Corollary 2.3.1.
Let \(V\) be a finite-dimensional vector space over a field \(F\text{,}\) and let \(U\) be a subspace of \(V.\) Then
\begin{equation*}
\dim (V/U) = \dim V - \dim U.
\end{equation*}
Section 2.3 Linear Maps out of quotients
Suppose we have vector spaces \(V,W\) and a subspace \(U\subseteq V\text{.}\) How should one define a linear map \(T:V/U
\to W?\) In general one should not (as least directly)! We want to show that every well-defined linear map \(T:V/U \to W\) arises in a natural way from a linear map \(T_0:V\to W.\)
Why such a fuss? How hard is it to define a map on cosets? While it isn’t that hard, each such definition requires an extra step. To define \(T(v+U)\) one must show that the definition is well-defined, meaning if \(v+U = v'+U\text{,}\) then \(T(v+U) =
T(v'+U).\) The method we shall propose will do this once and the result will apply to all maps.
To begin, we first note that there is a natural linear map (called a projection) \(\pi:V \to V/U\) defined by
\begin{equation*}
\pi(v) = v+U.
\end{equation*}
It is easy to check that this is a surjective linear map with \(\ker(\pi) = U.\)
As an immediate corollary of properties of the projection map, we deduce:
Proof.
Consider the projection map \(\pi: V\to V/U\) (given by \(\pi(v) = v+U)\text{.}\) We have already stated that \(\pi\) is a surjective linear map with \(\ker \pi = U.\) So by the rank-nullity theorem we have that
\begin{equation*}
\dim V = \dim (V/U) + \dim U,
\end{equation*}
from which the result follows.
So now we suppose that we have a linear map \(T:V\to W\text{,}\) and a subspace \(U\subseteq V\text{.}\) We want to know when we can induce a linear map \(T_*: V/U\to W\) which makes the diagram below commute. What that means is that starting with a vector \(v\in V\text{,}\) following either path to \(W\) produces the same result. In terms of the functions, this means that
\begin{equation*}
T(v) = (T_*\circ \pi) (v) = T_*(v+U).
\end{equation*}
It is evident that if such a linear map exists, it can have only one definition:
\begin{equation*}
T_*(v+U) = T(v),
\end{equation*}
and it is here we confront and deal with the issue of \(T_*\) being well-defined. If \(v+U = v'+U\text{,}\) we need that \(T(v) =
T(v')\text{.}\)
By definition, the condition \(v+U = v'+U\) is equivalent to \(v-v'\in U\text{,}\) say \(v = v' + u\) for some \(u\in U.\) The requirement that \(T(v) =
T(v')\) demands that
\begin{equation*}
T(v) = T(v'+u) = T(v') + T(u) = T(v'),
\end{equation*}
so we must have \(T(u)=0.\) Thus a necessary and sufficient condition that the map be well-defined is that \(U\subseteq \ker(T).\)
Theorem 2.3.3. Fundamental theorem on linear maps.
Let \(T:V\to W\) be a linear map and \(U\) a subspace of \(V\) with \(U \subseteq \ker(T).\) Then there is a unique linear map \(T_*:V/U \to W\) (defined by \(T_*(v+U) = T(v)\)) with \(\Im(T_*)=\Im(T)\text{,}\) and with \(\ker(T_*) = \ker(T)/U.\)
Corollary 2.3.4.
If \(T:V\to W\) is a linear map, then the induced map \(T_*:V/\ker T \to W\) is injective.
What this says is that if the original map \(T\) is not injective (many things in \(V\) mapping to the same element in \(W\)), the coset \(v+\ker T\) collects together all the elements of \(V\) which map to the element \(T(v)\) in \(W.\) In this way we obtain an isomorphic copy of \(V/\ker T\) inside of \(W.\)
Corollary 2.3.5. First Isomorphism theorem.
Let \(T:V\to W\) be a linear map. Then \(V/\ker(T) \cong
\Im(T)\) via the map \(T_*(v+\ker(T)) = T(v).\) In particular, if \(T\) is a surjective map then \(V/\ker(T)\cong
W.\)
In the interest of full disclosure, the first isomorphism theorem (and its corollaries) are not that robust in linear algebra since vector spaces are classified up to isomophism by their dimension. They become much more important in group and ring theory, but let’s try to give a sense of what they accomplish.
Example 2.3.6.
Let’s take a simple example from multivariable calculus. Let \(a,b,c\) be real numbers, not all zero. Consider the linear map \(T:\R^3 \to\R\) given by \(T(x,y,z) = ax + by + cz\text{.}\) This is certainly linear as it is of the form \(\mathbf x \mapsto
A\mathbf x\) where \(A\) is the \(1\times 3\) matrix \([a\
b\ c]\text{.}\)
Let \(P =\ker T\text{,}\) that is \(P\) is the plane through the origin \(ax+by+cz=0.\) How do we characterize the quotient space \(\R^3/P?\)
It is trivial to check that the map \(T\) is surjective, so the first isomorphism theorem says that \(\R^3/P \cong \R.\)
Alternatively, we know that \(\dim P = 2\text{,}\) so \(\dim
(\R^3/P) = 3-2=1\text{,}\) and any two spaces of the same (finite) dimension are isomorphic, though the orignial map \(T\) is more intrinsic to the problem.
