Confusion with Pseudovectors This is probably really silly - I must have some basic misunderstanding of what pseudovectors are supposed to do...
So a simple example: consider the three unit vectors $i$,$j$ and $k$ in a right handed coordinate system with axes $x$,$y$ and $z$. We know that in this coordinate system, $i\times j=k$. We can see this using $\begin{pmatrix} 1 \\ 0 \\  0\end{pmatrix}\times \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ or by visualizing rotating $i$ into $j$ through the smallest possible angle, which will be anticlockwise, and so the direction of the result has to be $k$.
Now invert all the coordinate axes, so $x \rightarrow -x$,$y \rightarrow -y$ and $z \rightarrow -z$. Then our original unit vectors become $i'=-i$,$j'=-j$ and $k'=-k$. Now calculate the same vector product $(i') \times (j')=i \times j = k$ and not $-k$ as I would expect - $k$ points along the new $-z$ axis and so is in the opposite direction to the old cross product. This contradicts the visualization method because the two vectors don't move in space, so the cross product doesn't move in space.
Can anybody de-confuse me? Thanks :)
 A: I like the verb "de-confuse", and that is certainly something I would like to be able to perform!
On the lowest algebraic level we have
$$
\left|\begin{matrix}a&c\\b&d\end{matrix}\right|=ad-bc=(-a)(-d)-(-b)(-c)=\left|\begin{matrix}-a&-c\\-b&-d\end{matrix}\right|
$$
for each of the three determinants in the cross product which then implies $(-\mathbf i)\times(-\mathbf j)=\mathbf k$. More generally, it can be proven that
$$
(r\mathbf a)\times\mathbf b=\mathbf a\times(r\mathbf b)=r(\mathbf a\times\mathbf b)
$$
so applying this twice with $r=-1$ to $\mathbf i\times\mathbf j$ yields the same conclusion again, namely $(-\mathbf i)\times(-\mathbf j)=\mathbf k$.

Regarding the geometric level, we have the right-hand-rule seen in the picture here. Try using this for $\mathbf i\times\mathbf j=\mathbf k$ and then flip over $\mathbf i$ to $-\mathbf i$ first by fixing middle finger and flipping the index finger to the opposite direction. Now your thumb is pointing in the opposite direction too. Then fix the index finger and flip the middle finger corresponding to changing $\mathbf j$ to $-\mathbf j$, and see how your thumb, the cross product, is pointing upwards again.
