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Section 5.4 Row and Column rank

We show that the row and the column rank of a matrix is always the same. Let \(A=(a_{ij})\) be an \(m\times n\) matrix over a field \(F\text{.}\) Let \(\{e_i:1\leq i\leq m\}\) be the standard basis of \(F^m\) and \(\{f_i^t=(0,\ldots,0,1,0,\ldots,0)^t:1\leq i\leq n\}\) be a basis for \(M_{n\times 1}(F)\text{.}\) We make following observations.

Left multiplying \(A\) by \(e_i\) gives the \(i\)-th row of \(A\text{.}\)

\begin{equation} (0,\ldots,0,1,0,\ldots,0)_{1\times m}\cdot\begin{pmatrix}a_{11}\amp a_{12}\amp\cdots\amp a_{1n}\\a_{21}\amp a_{22}\amp\cdots\amp a_{2n}\\\vdots\amp\vdots\amp\ddots\amp\vdots\\a_{m1}\amp a_{m2}\amp\cdots\amp a_{mn} \end{pmatrix}_{m\times n}=(a_{i1},a_{i2},\ldots,a_{in})\label{left-multi-row}\tag{5.5} \end{equation}

Right multiplying \(A\) by \(f_i^t\) gives the \(i\)-th column of \(A\text{.}\)

\begin{equation} \begin{pmatrix}a_{11}\amp a_{12}\amp\cdots\amp a_{1n}\\a_{21}\amp a_{22}\amp\cdots\amp a_{2n}\\\vdots\amp\vdots\amp\ddots\amp\vdots\\a_{m1}\amp a_{m2}\amp\cdots\amp a_{mn} \end{pmatrix}_{m\times n}\cdot\begin{pmatrix}0\\\vdots\\0\\1\\0\\\vdots\\0\end{pmatrix}_{n\times 1}=\begin{pmatrix}a_{1i}\\a_{2i}\\\vdots\\a_{mi}\end{pmatrix}\label{right-multi-column}\tag{5.6} \end{equation}
Definition 5.4.1. (Row space).
Let \(A=(a_{ij})\) be an \(m\times n\) matrix over a field \(F\text{.}\) The row space of \(A\) is \(\langle e_1A,e_2A,\ldots,e_mA\rangle\subset F^m\text{.}\)

The dimension of the row space is called row rank of \(A\text{,}\) and it is denoted by \(rrk(A)\text{.}\)

Definition 5.4.2. (Column space).
Let \(A=(a_{ij})\) be an \(m\times n\) matrix over a field \(F\text{.}\) The column space of \(A\) is \(\langle Af_1^t,Af_2^t,\ldots,Af_n^t\rangle\subset M_{n\times 1}(F)\text{.}\)

The dimension of the column space is called column rank of \(A\text{,}\) and it is denoted by \(crk(A)\text{.}\)

We only prove the result for one of the elementary matrices. Recall from Exercise 1.2.3 that \(T_{pq}(\alpha)\) is the matrix obtained by multiplying \(q\)-th row of \(I_n\) by \(\alpha\in F\) and adding this row to \(p\)-th row of \(I_n\text{.}\) For simplicity we further assume that \(p\) is less than \(q\text{.}\)

Thus, the matrix \(AT_{pq}(\alpha)\) is obtained by multiplying \(q\)-th column of \(A\) by \(\alpha\in F\) and adding this column to \(p\)-th column of \(A\text{.}\) Therefore, using notation of eq. (5.6), the column space of \(AT_{pq}(\alpha)\) is

\begin{equation*} \big\langle Af_1^t,\ldots,Af_{p-1}^t,Af_p^t+\alpha Af_q^t,Af_{p+1}^t,\ldots,Af_q^t,Af_n^t\big\rangle. \end{equation*}

Since, \(Af_p^t=(Af_p^t+\alpha Af_q^t)-\alpha Af_q^t\) we get

\begin{equation*} \big\langle Af_1^t,\ldots,Af_{p-1}^t,Af_p^t+\alpha Af_q^t,Af_{p+1}^t,\ldots,Af_q^t,Af_n^t\big\rangle=\langle Af_i^t:1\leq i\leq n\rangle. \end{equation*}

