Let [math]T:\mathbb{R}^n\to\mathbb{R}^m[/math] be a linear transformation and [math]A[/math] be the m x n matrix such that [math]T(x)=Ax[/math] for any vector [math]x[/math] in [math]\mathbb{R}^n[/math]. Suppose [math]A=\left(\mathbf{a}_1 \ \mathbf{a}_2 \ \cdots \ \mathbf{a}_n\right)[/math], where [math]\mathbf{a}_1, \mathbf{a}_2, \ldots, \mathbf{a}_n[/math] are the column vectors in [math]A[/math]. Then we have the following definitions:[br][br][u]Definition[/u]: The [b]column space[/b] of [math]A[/math], denoted by [math]\text{Col} \ A[/math], is [math]\text{Span}\left\{\mathbf{a}_1,\mathbf{a}_2,\ldots,\mathbf{a}_n\right\}[/math], or equivalently, [math]\text{Im}(T)[/math].[br][br][u]Definition[/u]: The [b]null space[/b] of [math]A[/math], denoted by [math]\text{Nul} \ A[/math], is [math]\left\{x\in\mathbb{R}^n \ | \ Ax=0\right\}[/math], or equivalently, [math]\text{Ker}(T)[/math].[br][br][br]Finding the null space of a matrix [math]A[/math] means solving the homogeneous equation [math]Ax=0[/math]. When A is row reduced to the matrix in reduced echelon form, [math]\dim(\text{Nul} \ A)[/math] is the number of non-pivot positions (or free variables). By the rank-nullity theorem, [math]n=\dim(\text{Ker}(T))+\dim(\text{Im}(T))=\dim(\text{Nul} \ A)+\dim(\text{Col} \ A)[/math]. Hence, [math]\dim(\text{Col} \ A)[/math] is the number of pivot positions (or basic variables) in the reduced echelon form.[br][br][u]Definition[/u]: The [b]rank[/b] (or [b]column rank[/b]) of [math]A[/math] is [math]\dim(\text{Col} \ A)[/math].[br][br]It is obvious that the rank of a m x n matrix must be smaller or equal to [math]\min(m,n)[/math]. A matrix is said to have [b]full rank[/b] if its rank equals [math]\min(m,n)[/math].[br][br]

We consider the following 4 x 5 matrix [math]A[/math]:[br][br][math]A=\begin{pmatrix}-2&-5&8&0&-17\\1&3&-5&1&5\\3&11&-19&7&1\\1&7&-13&5&-3\end{pmatrix}[/math][br][br]Find the basis for [math]\text{Nul} \ A[/math] and [math]\text{Col} \ A[/math].[br][br][u]Solution[/u]: Using the row reduction algorithm, A is row reduced to R in reduced echelon form:[br][br][math]R=\begin{pmatrix}1&0&1&0&1\\0&1&-2&0&3\\0&0&0&1&-5\\0&0&0&0&0\end{pmatrix}[/math][br][br]To find [math]\text{Nul} \ A[/math], we need to solve the homogeneous equation [math]Ax=0[/math], which is equivalent to [math]Rx=0[/math]. The following is the corresponding linear system:[br][br][math]\left\{\begin{eqnarray}x_1+\quad \quad x_3\quad\quad+ \ x_5&=&0\\ \quad \quad x_2-2x_3\quad \ \ +3x_5&=&0\\ x_4-5x_5&=&0\end{eqnarray}\right.[/math][br][br]Then [math]x_3,x_5[/math] are free variables. To express the solutions in parametric vector form, we set [math]x_3=s,x_5=t[/math] and get[br][br][math] \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{pmatrix}= \begin{pmatrix}-s-t \\ 2s-3t \\ s \\ 5t \\ t \end{pmatrix}=s\begin{pmatrix}-1 \\ 2 \\ 1 \\ 0 \\ 0\end{pmatrix}+t\begin{pmatrix}-1 \\ -3 \\ 0 \\ 5 \\ 1\end{pmatrix}[/math][br][br]Hence the basis for [math]\text{Nul} \ A[/math] is [math]\left\{\begin{pmatrix}-1 \\ 2 \\ 1 \\ 0 \\ 0\end{pmatrix},\begin{pmatrix}-1 \\ -3 \\ 0 \\ 5 \\ 1\end{pmatrix}\right\}[/math]. And [math]\dim(\text{Nul} \ A)=2[/math].[br][br][br]As for [math]\text{Col} \ A[/math], we already know that [math]\dim(\text{Col} \ A)=5-\dim(\text{Nul} \ A)=5-2=3[/math]. Therefore, we just need to select three column vectors in the given matrix [math]A[/math] that are linearly independent.