Kernel and Image (2)

The following facts are immediate consequences of the rank-nullity theorem.

Corollary 4.28 (Related exercises: Exercise 4.18)

Let \(f : V \to W\) be a linear map between finite-dimensional vector spaces.

  1. If \(f\) is injective then \(\dim V \le \dim W\) (since then \(\ker f = \{0\}\), i.e., \(\dim \ker f = 0\)).

  2. If \(f\) is surjective then \(\dim V \ge \dim W\) (since them \({\operatorname{im}\ } f = W\), so \(\dim {\operatorname{im}\ } f = \dim W\)).

  3. If \(f\) is bijective then \(\dim V = \dim W\).

  4. The preceding three statements can in general not be reversed: if, say, \(\dim V \le \dim W\), \(f\) need not be injective. For example the zero map \(V \to W\), \(v \mapsto 0\) is never injective if \(V \ne \{ 0\}\).

  5. Suppose in addition that \(\dim V = \dim W\). In this case \(f\) is injective precisely if \(f\) is surjective. (If \(f\) is injective, then \(\dim {\operatorname{im}\ } f = \dim V = \dim W\), so that \({\operatorname{im}\ } f = W\) by Theorem 3.713.. Similarly, if \(f\) is surjective, then \(\dim {\operatorname{im}\ } f = \dim W = \dim V\), so \(\dim \ker f = 0\), so that \(\ker f = \{ 0\}\).)

An important case of this theorem is where \(f : {\bf R}^n \to {\bf R}^m\) is the linear map given by multiplication with a fixed \(m \times n\)-matrix \(A\). We call the rank of \(A\), resp. the nullity the rank, resp. nullity of that linear map. The rank is denoted by \({\operatorname {rk}} A\). These are two highly important numbers associated to a matrix, so we want to have a device for computing them. This is based on the following computation: recall from Example 3.62 the standard basis vectors

\[ e_1 = (1, 0, \dots, 0), e_2 = (0, 1, 0, \dots 0), \dots, e_n = (0, \dots, 0, 1) \in {\bf R}^n. \]

We will in the sequel write them as column vectors, so \(e_1 = \left ( \begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right )\) etc. Then we have

\[ f(e_i) = A e_i = \left ( \begin{array}{c} a_{11} \cdot 0 + \dots + a_{1i} \cdot 1 + \dots + a_{1n} \cdot 0 \\ \vdots \\ a_{m1} \cdot 0 + \dots + a_{mi} \cdot 1 + \dots + a_{mn} \cdot 0 \end{array} \right ) = \left ( \begin{array}{c} a_{1i} \\ \vdots \\ a_{mi} \end{array} \right ). \]
\[ A e_i = \left ( \begin{array}{c} a_{1i} \\ \vdots \\ a_{mi} \end{array} \right ). \]

(4.29)

In other words, the product \(A e_i\) is precisely the \(i\)-th column of the matrix \(A\)!

Since any vector \(v \in {\bf R}^n\) is a linear combination of the \(e_i\), we have, for appropriate \(b_1, \dots, b_n \in {\bf R}\)

\[ f(v) = f(\sum_{i=1}^n b_i e_i) = \sum_{i=1}^n b_i f(e_i). \]

Thus, \(f(v)\) is a linear combination of the columns of \(A\). This proves the following statement:

Proposition 4.30 (Related exercises: Exercise 4.17)

Let \(A\) be an \(m \times n\)-matrix and \(f : {\bf R}^n \to {\bf R}^m\) the linear map given by multiplication with \(A\). We write

\[ A = (c_1 \ c_2 \ \dots c_n), \]

i.e., the \(c_i (\in {\bf R}^m)\) is the \(i\)-th column of \(A\). Then

\[ {\operatorname{im}\ } f = L(c_1, \dots, c_n). \]

This subspace of \({\bf R}^m\) is also called the column space of \(A\).

Definition 4.31

The row space of \(A\) is the subspace of \({\bf R}^n\) spanned by the rows of the matrix \(A\).

We can compute the rank of \(A\), i.e., \(\dim {\operatorname{im}\ } f\), as follows:

Proposition 4.32 (Related exercises: Exercise 4.10, Exercise 4.13, Exercise 4.16, Exercise 4.3, Exercise 4.4, Exercise 4.7, Exercise 4.14, Exercise 4.18)

Let \(A\) be an \(m \times n\)-matrix. Suppose \(B\) is a (possibly non-reduced) row-echelon matrix obtained from \(A\) by means of elementary row operations (Definition 2.28).

  1. Then the non-zero rows of \(B\) form a basis of the row space of \(A\).

  2. If the leading ones of \(B\) lie in the columns \(j_1, \dots, j_r\), then these columns of \(A\) form a basis of the column space of \(A\).

  3. The following numbers are all the same: a) \({\operatorname {rk}} A\), b) the dimension of the column space, c) the dimension of the row space of \(A\), d) the number of leading ones in \(B\).

