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. \(\blacksquare\)