Rank of a Transpose

Rank and Transposition

The rank of a matrix measures how many independent directions its columns span. Transposing a matrix turns its rows into columns, so understanding the rank of a transpose requires connecting the row space of the original matrix to its column space. The key fact is that these spaces have the same dimension. This lets us conclude that transposition changes the shape of a matrix but not its rank.

Let \(\mathbf{A}\in\mathbb{R}^{m\times n}\). Its column space, \(\operatorname{Col}\mathbf{A}\), is the subspace of \(\mathbb{R}^m\) spanned by its columns. Its row space can be viewed as a subspace of \(\mathbb{R}^n\): we treat each row as a column vector by transposing it before taking the span. The rank of \(\mathbf{A}\) is the dimension of its column space. The row rank is the dimension of its row space.

The Row-Rank/Column-Rank Theorem was established in the earlier material on row rank and column rank. In particular, it applies to rectangular matrices; it does not require a matrix to be square. We will use it to deduce the rank-of-a-transpose result. First, the relation between the two spaces can be stated precisely.

Proof. Write the rows of \(\mathbf{A}\) as \(\mathbf{r}_1^{\top},\dots,\mathbf{r}_m^{\top}\), where each \(\mathbf{r}_i\) is a column vector in \(\mathbb{R}^n\). By the definition of transpose, the columns of \(\mathbf{A}^{\top}\) are precisely \(\mathbf{r}_1,\dots,\mathbf{r}_m\). Therefore \(\operatorname{Col}(\mathbf{A}^{\top})=\operatorname{span}\{\mathbf{r}_1,\dots,\mathbf{r}_m\}\), which is the row space of \(\mathbf{A}\) under our column-vector convention. Taking dimensions gives the stated equality. \(\square\)

The Rank-of-a-Transpose Theorem

Proof. By definition, \(\operatorname{rank}(\mathbf{A}^{\top})=\dim(\operatorname{Col}(\mathbf{A}^{\top}))\). By the lemma, this is the row rank of \(\mathbf{A}\). The Row-Rank/Column-Rank Theorem says that the row rank of \(\mathbf{A}\) equals its column rank, which by definition is \(\operatorname{rank}\mathbf{A}\). Combining these equalities gives \(\operatorname{rank}(\mathbf{A}^{\top})=\operatorname{rank}\mathbf{A}\). \(\square\)

The theorem compares dimensions of spaces, not the shapes of the matrices. If \(\mathbf{A}\) is \(m\times n\), then \(\mathbf{A}^{\top}\) is \(n\times m\), but both have the same rank. In a row-reduction calculation, the number of pivots gives the rank. One may row-reduce either matrix to compute that common number; the pivot positions themselves need not correspond, because the two matrices have different sizes and different entries.

Checking the Result on Rectangular Matrices

Rank Is Bounded by Both Dimensions

The equality of row rank and column rank also gives a useful bound that applies to every rectangular matrix. A matrix with \(m\) rows cannot have more than \(m\) independent columns, and its \(n\) rows cannot span a space of dimension greater than \(n\). Therefore the rank is limited by whichever dimension is smaller.

Proof. The column space is a subspace of \(\mathbb{R}^m\), so its dimension is at most \(m\). To see this bound, a basis of the column space is an independent list in \(\mathbb{R}^m\); placing its vectors as columns gives a matrix with a pivot in every column, so the number of basis vectors cannot exceed the number \(m\) of rows. Thus \(\operatorname{rank}\mathbf{A}\leq m\). The row space is a subspace of \(\mathbb{R}^n\), so the same argument gives that its dimension is at most \(n\). By the Row-Rank/Column-Rank Theorem, that dimension is \(\operatorname{rank}\mathbf{A}\), hence \(\operatorname{rank}\mathbf{A}\leq n\) as well. A dimension is nonnegative, so combining the three inequalities gives the result. \(\square\)

For a nonzero \(m\times n\) matrix, the rank may reach \(\min(m,n)\), but it cannot exceed it. In the previous example, \(\mathbf{C}\) has two columns and three rows, and its rank is \(2=\min(3,2)\). Its transpose has three columns and two rows, with the same maximum possible rank, \(2\). This is one reason it matters to distinguish rank from the number of rows or columns: rectangular matrices can have either dimension larger than their rank.

Common Pitfall: Counting Rows as Rank

A frequent mistake is to count the rows of the transpose and call that number its rank. For the first worked example, \(\mathbf{A}^{\top}\) has three rows. The incorrect calculation is “\(\operatorname{rank}(\mathbf{A}^{\top})=3\), because it has three rows.” But three rows only place an upper bound on rank; they do not guarantee three independent directions. In fact, row reduction gives

There are only two pivots, so the corrected result is \(\operatorname{rank}(\mathbf{A}^{\top})=2\). The theorem gives the same conclusion from \(\operatorname{rank}\mathbf{A}=2\). In general, row counts and column counts give bounds; pivots or the dimensions of the corresponding spaces determine the actual rank.

What to Carry Forward

The rank-of-a-transpose theorem follows from two facts: the columns of \(\mathbf{A}^{\top}\) are the rows of \(\mathbf{A}\), and the row rank of a matrix equals its column rank. The first identifies the relevant spaces; the second equates their dimensions with rank. No square-matrix assumption is needed, and no special case is needed when the rank is zero. The zero matrix, for example, has only zero columns and rows, so both it and its transpose have rank zero.

When checking a calculation, keep track of the ambient spaces as well as the number of pivots. The columns of an \(m\times n\) matrix lie in \(\mathbb{R}^m\), whereas the columns of its transpose lie in \(\mathbb{R}^n\). Their spans may live in different spaces and have different vectors, yet the theorem guarantees equal dimensions. That distinction—different spaces, equal dimension—is the central point of the result.