Algebraic and orthogonal complements

Question

I have been trying to understand this Encyclopedia of mathematics article. Specifically, in the comments section there is the following comment:

The codimension of a subspace $L$ of a vector space $V$ is equal to the dimension of any complement of $L$ in $V$, since all complements have the same dimension (as the orthogonal complement).

In a comment, Nate Eldredge pointed out to me that this statement is poorly written.

He provides an example of a space and a map such that the orthogonal complement of the image of the map is $0$ but the (algebraic) complement has infinite dimension.

Doesn't this mean that the comment in the Encyclopedia is not ''poorly written'' but rather plain wrong?

I apologize for hunting down the details so meticulously but I'm just a learner, not a mathematician yet and I need to know whether I really understand every detail.

Edit

For the sake of the question let's assume we have an inner product. Obviously, if we don't have one the comment is wrong. But if we do have one then the situation is less clear to me.

Answer 1

That is poorly written indeed. There is no orthogonal complement if there is no inner product defined on the vector space. Algebraically, any two complements of a subspace are isomorphic, so they will have the same dimension. If $Y\subset X$ is a subspace and $Z_1$ and $Z_2$ are two complements, then consider the maps $J_1:Z_1\to X$ which is just the inclusion and $P_2:X \to Z_2$ defined as $P_2(y+z_2)=z_2$ (the algebraic projection onto $Z_2$). It is not hard to show that $P_2\circ J_1$ is an isomorphism from $Z_1$ onto $Z_2$.

If $X$ is an inner product space, then the orthogonal complement is not necessarily an algebraic complement. For instance, consider $X$ to be a Hilbert space and $(e_n)$ an orthonormal basis. Keep in mind, an orthogonal basis is not a vector space basis for $X$. Take $Y$ to be the linear span of $(e_n)$, so all finite linear combinations of the orthonormal basis. Then the orthogonal complement of $Y$ is just $\{0\}$, while the algebraic complement is quite large, in fact it does not even have a countable basis.