Opening the book…
In this section we use the concept of dimension to clarify the definition of the rank of a matrix given in Section 1_2, and to study its properties. This requires that we deal with rows and columns in the same way. While it has been our custom to write the \(n\)-tuples in \(\mathbb{R}^n\) as columns, in this section we will frequently write them as rows. Subspaces, independence, spanning, and dimension are defined for rows using matrix operations, just as for columns. If \(A\) is an \(m \times n\) matrix, we define:
- The column space, \(\text{col} A\), of \(A\) is the subspace of \(\mathbb{R}^m\) spanned by the columns of \(A\).
- The row space, \(\text{row} A\), of \(A\) is the subspace of \(\mathbb{R}^n\) spanned by the rows of \(A\).
Much of what we do in this section involves these subspaces. We begin with:
Let \(A\) and \(B\) denote \(m \times n\) matrices.\begin{enumerate}
- If \(A \to B\) by elementary row operations, then \(\text{row} A = \text{row} B\).
- If \(A \to B\) by elementary column operations, then \(\text{col} A = \text{col} B\).
We prove (1); the proof of (2) is analogous. It is enough to do it in the case when \(A \to B\) by a single row operation. Let \(R_{1}, R_{2}, \dots, R_{m}\) denote the rows of \(A\). The row operation \(A \to B\) either interchanges two rows, multiplies a row by a nonzero constant, or adds a multiple of a row to a different row. We leave the first two cases to the reader. In the last case, suppose that \(a\) times row \(p\) is added to row \(q\) where \(p < q\). Then the rows of \(B\) are \(R_{1}, \dots, R_{p}, \dots, R_{q} + aR_p, \dots, R_{m}\), and Theorem 5.1.1 shows that
\[ \text{span}\{ R_1, \dots, R_p, \dots, R_q, \dots, R_m \} = \text{span}\{ R_1, \dots, R_p, \dots, R_q + aR_p, \dots, R_m \} \nonumber\]
That is, \(\text{row} A = \text{row} B\).
\(\square\)
If \(A\) is any matrix, we can carry \(A \to R\) by elementary row operations where \(R\) is a row-echelon matrix. Hence \(\text{row} A = \text{row} R\) by Lemma 5.4.1; so the first part of the following result is of interest.
If \(R\) is a row-echelon matrix\index{matrix!row-echelon matrix}, then
- The nonzero rows of \(R\) are a basis of \(\text{row} R\).
- The columns of \(R\) containing leading ones are a basis of \(\text{col} R\).
The rows of \(R\) are independent by Example 5.2.6, and they span \(\text{row} R\) by definition. This proves (1).
Let \(\textbf{c}_{j_{1}}, \textbf{c}_{j_{2}}, \dots, \textbf{c}_{j_{r}}\) denote the columns of \(R\) containing leading \(1\)s. Then \(\{\textbf{c}_{j_{1}}, \textbf{c}_{j_{2}}, \dots, \textbf{c}_{j_{r}}\}\) is independent because the leading \(1\)s are in different rows (and have zeros below and to the left of them). Let \(U\) denote the subspace of all columns in \(\mathbb{R}^m\) in which the last \(m - r\) entries are zero. Then \(\text{dim}U = r\) (it is just \(\mathbb{R}^r\) with extra zeros). Hence the independent set \(\{\textbf{c}_{j_{1}}, \textbf{c}_{j_{2}}, \dots, \textbf{c}_{j_{r}}\}\) is a basis of \(U\) by Theorem 5.2.7. Since each \(\textbf{c}_{j_{i}}\) is in \(\text{col} R\), it follows that \(\text{col} R = U\), proving (2).
\(\square\)
With Lemma 5.4.2 we can fill a gap in the definition of the rank of a matrix given in Chapter 1. Let \(A\) be any matrix and suppose \(A\) is carried to some row-echelon matrix \(R\) by row operations. Note that \(R\) is not unique. In Section 1.2 we defined the \(\textbf{rank}\) of \(A\), denoted \(\text{rank} A\), to be the number of leading \(1\)s in \(R\), that is the number of nonzero rows of \(R\). The fact that this number does not depend on the choice of \(R\) was not proved in Section 1.2. However part 1 of Lemma 5.4.2 shows that
\[\text{rank } A = \text{dim} (\text{row } A) \nonumber\]
and hence that \(\text{rank} A\) is independent of \(R\).
