Rank

In linear algebra, the rank of a matrix is the dimension of its row space or column space. It is an important fact that the row space and column space of a matrix have equal dimensions.

Let \(A\) be a matrix. If \(R(A)\) denotes the row space of \(A\) and \(C(A)\) denotes the column space of \(A\), then \(\dim\big(R(A)\big) = \dim\big(C(A)\big)\). This quantity is called the rank of \(A\), and denoted \(\text{rk}(A)\).

Intuitively, the rank measures how far the linear transformation represented by a matrix is from being injective or surjective. Suppose \(A\) is an \(m\)-by-\(n\) matrix representing a linear transformation \(T: \mathbb{R}^n \to \mathbb{R}^m\). Since \(C(A) = \text{Im}(T)\), the rank of \(A\) is an integer between \(0\) and \(m\), and equals \(m\) precisely when \(T\) is surjective. A similar interpretation exists for \(R(A)\) in terms of injectivity: the rank of \(A\) is an integer between \(0\) and \(n\), and equals \(n\) precisely when \(T\) is injective.

Let \(A\) be an \(m\)-by-\(n\) matrix, representing a linear transformation \(T: \mathbb{R}^n \to \mathbb{R}^m\). We define the row rank of \(A\) to be \(\dim\big(R(A)\big)\), and similarly the column rank \(\dim\big(C(A)\big)\). A priori, these numbers are not necessarily equal, but we will prove they actually are.

Suppose the row rank of \(A\) is \(r\). Our approach will be to produce \(r\) linearly independent vectors in \(C(A)\). This will prove \(\dim\big(C(A)\big) \ge \dim\big(R(A)\big)\). By applying the same argument to the transpose of \(A\), we obtain the reverse inequality \(\dim\big(R(A)\big) \ge \dim\big(C(A)\big)\), hence the result; this works because taking the transpose simply swaps row and column spaces.

How could one produce \(r\) linearly independent vectors in \(C(A) = \text{Im}(T) \subset \mathbb{R}^m?\) A possible approach is to choose \(r\) linearly independent vectors in \(\mathbb{R}^n\), and then apply \(T\) to each of them, hoping the resultant \(r\) vectors in \(\mathbb{R}^m\) will themselves be linearly independent. To ensure this works, the set of \(r\) linearly independent vectors in \(\mathbb{R}^n\) should have some additional structure we can work with; they cannot just be \(r\) random vectors. But there is a very natural set of \(r\) linearly independent vectors in \(\mathbb{R}^n\): a basis for the row space \(R(A)!\)

Accordingly, let \(x_1, \ldots, x_r\) be a basis for \(R(A)\). We must prove \(Tx_1, \ldots, Tx_r\) are linearly independent. Suppose they aren't; then there are \(c_i \in \mathbb{R}\) (not all zero) such that

\[0 = \sum_{i=1}^{r} c_i Tx_i = T(c_1 x_1 + \cdots + c_r x_r).\]

By construction, we know \(v= c_1 x_1 + \cdots + c_r x_r\) is in the row space of \(A\). Furthermore, since \(Tv = 0\), a computation done in the article on row and column spaces implies \(v\) is orthogonal to every row of \(A\). But \(v\) is a linear combination of these rows, so \(v\) is orthogonal to itself; thus, \(v = 0\). This implies the relation \(c_1 x_1 + \cdots + c_r x_r = 0\), which is absurd on account of linear independence of the collection \(\{x_i\}\). \(_\square\)

What is the rank of the matrix

\[A = \begin{pmatrix} 2 & 1 & 0 \\ 3& -1 & 2 \end{pmatrix}?\]

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Note that the first two columns of \(A\) are linearly independent, since they are not multiples of each other. Thus, \(C(A) = \mathbb{R}^2 \implies \text{rank}(A) = 2\). \(_\square\)

Let \(A\) be an \(n\)-by-\(n\) square matrix. Prove \(A\) has rank \(n\) if and only if \(A\) is invertible.

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\(A\) has rank \(n\) if and only if the row space \(R(A)\) equals \(\mathbb{R}^n\). But the kernel of \(A\) is precisely the orthogonal complement of the row space, and hence \(R(A) = \mathbb{R}^n \iff \text{Ker}(A) = \{0\}\). We know \(A\) is invertible if and only if its kernel is zero, so we are done. \(_\square\)