Rank of a Sum and Its Sharp Bound

How the Columns of a Sum Are Constrained

Adding two matrices combines corresponding columns. Each resulting column is therefore made from one column of the first matrix and one column of the second. This places a limit on the number of independent columns the sum can have: its columns must lie in the space generated by the columns available from both matrices. The resulting rank bound is simple to state, but its inequality can be strict—even when both matrices have positive rank.

Throughout, let \(\mathbb{F}\) be either \(\mathbb{R}\) or \(\mathbb{C}\), and let \(\mathbf{A},\mathbf{B}\in\mathbb{F}^{m\times n}\). The matrices must have the same numbers of rows and columns for their sum to be defined. The column space \(\operatorname{Col}\mathbf{A}\) is the subspace spanned by the columns of \(\mathbf{A}\), and the rank is \(\operatorname{rank}\mathbf{A}=\dim(\operatorname{Col}\mathbf{A})\). For subspaces \(U,W\) of the same vector space, their sum is \(U+W=\{\mathbf{u}+\mathbf{w}:\mathbf{u}\in U,\ \mathbf{w}\in W\}\).

Proof. Let \(\mathbf{a}_j\) and \(\mathbf{b}_j\) denote column \(j\) of \(\mathbf{A}\) and \(\mathbf{B}\), respectively. Column \(j\) of \(\mathbf{A}+\mathbf{B}\) is \(\mathbf{a}_j+\mathbf{b}_j\), which belongs to \(\operatorname{Col}\mathbf{A}+\operatorname{Col}\mathbf{B}\). Since the column space is the span of the columns, this gives \(\operatorname{Col}(\mathbf{A}+\mathbf{B})\subseteq\operatorname{Col}\mathbf{A}+\operatorname{Col}\mathbf{B}\). Choose a basis of each of \(\operatorname{Col}\mathbf{A}\) and \(\operatorname{Col}\mathbf{B}\). The list formed by putting those two bases together spans their sum: every vector in the sum is a sum of combinations of vectors from the two bases. This list has \(\operatorname{rank}\mathbf{A}+\operatorname{rank}\mathbf{B}\) vectors. A finite spanning list can be reduced to a basis by removing vectors that are combinations of the others, so the dimension of the sum is at most the length of this list. Finally, a subspace cannot have larger dimension than a space containing it. Therefore \(\operatorname{rank}(\mathbf{A}+\mathbf{B})\leq\dim(\operatorname{Col}\mathbf{A}+\operatorname{Col}\mathbf{B})\leq\operatorname{rank}\mathbf{A}+\operatorname{rank}\mathbf{B}\). \(\square\)

When the Bound Is Attained

The theorem gives an upper bound, not an automatic formula for the rank of a sum. Equality is possible when the columns contributed by the two matrices provide separate independent directions. The next construction shows that the bound is sharp for every pair of ranks that can fit into the available row and column dimensions.

Proof. Define \(\mathbf{A}\) to have a \(1\) in diagonal positions \((i,i)\) for \(1\leq i\leq r\), and zeros elsewhere. Define \(\mathbf{B}\) to have a \(1\) in diagonal positions \((i,i)\) for \(r+1\leq i\leq r+s\), and zeros elsewhere. The indicated positions exist because \(r+s\leq\min(m,n)\). The nonzero columns of \(\mathbf{A}\) are the first \(r\) standard coordinate vectors in \(\mathbb{F}^m\), so they are independent and \(\operatorname{rank}\mathbf{A}=r\); when \(r=0\), \(\mathbf{A}\) is zero and has rank \(0\). The nonzero columns of \(\mathbf{B}\) are the next \(s\) standard coordinate vectors, so \(\operatorname{rank}\mathbf{B}=s\), including the zero case \(s=0\). Their nonzero columns occupy distinct positions, and the sum has \(r+s\) independent nonzero columns. Hence \(\operatorname{rank}(\mathbf{A}+\mathbf{B})=r+s\), as claimed. \(\square\)

Why the Inequality Can Be Strict

The column-space argument allows the two spaces to overlap. If some of the directions available in \(\operatorname{Col}\mathbf{A}\) are already available in \(\operatorname{Col}\mathbf{B}\), the combined space need not have dimension equal to the sum of the two dimensions. Cancellation can reduce the rank even further: corresponding columns may add to zero.

In this example, both column spaces are the same line, namely \(\operatorname{span}\{(1,0)^\top\}\). The matrices have rank one each, but adding them does not produce two independent column directions. Thus the sum of the ranks counts available dimensions separately, even when those dimensions describe the same direction.

Common Mistake: Treating the Bound as an Equality

A tempting but incorrect calculation is to replace the rank bound with the claim \(\operatorname{rank}(\mathbf{A}+\mathbf{B})=\operatorname{rank}\mathbf{A}+\operatorname{rank}\mathbf{B}\) for every pair of matrices. For the rank-one matrices in the previous example, that calculation would give rank \(2\) for the sum. But the actual sum is \(\begin{bmatrix}1&1\\0&0\end{bmatrix}\), whose two columns are equal and whose rank is \(1\). The correction is to use the theorem as an inequality, then compute or otherwise establish the rank of the particular sum. Equality requires that the sum have enough independent column directions; it is not guaranteed merely because the two input ranks are known.

Another useful check is the column-space inclusion from the proof. A proposed answer that gives \(\operatorname{rank}(\mathbf{A}+\mathbf{B})\) larger than either the ambient limit \(\min(m,n)\) or the bound \(\operatorname{rank}\mathbf{A}+\operatorname{rank}\mathbf{B}\) cannot be correct. Conversely, a small rank for the sum is entirely possible, as cancellation shows.

Using the Bound in Practice

The inequality is useful when the matrices themselves are complicated but their ranks are already known or bounded. It gives an immediate ceiling on the rank of a sum without requiring a full row reduction. For example, if each of two \(m\times n\) matrices has rank at most \(k\), then the rank of their sum is at most \(2k\). This estimate may not be exact, because the column spaces can overlap or cancel, but it rules out ranks above the stated ceiling.

When deciding whether equality holds in a concrete problem, keep the proof’s mechanism in view: the columns of the sum lie in the sum of the two column spaces. If those spaces contribute independent directions, equality can occur. If they overlap, or if corresponding column contributions cancel, the rank can be smaller. The examples show both possibilities and verify why the theorem gives a sharp universal upper bound rather than a general rank formula.