In the book of Introductory Real Analysis of Kolmogorov and Fomin (page 122), there is some theorem about the dimension of quotient spaces. Let $L$ be vector space and $L'$ be a vector subspace. It says that $L/L'$ (the quotient space) is finite dimensional (with dimension $n$) if and only if any $x \in L$ can be written as:
$$x= \mu_1 x_1 + … \mu_n x_n + y$$
where $x_i \notin L'$, $y \in L'$.
My question is then the following: Is this equivalent to saying that the quotient space is finite dimensional if and only if $L-L'$ (the complement of $L'$) is finite dimensional? It seems right but I am struggling proving it. Thanks!
Best Answer
This is true depending on your definitions. If you mean the set theoretic complement $L-L'$, which is to say the set of all vectors $v \in L$ which are not in $L'$, then no. The set theoretic complement is not a vector subspace of $L$, and hence cannot be assigned a vector space dimension. If you mean a vector space complement of $L'$ in $L$, which is to say a subspace $L''$ of $L$ such that $L$ is the direct sum of $L'$ and $L''$, then this is true.
Note that it is really more correct to say a (vector space) complement of a subspace rather than the complement. This is because complements are far from unique. For example, if $L$ is the plane $\mathbb R^2$, and $L'$ is a one-dimensional subspace of $L$ (that is, a line passing through the origin), then every other line passing through the origin is a complement of $L'$ in $L$.
Anyway, if $L''$ is a complement of $L'$ in $L$, then $L''$ is isomorphic to the quotient vector space $L/L'$, an isomorphism being given by sending any $v \in L''$ to the coset $v + L'$ in $L/L'$. Hence $L/L'$ is finite dimensional if and only if $L''$ is.