---
title: "fundamental theorem of linear maps"
type: concept
related: [Linear Map, Subspace, Homogeneity, Null Space, Basis]
source: https://www.jemoka.com/posts/kbhfundamental_theorem_of_linear_maps/
confidence: high
status: active
---

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&rsquo;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&rsquo;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, \(n &lt; \infty\) 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. \(\blacksquare\)