Lemma 5.4.2 can be used to find bases of subspaces of \(\mathbb{R}^n\) (written as rows). Here is an example.
Find a basis of \(U = \text{span}\{(1, 1, 2, 3), (2, 4, 1, 0), (1, 5, -4, -9)\}\).
Solution
\(U\) is the row space of
\[\left[ \begin{array}{rrrr}
1 & 1 & 2 & 3 \\
2 & 4 & 1 & 0 \\
1 & 5 & -4&-9
\end{array} \right] \nonumber\]
This matrix has row-echelon form
\[\left[ \begin{array}{rrrr}
1 & 1 & 2 & 3 \\
0 & 1 & -\frac{3}{2} & -3 \\
0 & 0 & 0 & 0
\end{array} \right] \nonumber\]
so \(\{(1, 1, 2, 3), (0, 1, - \frac{3}{2}, -3)\}\) is basis of \(U\)
by Lemma 5.4.2.
Note that \(\{(1, 1, 2, 3), (0, 2, -3, -6)\}\) is another basis that avoids fractions.
Lemmas 5.4.1 and 5.4.2 are enough to prove the following fundamental theorem.
Let \(A\) denote any \(m \times n\) matrix of rank \(r\). Then\index{rank!theorem}
\[\text{dim}(\text{col }A ) = \text{dim} (\text{row } A) = r \nonumber\]
Moreover, if \(A\) is carried to a row-echelon matrix \(R\) by row operations, then
\item The \(r\) nonzero rows of \(R\) are a basis of \(\text{row} A\).
\item If the leading \(1\)s lie in columns \(j_{1}, j_{2}, \dots, j_{r}\) of \(R\), then columns \(j_{1}, j_{2}, \dots, j_{r}\) of \(A\) are a basis of \(\text{col} A\).
We have \(\text{row} A = \text{row} R\) by Lemma 5.4.1, so (1) follows from Lemma 5.4.2. Moreover, \(R = UA\) for some invertible matrix \(U\) by Theorem 2.5.1. Now write
\[A = \left[ \begin{array}{cccc} \textbf{c}_1 & \textbf{c}_2 & \dots & \textbf{c}_n \end{array} \right] \nonumber\]
where \(\textbf{c}_{1}, \textbf{c}_{2}, \dots, \textbf{c}_{n}\) are the columns of \(A\). Then
\[R = UA = U \left[ \begin{array}{cccc} \textbf{c}_1 & \textbf{c}_2 & \cdots & \textbf{c}_n \end{array} \right] = \left[ \begin{array}{cccc} U\textbf{c}_1 & U\textbf{c}_2 & \cdots & U\textbf{c}_n \end{array} \right] \nonumber\]
Thus, in the notation of (2), the set \(B = \{U\textbf{c}_{j_{1}}, U\textbf{c}_{j_{2}}, \dots, U\textbf{c}_{j_{r}}\}\) is a basis of \(\text{col} R\) by Lemma 5.4.2. So, to prove (2) and the fact that \(\text{dim}(\text{col} A) = r\), it is enough to show that \(D = \{\textbf{c}_{j_{1}}, \textbf{c}_{j_{2}}, \dots, \textbf{c}_{j_{r}}\}\) is a basis of \(\text{col} A\). First, \(D\) is linearly independent because \(U\) is invertible (verify), so we show that, for each \(j\), column \(\textbf{c}_{j}\) is a linear combination of the \(\textbf{c}_{j_{i}}\). But \(U\textbf{c}_{j}\) is column \(j\) of \(R\), and so is a linear combination of the \(U\textbf{c}_{j_{i}}\), say \(U\textbf{c}_{j} = a_{1}U\textbf{c}_{j_{1}} + a_{2}U\textbf{c}_{j_{2}} + \dots + a_{r}U\textbf{c}_{j_{r}}\) where each \(a_{i}\) is a real number.
