According to Wikipedia:
a pseudovector (or axial vector) is a quantity that transforms like a vector under a proper rotation, but in three dimensions gains an additional sign flip under an improper rotation such as a reflection.
It seems that pseudovectors "are not real vectors". But if you think about it, every vector in $\mathbb{R}^3$ can be written as a cross product of two vectors. Let $\vec{v}_1 = (a,b,c)$, then $\vec{v}_2 = \frac{1}{\sqrt{a}}(-c,0,a), \vec{v}_3 = \frac{1}{\sqrt{a}}(-b,a,0)$ satisfy
$$
\vec{v}_3\times\vec{v}_2 = (a,b,c)
$$
So we get that $\vec{v}_1$ is a pseudovector. But this means every vector is a pseudovector, so this definition seems empty to me in 3D.
Best Answer
You're right that $\vec{v}_1$ has an equal claim to being a vector and a pseudovector. But the conclusion is not that it is both. It is neither! Look at how you introduced it:
What does that mean? If you fix three real numbers $a,b,c$, for example the first three prime numbers $2,3,5$, then the quantity $(a,b,c)=(2,3,5)\in\mathbb R^3$ is neither a vector nor a pseudovector with respect to rotations in three-dimensional physical space. It is a scalar, because it doesn't transform under rotations. That scalar just happens to have an index in a three-dimensional trivial representation of the spatial symmetry group.
It is healthier to consider that vectors and pseudovectors live in separate spaces, endowed with different representations of the symmetry group, in this case $O(3)$. Both spaces are three-dimensional, so there exists an isomorphism between them (by fixing a coordinate frame). But they are not naturally isomorphic to each other, nor are they naturally isomorphic to $\mathbb R^3$.