[br][br]Claim: The column vectors in [math]A[/math] corresponding to the pivot positions are linearly independent.[br][br]Let [math]E_1,E_2,\ldots, E_r[/math] be the elementary matrices corresponding to the row reduction steps. Let [math]\mathbf{a_1},\mathbf{a_2},\mathbf{a_4}[/math] be the column vectors in [math]A[/math] corresponding to the pivot positions. Then we have[br][br][math]E_r\cdots E_1\mathbf{a_1}=\begin{pmatrix}1\\0\\0\\0\end{pmatrix},E_r\cdots E_1\mathbf{a_2}=\begin{pmatrix}0\\1\\0\\0\end{pmatrix},E_r\cdots E_1\mathbf{a_4}=\begin{pmatrix}0\\0\\1\\0\end{pmatrix}[/math][br][br]Let [math]E=E_r\cdots E_1[/math]. [math]E[/math] is clearly invertible i.e. the corresponding linear transformation is an isomorphism from [math]\mathbb{R}^4[/math] to [math]\mathbb{R}^4[/math]. In other words, the inverse transformation corresponding to [math]E^{-1}[/math] maps the set [math]\left\{\begin{pmatrix}1\\0\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\0\end{pmatrix},\begin{pmatrix}0\\0\\1\\0\end{pmatrix}\right\}[/math] to the set [math]\left\{\mathbf{a_1},\mathbf{a_2},\mathbf{a_4}\right\}[/math]. Since any isomorphism maps a linearly independent set to a linearly independent set (Why?), [math]\left\{\mathbf{a_1},\mathbf{a_2},\mathbf{a_4}\right\}[/math] is linearly independent.[br][br]Hence the basis for [math]\text{Col} \ A[/math] is [math]\left\{\begin{pmatrix}-2\\1\\3\\1\end{pmatrix},\begin{pmatrix}-5\\3\\11\\7\end{pmatrix},\begin{pmatrix}0\\1\\7\\5\end{pmatrix}\right\}[/math].[br][br][br]

Given an m x n matrix [math]A[/math], we can consider its rows instead of columns. Each row can be regarded as a vector in [math]\mathbb{R}^n[/math]. Let [math]\mathbf{r}_1,\mathbf{r}_2,\ldots,\mathbf{r}_m[/math] be the [b]row vectors[/b] of [math]A[/math]. In the previous example, the row vectors for the 4 x 5 matrix are[br][math]\begin{eqnarray}\mathbf{r}_1&=&(-2 \ -5 \ 8 \ 0 \ -17)\\\mathbf{r}_2&=&(1 \ 3 \ -5 \ 1 \ 5)\\\mathbf{r}_3&=&(3 \ 11 \ -19 \ 7 \ 1)\\\mathbf{r}_4&=&(1 \ 7 \ -13 \ 5 \ -3)\end{eqnarray}[/math][br][br][u]Definition[/u]: The [b]row space[/b] of [math]A[/math], denoted by [math]\text{Row} \ A[/math], is [math]\text{Span}\left\{\mathbf{r}_1,\mathbf{r}_2,\ldots,\mathbf{r}_m\right\}[/math].[br][br]How can we find a basis for [math]\text{Row} \ A[/math]?[br][br]Again, we consider the previous example. By the row reduction algorithm, we have [math]EA=E_r\cdots E_1A=R[/math]. Notice that any row operation does not change the row space. (Why?) Therefore, we have [math]\text{Row} \ A=\text{Row} \ R[/math]. In [math]R[/math], we can discard the row of zeros and collect the first three row vectors that span the row space. Obviously, they are linearly independent. [br][br]Hence a basis for [math]\text{Row} \ A[/math] is [math]\left\{(1 \ 0 \ 1 \ 0 \ 1), (0 \ 1 \ -2 \ 0 \ 3), (0 \ 0 \ 0 \ 1 \ -5)\right\}[/math].[br][br][u]Remark[/u]: The above method still works if [math]R[/math] is in echelon form instead of reduced echelon form i.e.[br][br][math]R=\begin{pmatrix}1&3&-5&1&5\\0&1&-2&2&-7\\0&0&0&-4&20\\0&0&0&0&0\end{pmatrix}[/math][br][br]Then a basis for [math]\text{Row} \ A[/math] is [math]\left\{(1 \ 3 \ -5 \ 1 \ 5), (0 \ 1 \ -2 \ 2 \ -7), (0 \ 0 \ 0 \ -4 \ 20)\right\}[/math].[br][br](Note: In general, there are many bases for a vector space so we won't expect the answer is unique.)[br][br][br]In general, for any matrix [math]A[/math], [math]\dim(\text{Col} \ A)=\dim(\text{Row} \ A)[/math] because they both equal the number of pivot positions in [math]A[/math].[br]