Since \(U\) is invertible, it follows that
\[\textbf{c}_{j} = a_{1}\textbf{c}_{j_{1}} + a_{2}\textbf{c}_{j_{2}} + \dots + a_{r}\textbf{c}_{j_{r}} \nonumber\]
and the proof is complete.
\(\square\)
Compute the rank of
\[A = \left[ \begin{array}{rrrr}
1 & 2 & 2 & -1 \\
3 & 6 & 5 & 0 \\
1 & 2 & 1 & 2
\end{array} \right] \nonumber\]
and find bases for \(\text{row} A\) and \(\text{col} A\)
Solution
The reduction of \(A\) to row-echelon form is as follows:
\[\left[ \begin{array}{rrrr}
1 & 2 & 2 & -1 \\
3 & 6 & 5 & 0 \\
1 & 2 & 1 & 2
\end{array} \right]
\rightarrow
\left[ \begin{array}{rrrr}
1 & 2 & 2 & -1 \\
0 & 0 & -1 & 3 \\
0 & 0 & -1 & 3
\end{array} \right]
\rightarrow
\left[ \begin{array}{rrrr}
1 & 2 & 2 & -1 \\
0 & 0 & -1 & 3 \\
0 & 0 & 0 & 0
\end{array} \right] \nonumber\]
Hence \(\text{rank} A\) = 2, and
\[ \{
\left[ \begin{array}{cccc}
1 & 2 & 2 & -1
\end{array} \right],
\left[ \begin{array}{cccc}
0 & 0 & 1 & -3
\end{array} \right] \} \nonumber\]
is a basis of \(\text{row} A\) by Lemma 5.4.2. Since the leading \(1\)s are in columns 1 and 3 of the row-echelon matrix, Theorem 5.4.1 shows that columns 1 and 3 of \(A\) are a basis
\[\left\{
\left[ \begin{array}{r}
1\\
3\\
1
\end{array} \right], \left[ \begin{array}{r}
2\\
5\\
1
\end{array} \right]
\right\} \nonumber\]
of \(\text{col} A\).
Theorem 5.4.1 has several important consequences. The first, Corrollary 5.4.1 below, follows because the rows of \(A\) are independent (respectively span \(\text{row} A\)) if and only if their transposes are independent (respectively span \(\text{col} A\)).
If \(A\) is any matrix, then \(\text{rank} A = \text{rank}(A^{T})\).
If \(A\) is an \(m \times n\) matrix, we have \(\text{col} A \subseteq \mathbb{R}^m\) and \(\text{row} A \subseteq \mathbb{R}^n\). Hence Theorem 5.2.8 shows that \(\text{dim}(\text{col} A) \leq \text{dim}(\mathbb{R}^m) = m\) and \(\text{dim}(\text{row} A) \leq \text{dim}(\mathbb{R}^n) = n\). Thus Theorem 5.4.1 gives:
If \(A\) is an \(m \times n\) matrix, then \(\text{rank }A \leq m\) and \(\text{rank }A \leq n\).
\(\text{rank} A = \text{rank}(UA) = \text{rank}(AV)\) whenever \(U\) and \(V\) are invertible.
Lemma 5.4.1 gives \(\text{rank} A = \text{rank}(UA)\). Using this and Corrollary 5.4.1 we get
\[\text{rank} (AV) = \text{rank}(AV)^T = \text{rank} (V^TA^T) = \text{rank}(A^T)= \text{rank } A \nonumber\]
\(\square\)
The next corollary requires a preliminary lemma.
Let \(A\), \(U\), and \(V\) be matrices of sizes \(m \times n\), \(p \times m\), and \(n \times q\) respectively.
- \(\text{col}(AV) \subseteq \text{col} A\), with equality if \(VV^{\prime}= I_{n}\) for some \(V^{\prime}\).