  4. The nullity of \(A\) equals \(n\) (the number of columns of \(A\)) minus any of the quantities in the previous point.

Proof. Parts 1. and 2. can be proven by showing that the row and column space of \(A\) do not change when one performs an elementary row operation to \(A\). We skip this part of the proof (e.g., see (Nicholson 1995, Lemma 5.4.1) for a proof).

3. follows from the first two: by definition, \({\operatorname {rk}} A = \dim {\operatorname{im}\ } f\) equals, by Proposition 4.30, the dimension of the column space. By the second statement, this is equal to the number of leading ones in \(B\). Since \(B\) is a row-echelon matrix, this is also the number of non-zero rows, i.e., by the first statement, the dimension of the row space.

Finally, 4. is a consequence of the rank-nullity-theorem. ◻

Example 4.33

Consider the matrix

\[ A = \left ( \begin{array}{cccc} 1 & 2 & 2 & -1 \\ 3 & 6 & 5 & 0 \\ 1 & 2 & 1 & 2 \end{array} \right ) \]

and the linear map

\[ f : {\bf R}^4 \to {\bf R}^3, v \mapsto Av. \]

The row space is the subspace of \({\bf R}^4\) spanned by the vectors \((1 \ 2 \ 2 \ -1)\) etc., while the column space is the subspace of \({\bf R}^3\) spanned by the vectors \(\left ( \begin{array}{c} 1 \\ 3 \\ 1 \end{array} \right )\) etc. We compute a basis of these two spaces as follows:

\[ A \leadsto \left ( \begin{array}{cccc} 1 & 2 & 2 & -1 \\ 0 & 0 & -1 & 3 \\ 0 & 0 & -1 & 3 \end{array} \right ) \leadsto \left ( \begin{array}{cccc} 1 & 2 & 2 & -1 \\ 0 & 0 & -1 & 3 \\ 0 & 0 & 0 & 0 \end{array} \right ) \leadsto \left ( \begin{array}{cccc} 1 & 2 & 2 & -1 \\ 0 & 0 & 1 & -3 \\ 0 & 0 & 0 & 0 \end{array} \right ). \]

Thus, the vectors \((1, 2, 2, -1)\) and \((0, 0, 1, -3)\) form a basis of the row space. In particular, its dimension is two. The first and third row of \(A\) form a basis of the column space of \(A\), i.e.,

\[ {\operatorname{im}\ } f = L(\left ( \begin{array}{c} 1 \\ 3 \\ 1 \end{array} \right ), \left ( \begin{array}{c} 2 \\ 5 \\ 1 \end{array} \right )). \]

Thus

\[ \dim {\operatorname{im}\ } f = {\operatorname {rk}} f = {\operatorname {rk}} A = 2. \]

According to the rank-nullity theorem (Theorem 4.26),

\[ \dim \ker f = \dim {\bf R}^4 - \dim {\operatorname{im}\ } f = 4 - 2 = 2, \]

(i.e., the nullity of \(f\) or of \(A\) is 2). In order to determine a basis of \(\ker f\), we denote the coordinates in \({\bf R}^4\) by \(x_1, \dots, x_4\). Then, according to Gaussian elimination, the variables \(x_2\) and \(x_4\) are free variables, and \(x_3 = 3 x_4\) from the second row above, and then \(x_1 = -2x_2 - 2x_3 + x_4 = -2x_2 -5x_4\). Thus a basis of \(\ker f\) is given by the vectors

\[ \left ( \begin{array}{c} -2 \\ 1 \\ 0 \\ 0 \end{array} \right ), \left ( \begin{array}{c} -5 \\ 0 \\ 3 \\ 1 \end{array} \right ). \]

This reconfirms that \(\dim \ker f = 2\).

Remark 4.34

Even though the dimension of the row space and the column space are the same, these vector spaces themselves are not the same. In fact, they are not even comparable, given that the row space is a subspace of \({\bf R}^n\), while the column space is a subspace of \({\bf R}^m\).

Here is another consequence of the rank-nullity theorem.

Theorem 4.35

(stated above in Theorem 3.77) Suppose \(A, B \subset V\) are two subspaces of a vector space. Then

\[ \dim (A \cap B) + \dim (A + B) = \dim A + \dim B. \]

(4.36)

Proof. The map

\[ f : A \oplus B \to V, (a, b) \mapsto a-b \]

is linear. Since for every \(b \in B\) also \(b' := -b\) is contained in \(B\), the image of this map is \(A + B\). The kernel of \(f\) consists of those vectors \((a,b) \in A \oplus B\) such that \(a - b = 0\), i.e., \(a = b\). This means that \(a \in A \cap B\). Therefore, the rank nullity theorem and Example 3.67 tell us

\[ \dim (A \cap B) + \dim (A + B) = \dim \ker f + \dim {\operatorname{im}\ } f = \dim (A \oplus B) = \dim A + \dim B. \]

◻

Nicholson, W. K. 1995. *Linear Algebra with Applications*. Mathematics Series. PWS Publishing Company. .