Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

6.5 The dimension formula

  • Page ID
    275
  • [ "article:topic", "vettag:vet4", "targettag:lower", "authortag:schilling", "authorname:schilling" ]

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    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.