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Mathematics LibreTexts

6.5 The dimension formula

The next theorem is the key result of this chapter. It relates the dimension of the kernel and range of a linear map.

Theorem 6.5.1.  Let \(V \) be a finite-dimensional vector space and \(T:V\to W \) be a linear map. Then \(\range(T) \) is a finite-dimensional subspace of \(W \) and
\[ \begin{equation} \label{eq:dim formula}
    \dim(V) = \dim(\kernel(T)) + \dim(\range(T)). \tag{6.5.1}
\end{equation}\]

Proof. 

Let \(V \) be a finite-dimensional vector space and \(T\in \mathcal{L}(V,W) \). Since \(\kernel(T) \) is a subspace of \(V \), we know that \( \kernel(T) \) has a basis \((u_1,\ldots, u_m) \). This implies that \(\dim(\kernel(T))=m \). By the Basis Extension Theorem, it follows that \( (u_1,\ldots,u_m) \) can be extended to a basis of \(V \), say \((u_1,\ldots,u_m,v_1,\ldots,v_n) \), so that \(\dim(V)=m+n \).

      The theorem will follow by showing that \((Tv_1,\ldots, Tv_n) \) is a basis of \(\range(T) \) since this would imply that \(\range(T) \) is finite-dimensional and \(\dim(\range(T))=n \), proving Equation 6.5.1.

      Since \((u_1,\ldots,u_m,v_1,\ldots,v_n) \) spans \(V \), every \(v\in V \) can be written as a linear combination of these vectors; i.e.,

\begin{equation*}
    v = a_1 u_1 + \cdots + a_m u_m + b_1 v_1 + \cdots + b_n v_n,
\end{equation*}
where \(a_i,b_j\in \mathbb{F} \). Applying \(T \) to \(v \), we obtain
\begin{equation*}
    Tv = b_1 T v_1 + \cdots + b_n T v_n,
\end{equation*}

where the terms \(Tu_i \) disappeared since \(u_i\in \kernel(T) \). This shows that \((Tv_1,\ldots, Tv_n) \) indeed spans \(\range(T) \).

     To show that \((Tv_1,\ldots, Tv_n) \) is a basis of \(\range(T) \), it remains to show that this list is linearly independent. Assume that \(c_1,\ldots, c_n \in \mathbb{F} \) are such that 

\[   c_1 T v_1 + \cdots + c_n T v_n =0.\]


By linearity of \(T \), this implies that

 \[ T(c_1 v_1 + \cdots + c_n v_n) = 0, \]

 

and so \(c_1 v_1 + \cdots + c_n v_n\in \kernel(T) \). Since \((u_1,\ldots,u_m) \) is a basis of \(\kernel(T) \), there must exist scalars \(d_1,\ldots,d_m\in\mathbb{F} \) such that 

\begin{equation*}
    c_1 v_1 + \cdots + c_n v_n = d_1 u_1 + \cdots + d_m u_m.
\end{equation*}

However, by the linear independence of \((u_1,\ldots, u_m,v_1,\ldots, v_n) \), this implies that all coefficients \(c_1=\cdots =c_n=d_1=\cdots =d_m=0 \). Thus, \((Tv_1,\ldots, Tv_n)\) is linearly independent, and we are done.

Example 6.5.2. Recall that the linear map \(T:\mathbb{R}^2 \to \mathbb{R}^2 \) defined by \(T(x,y)=(x-2y,3x+y) \) has \(\kernel(T)=\{0\} \) and \(\range(T)=\mathbb{R}^2 \). It follows that

\[   \dim(\mathbb{R}^2) = 2 = 0+2 =\dim(\kernel(T)) + \dim(\range(T)). \]

Corollary 6.5.3.  Let \(T\) in \(\mathcal{L}(V,W) \).

  1. If \(\dim(V)>\dim(W) \), then \(T \) is not injective.
  2. If \(\dim(V)<\dim(W) \), then \(T \) is not surjective.

 

Proof.

By Theorem 6.5.1, we have that

 \begin{equation*}
\begin{split}
    \dim(\kernel(T)) &= \dim(V) - \dim(\range(T))\\
                  &\ge \dim(V) - \dim(W)>0.
\end{split}
\end{equation*}
Since \(T \) is injective if and only if \(\dim(\kernel(T))=0 \), \(T \) cannot be injective.
Similarly,
\begin{equation*}
\begin{split}
    \dim(\range(T)) &= \dim(V) - \dim(\kernel(T))\\
                  &\le \dim(V)  < \dim(W),
\end{split}
\end{equation*}
and so \(\range(T) \) cannot be equal to \(W \). Hence, \(T \) cannot be surjective.