fundamental theorem of linear maps

The dimension of the null space plus the dimension of the range of a Linear Map equals the dimension of its domain.
This also implies that both the null space (but this one’s trivial b/c the null space is a subspace of the already finite-dimensional domain) and the range as well is finite-dimensional.
constituents T \in \mathcal{L}( V,W ) finite-dimensional V (otherwise commenting on computing its dimension doesn’t make sense) requirements \begin{equation} \dim V = \dim null\ T + \dim range\ T \end{equation}
for T \in \mathcal{L}(V,W)
proof We desire that \dim V = \dim null\ T + \dim range\ T for T \in \mathcal{L}(V,W).
Let us construct a basis of the null space of T, u_1, \dots u_{m}. This makes \dim null\ T = m.
We can extend this list to a basis of V, the domain, with some vectors v_1, \dots v_{n}. This makes the \dim V = m+n.
We now desire that \dim range\ T = n. We show this by showing Tv_{1}, \dots Tv_{n} is a basis of range\ T.
Recall that u_1, \dots u_{m}, v_1, \dots v_{n} is a basis of V the domain of T. This means that any element that can go into T takes the shape of:

\begin{equation} v = a_1u_1+ \dots +a_{m}u_{m} + b_{1}v_1 + \dots + b_{n}v_{n} \end{equation}

Recall also that the definition of the range of T is that:

\begin{equation} range\ T = \{Tv: v \in V\} \end{equation}

Therefore, every element of the range of T takes the shape of Tv: meaning:

\begin{equation} Tv = a_1Tu_1+ \dots +a_{m}Tu_{m} + b_{1}Tv_1 + \dots + b_{n}Tv_{n} \end{equation}

by additivity and homogeneity of Linear Maps.
Now, Tu_{j}=0, because each u_{j} is a basis (and so definitely at least an element of) the null space of T. This makes the above expression:

\begin{equation} Tv = 0 + b_{1}Tv_1 + \dots + b_{n}Tv_{n} = b_{1}Tv_1 + \dots + b_{n}Tv_{n} \end{equation}

Ok. Given that all elements of the range can be constructed by a linear combination of Tv_{1} \dots Tv_{n}, we declare that the list spans the range of T. Notably, as V is finite-dimensional and v_1, \dots v_{n} is a sublist of its basis, n < \infty and so the range of T is also finite-dimensional.
To finish showing Tv_{1}, \dots, Tv_{n} to be a basis of range\ T, we have to show that its linearly independent.
Suppose:

\begin{equation} c_1Tv_{1} + \dots + c_{n}Tv_{n} = 0 \end{equation}

By homogeneity and additivity, we have that:

\begin{equation} T(c_1v_{1} + \dots + c_{n}v_{n}) = 0 \end{equation}

this makes c_1v_1 + \dots a member of the null space of T. Recall that u_1, \dots u_{m} were a basis thereof, this means that the linear combination of v_{j} can be written as a linear combination of u_{j}:

\begin{equation} c_1 v_1 + \dots + c_{n}v_{n} = d_1 u_{1} + \dots + d_{m} u_{m} \end{equation}

Of course, the list u_1, \dots u_{m}, v_1, \dots v_{n} is linearly independent as it is a basis of V. This makes c_{j}=d_{j}=0 (to see this, move all the d_{j}u_{j} to the left and apply definition of linear independence).
We have therefore shown that, given

\begin{equation} c_1Tv_{1} + \dots + c_{n}Tv_{n} = 0 \end{equation}

c_1 = \dots = c_{n} =0, satisfying the definition of linear independence of the list of Tv_{j}.
Having shown that Tv_{j} to be a linearly independent spanning list of range\ T, we can conclude that it is indeed a basis of range\ T.
This makes the \dim range\ T = n, as desired. \blacksquare