Third Column

Three unit vectors in $\mathbb{R}^3$ , stacked as the columns of $Q$ :

$v_1 = \tfrac{1}{\sqrt{2}}(1, 1, 0) \qquad v_2 = \tfrac{1}{\sqrt{2}}(1, -1, 0) \qquad v_3 = \tfrac{1}{\sqrt{3}}(1, 1, 1)$

CS229 defines $Q$ as orthogonal when $Q^\top Q = I$ , which requires the columns to be pairwise orthogonal AND of unit norm.

Which entry of $Q^\top Q$ is wrong, and what is the smallest repair?

Nothing is wrong. All three vectors have unit norm, which is what orthonormal means.

$(Q^\top Q)_{13} = 2/\sqrt{6} \approx 0.816$ instead of 0. Replacing $v_3$ with $(0, 0, 1)$ fixes it, since $v_1$ and $v_2$ already span the plane $z = 0$ .

$(Q^\top Q)_{33} \neq 1$ , so $v_3$ needs to be divided by $\sqrt{3}$ to become a unit vector.

$(Q^\top Q)_{12} = 1/\sqrt{2}$ : $v_1$ and $v_2$ are the offending pair, because they share the same first coordinate.