We thus have \(crk(A)=crk(AT_{pq}(\alpha))\text{.}\)

We now show that an elementary column operation does not change the row rank. Consider the following \(F\)-linear map, right multiplication by \(E\text{:}\)

\begin{equation*} R_E\colon F^n\to F^n\quad\text{defined by}\quad v\mapsto v\cdot E. \end{equation*}

Since \(E\) is an elementary matrix it is invertible and \(F\)-linear map \(R_{E^{-1}}\) is the inverse of \(R_{E}\text{.}\) In particular, \(R_{E}\) is an \(F\)-isomorphism. By Lemma 5.1.6, we have the following.

\begin{equation*} \langle e_1A,\ldots,e_nA\rangle\simeq\langle e_1AE,\ldots,e_nAE\rangle. \end{equation*}

In particular, \(rrk(A)=rrk(AE)\text{.}\)

Remark 5.4.4.
The row rank and the column rank of the matrix obtained in Theorem 5.3.7 are the same.
\begin{equation*} rrk\begin{pmatrix}I_r\amp 0\\0\amp 0\end{pmatrix}=crk\begin{pmatrix}I_r\amp 0\\0\amp 0\end{pmatrix}=r. \end{equation*}

We consider \(L_A\colon M_{n\times 1}(F)\to M_{m\times 1}(F)\) defined in Example 4.2.1. Then, the matrix of \(L_A\) with respect to standard bases is \(A\) (see Example 3.6.1). By Theorem 5.3.7, there exists bases \(\mathfrak{B}\) and \(\mathfrak{C}\) of \(M_{n\times 1}(F)\) and \(M_{m\times 1}(F)\text{,}\) respectively such that

\begin{equation*} [T]_{\mathfrak{B}}^{\mathfrak{C}}=\begin{pmatrix}I_r\amp 0\\0\amp 0\end{pmatrix}. \end{equation*}

Note that \(rrk\begin{pmatrix}I_r\amp 0\\0\amp 0\end{pmatrix}=crk\begin{pmatrix}I_r\amp 0\\0\amp 0\end{pmatrix}=r\text{.}\) By Theorem 5.3.5, there are invertible matrices \(P, Q\) such that

\begin{equation*} A=P\cdot [T]_{\mathfrak{B}}^{\mathfrak{C}}\cdot Q \end{equation*}

Since, \(P,Q\) are invertible matrices, by a part of the proof of Lemma 5.4.3, we get that

\begin{equation*} rrk(A)=r=crk(A). \end{equation*}

In view of the above Theorem 5.4.6 we can define a notion of the rank of a matrix.

Definition 5.4.7. (Rank of a matrix).
Let \(A\in M_{m\times n}(F)\text{.}\) The rank of \(A\) is the row rank (=column rank) of \(A\text{.}\) We denote the rank of \(A\) by \(rk(A)\text{.}\)
Since the row space of \(A\) is the column space of \(A^t\) we get the result by Theorem 5.4.6.
Remark 5.4.9.
By Example 5.6.4, \(A^t\) is a matrix representation of the dual transformation with respect to a dual basis.
Let \(A\in M_{\ell\times m}(F)\) and \(B\in M_{m\times n}(F)\) be matrices. Then
\begin{equation*} rk(AB)\leq\min\{rk(A),rk(B)\}. \end{equation*}

We have column space of \(AB\) is a subspace of the column space of \(A\text{,}\) hence \(rk(AB)\leq rk(A)\text{.}\) Now assume that \(rk(B)<rk(AB)\text{.}\) Let \(\{y_1,\ldots,y_r\}\) be a basis for the column space of \(AB\text{.}\) Thus, there exists \(x_i\) in the column space of \(B\) such that \(y_i=Ax_i\text{.}\) Since, \(rk(B)<r\) the set \(\{x_1,\ldots,x_r\}\) is linearly dependent. Thus, there exists scalars \(\alpha_i\text{,}\) not all zero, such that \(\sum_i\alpha_ix_i=0\text{.}\) This implies that

\begin{equation*} 0=\sum\alpha_iA(x_i)=\sum\alpha_iy_i. \end{equation*}

This is a contradiction and hence we get the result.