- \(\text{row}(UA) \subseteq \text{row} A\), with equality if \(U^{\prime}U=I_{m}\) for some \(U^{\prime}\).
For (1), write \(V = \left[ \textbf{v}_{1}, \textbf{v}_{2}, \dots, \textbf{v}_{q}\right]\) where \(\textbf{v}_{j}\) is column \(j\) of \(V\). Then we have
\[AV = \left[ A\textbf{v}_{1}, A\textbf{v}_{2}, \dots, A\textbf{v}_{q}\right], \nonumber\]
and each \(A\textbf{v}_{j}\) is in \(\text{col} A\) by Definition 2.5. It follows that \(\text{col}(AV) \subseteq \text{col} A\). If \(VV^{\prime}=I_{n}\), we obtain \(\text{col} A = \text{col}\left[ (AV)V^{\prime} \right] \subseteq \text{col}(AV)\) in the same way. This proves (1).
As to (2), we have \(\text{col}\left[ (UA)^{T}\right] = \text{col}(A^{T}U^{T}) \subseteq \text{col}(A^{T})\) by (1), from which \(\text{row}(UA) \subseteq \text{row} A\). If \(U^{\prime}U=I_{m}\), this is equality as in the proof of (1).
\(\square\)
If \(A\) is \(m \times n\) and \(B\) is \(n \times m\), then \(\text{rank} AB \leq \text{rank} A\) and \(\text{rank} AB \leq \text{rank} B\).
By Lemma 5.4.3, \(\text{col}(AB) \subseteq \text{col} A\) and \(\text{row}(BA) \subseteq \text{row} A\), so Theorem 5.4.1 applies.
\(\square\)
In Section 5.1 we discussed two other subspaces associated with an \(m \times n\) matrix \(A\): the null space \(\text{null}(A)\) and the image space \(\text{im}(A)\)
\[\text{null} (A) = \{\textbf{x} \mbox{ in } \mathbb{R}^n \mid A\textbf{x} = \textbf{0} \}\mbox{ and } \text{im} (A) = \{A\textbf{x} \mid \textbf{x} \mbox{ in } \mathbb{R}^n \} \nonumber\]
Before we proceed to an important theorem, we first define what is meant by the nullity of a matrix.
The dimension of the null space of a matrix is called the nullity, denoted by \(\text{dim}[\text{null}(A)]\).
We will see shortly that the rank and the nullity of an \(m \times n\) matrix \(A\) add up to \(n\) (no matter m!), this is part (1) of the following theorem.
Recall that \(\text{im}(A)= \text{col}(A)\) by Example 5.1.8. So if A has rank r, we have \(\text{dim}[\text{im}(A)]= \text{dim}[\text{col} (A)]=r\). Hence Theorem 5.4.1 provides a method of finding a basis of \(\text{im}(A)\). This is recorded as part (2) of the following theorem.
Let \(A\) denote an \(m \times n\) matrix of rank \(r\). Then
- The \(n - r\) basic solutions to the system \(A\textbf{x} = \textbf{0}\) provided by the gaussian algorithm are a basis of \(\text{null}(A)\), so \(\text{dim}[\text{null}(A)] = n - r\).
- Theorem 5.4.1 provides a basis of \(\text{im}(A) = \text{col}(A)\), and \(\text{dim}[\text{im}(A)] = r\).
It remains to prove (1). We already know (Theorem 2.2.1) that \(\text{null}(A)\) is spanned by the \(n - r\) basic solutions of \(A\textbf{x} = \textbf{0}\). Hence using Theorem 5.2.7, it suffices to show that \(\text{dim}[\text{null}(A)] = n - r\). So let \(\{\textbf{x}_{1}, \dots, \textbf{x}_{k}\}\) be a basis of \(\text{null}(A)\), and extend it to a basis \(\{\textbf{x}_{1}, \dots, \textbf{x}_{k}, \textbf{x}_{k+1}, \dots, \textbf{x}_{n}\}\) of \(\mathbb{R}^n\) (by Theorem 5.2.6). It is enough to show that \(\{A\textbf{x}_{k+1}, \dots, A\textbf{x}_{n}\}\) is a basis of \(\text{im}(A)\); then \(n - k = r\) by the above and so \(k = n - r\) as required.
Spanning
Choose \(A\textbf{x}\) in \(\text{im}(A)\), \(\textbf{x}\) in \(\mathbb{R}^n\), and write \(\textbf{x} = a_{1}\textbf{x}_{1} + \dots + a_{k}\textbf{x}_{k} + a_{k+1}\textbf{x}_{k+1} + \dots + a_{n}\textbf{x}_{n}\) where the \(a_{i}\) are in \(\mathbb{R}\). Then \(A\textbf{x} = a_{k+1}A\textbf{x}_{k+1} + \dots + a_{n}A\textbf{x}_{n}\) because \(\{\textbf{x}_{1}, \dots, \textbf{x}_{k}\} \subseteq \text{null}(A)\).
Independence
Let \(t_{k+1}A\textbf{x}_{k+1} + \dots + t_{n}A\textbf{x}_{n} = \textbf{0}\), \(t_{i}\) in \(\mathbb{R}\). Then \(t_{k+1}\textbf{x}_{k+1} + \dots + t_{n}\textbf{x}_{n}\) is in \(\text{null} A\), so \(t_{k+1}\textbf{x}_{k+1} + \dots + t_{n}\textbf{x}_{n} = t_{1}\textbf{x}_{1} + \dots + t_{k}\textbf{x}_{k}\) for some \(t_{1}, \dots, t_{k}\) in \(\mathbb{R}\). But then the independence of the \(\textbf{x}_{i}\) shows that \(t_{i} = 0\) for every \(i\).
\(\square\)
If \[A = \left[ \begin{array}{rrrr} 1 & -2 & 1 & 1 \\-1 & 2 & 0 & 1 \\2 & -4 & 1 & 0 \end{array} \right] \nonumber\]
find bases of \(\text{null}(A)\) and \(\text{im}(A)\0, and so find their dimensions.
Solution
If \(\textbf{x}\) is in \(\text{null}(A)\), then \(A\textbf{x} = \textbf{0}\), so \(\textbf{x}\) is given by solving the system \(A\textbf{x} = \textbf{0}\). The reduction of the augmented matrix to reduced form is
\[
\left[ \begin{array}{rrrr|r}
1 & -2 & 1 & 1 & 0 \\
-1 & 2 & 0 & 1 & 0 \\
2 & -4 & 1 & 0 & 0
\end{array} \right]
\rightarrow
\left[ \begin{array}{rrrr|r}
1 & -2 & 0 & -1 & 0 \\
0 & 0 & 1 & 2 & 0 \\
0 & 0 & 0 & 0 & 0
\end{array} \right]
\nonumber \]
Hence \(r = \text{rank}(A) = 2\). Here, \(\text{im}(A) = \text{col}(A)\) has basis
\[\left\{
\left[ \begin{array}{r}
1\\
-1\\
2
\end{array} \right], \left[ \begin{array}{r}
1\\
0\\
1
\end{array} \right]\right\} \nonumber\]
by Theorem 5.4.1 because the leading \(1\)s are in columns 1 and 3. In particular, \(\text{dim}[\text{im}(A)] = 2 = r\) as in Theorem 5.4.2.
Turning to \(\text{null}(A)\), we use gaussian elimination. The leading variables are \(x_{1}\) and \(x_{3}\), so the nonleading variables become parameters: \(x_{2} = s\) and \(x_{4} = t\). It follows from the reduced matrix that \(x_{1} = 2s + t\) and \(x_{3} = -2t\), so the general solution is
\[\textbf{x} =
\left[ \begin{array}{r}
x_1\\
x_2\\
x_3\\
x_4
\end{array} \right] = \left[ \begin{array}{c}
2s + t\\
s\\
-2t\\
t
\end{array} \right] = s\textbf{x}_1 + t\textbf{x}_2 \mbox{ where } \textbf{x}_1 =
\left[ \begin{array}{r}
2\\
1\\
0\\
0
\end{array} \right], \mbox{ and } \textbf{x}_2 =
\left[ \begin{array}{r}
1\\
0\\
-2\\
1
\end{array} \right]. \nonumber\]
Hence \(\text{null}(A)\). But \(\textbf{x}_{1}\) and \(\textbf{x}_{2}\) are solutions (basic), so
\[\text{null}(A) = \text{span}\{\textbf{x}_1, \textbf{x}_2 \} \nonumber\]
However Theorem 5.4.2 asserts that \(\{\textbf{x}_{1}, \textbf{x}_{2}\}\) is a basis of \(\text{null}(A)\). (In fact it is easy to verify directly that \(\{\textbf{x}_{1}, \textbf{x}_{2}\}\) is independent in this case.) In particular, \(\text{dim}[\text{null}(A)] = 2 = n - r\), as Theorem 5.4.2 asserts.
Let \(A\) be an \(m \times n\) matrix. Corollary 5.4.2 of Theorem 5.4.1 asserts that \(\text{rank} A \leq m\) and \(\text{rank} A \leq n\), and it is natural to ask when these extreme cases arise. If \(\textbf{c}_{1}, \textbf{c}_{2}, \dots, \textbf{c}_{n}\) are the columns of \(A\), Theorem 5.2.2 shows that \(\{\textbf{c}_{1}, \textbf{c}_{2}, \dots, \textbf{c}_{n}\}\) spans \(\mathbb{R}^m\) if and only if the system \(A\textbf{x} = \textbf{b}\) is consistent for every \(\textbf{b}\) in \(\mathbb{R}^m\), and that \(\{\textbf{c}_{1}, \textbf{c}_{2}, \dots, \textbf{c}_{n}\}\) is independent if and only if \(A\textbf{x} = \textbf{0}\), \(\textbf{x}\) in \(\mathbb{R}^n\), implies \(\textbf{x} = \textbf{0}\). The next two useful theorems improve on both these results, and relate them to when the rank of \(A\) is \(n\) or \(m\).
The following are equivalent for an \(m \times n\) matrix \(A\):
- \(\text{rank} A = n\).
- The rows of \(A\) span \(\mathbb{R}^n\).
- The columns of \(A\) are linearly independent in \(\mathbb{R}^m\).
- The \(n \times n\) matrix \(A^{T}A\) is invertible.
- \(CA = I_{n}\) for some \(n \times m\) matrix \(C\).
- If \(A\textbf{x} = \textbf{0}\), \(\textbf{x}\) in \(\mathbb{R}^n\), then \(\textbf{x} = \textbf{0}\).
(1) \(\Rightarrow\) (2). We have \(\text{row} A \subseteq \mathbb{R}^n\), and \(\text{dim}(\text{row} A) = n\) by (1), so \(\text{row} A = \mathbb{R}^n\) by Theorem 5.2.8. This is (2).
(2) \(\Rightarrow\) (3). By (2), \(\text{row} A = \mathbb{R}^n\), so \(\text{rank} A = n\). This means \(\text{dim}(\text{col} A) = n\). Since the \(n\) columns of \(A\) span \(\text{col} A\), they are independent by Theorem 5.2.7.
(3) \(\Rightarrow\) (4). If \((A^{T}A)\textbf{x} = \textbf{0}\), \(\textbf{x}\) in \(\mathbb{R}^n\), we show that \(\textbf{x} = \textbf{0}\) (Theorem 2.4.5). We have
\[|| A\textbf{x} ||^2 = (A\textbf{x})^TA\textbf{x} = \textbf{x}^TA^TA\textbf{x} = \textbf{x}^T\textbf{0} = \textbf{0} \nonumber\]
Hence \(A\textbf{x} = \textbf{0}\), so \(\textbf{x} = \textbf{0}\) by (3) and Theorem 5.2.2.
(4) \(\Rightarrow\) (5). Given (4), take \(C = (A^{T}A)^{-1}A^{T}\).
(5) \(\Rightarrow\) (6). If \(A\textbf{x} = \textbf{0}\), then left multiplication by \(C\) (from (5)) gives \(\textbf{x} = \textbf{0}\).
(6) \(\Rightarrow\) (1). Given (6), the columns of \(A\) are independent by Theorem 5.2.2. Hence \(\text{dim}(\text{col} A) = n\), and (1) follows.
\(\square\)
The following are equivalent for an \(m \times n\) matrix \(A\):
- \(\text{rank} A = m\).
- The columns of \(A\) span \(\mathbb{R}^m\).
- The rows of \(A\) are linearly independent in \(\mathbb{R}^n\).
- The \(m \times m\) matrix \(AA^T\) is invertible.
- \(AC = I_m\) for some \(n \times m\) matrix \(C\).
- The system \(A\textbf{x} = \textbf{b}\) is consistent for every \(\textbf{b}\) in \(\mathbb{R}^m\).
(1) \(\Rightarrow\) (2). By (1), \(\text{dim}(\text{col} A = m\), so \(\text{col} A = \mathbb{R}^m\) by Theorem 5.2.8.
(2) \(\Rightarrow\) (3). By (2), \(\text{col} A = \mathbb{R}^m\), so \(\text{rank} A = m\). This means \(\text{dim}(\text{row} A) = m\). Since the \(m\) rows of \(A\) span \(\text{row} A\), they are independent by Theorem 5.2.7.
(3) \(\Rightarrow\) (4). We have \(\text{rank} A = m\) by (3), so the \(n \times m\) matrix \(A^{T}\) has rank \(m\). Hence applying Theorem 5.4.3 to \(A^{T}\) in place of \(A\) shows that \((A^{T})^{T}A^{T}\) is invertible, proving (4).
(4) \(\Rightarrow\) (5). Given (4), take \(C = A^{T}(AA_{T})^{-1}\) in (5).
(5) \(\Rightarrow\) (6). Comparing columns in \(AC = I_{m}\) gives \(A\textbf{c}_{j} = \textbf{e}_{j}\) for each \(j\), where \(\textbf{c}_{j}\) and \(\textbf{e}_{j}\) denote column \(j\) of \(C\) and \(I_{m}\) respectively. Given \(\textbf{b}\) in \(\mathbb{R}^m\), write \(\textbf{b} = \sum_{j=1}^{m} r_j\textbf{e}_j\), \(r_{j}\) in \(\mathbb{R}\). Then \(A\textbf{x} = \textbf{b}\) holds with \(\textbf{x}= \sum_{j=1}^{m} r_j\textbf{c}_j \) as the reader can verify.
(6) \(\Rightarrow\) (1). Given (6), the columns of \(A\) span \(\mathbb{R}^m\) by Theorem 5.2.2. Thus \(\text{col} A = \mathbb{R}^m\) and (1) follows.
\(\square\)
Show that
\[ \left[ \begin{array}{cc}
3 & x + y + z \\
x + y + z & x^2 +y^2 + z^2
\end{array} \right] \nonumber\]
is invertible if \(x\), \(y\), and \(z\) are not all equal.
Solution
The given matrix has the form \(A^{T}A\) where
\[A =
\left[ \begin{array}{rr}
1 & x\\
1 & y \\
1 & z
\end{array} \right] \nonumber\]
has independent columns because \(x\), \(y\), and \(z\) are not all equal (verify). Hence Theorem 5.4.3 applies.
Theorem 5.4.3 and Theorem 5.4.4 relate several important properties of an \(m \times n\) matrix \(A\) to the invertibility of the square, symmetric matrices \(A^{T}A\) and \(AA^{T}\). In fact, even if the columns of \(A\) are not independent or do not span \(\mathbb{R}^m\), the matrices \(A^{T}A\) and \(AA^T\) are both symmetric and, as such, have real eigenvalues as we shall see. We return to this in Chapter 7.
This page titled 5.4: Rank of a Matrix was last modified on Sat, 17 Jan 2026 00:22:24 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by W. Keith Nicholson via source content that was edited to the style and standards of the LibreTexts platform.
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