Matrix Rank From Nonzero Singular Values

Rank as a Count of Active Directions

The singular value decomposition separates a matrix into changes of coordinates and a diagonal scaling. The unitary factors change the input and output coordinates, while the diagonal factor records which directions are scaled by a nonzero amount. This makes the diagonal factor a natural place to read off rank. The key is to justify why changing coordinates does not alter rank, and then to account carefully for the rectangular shape of the diagonal factor.

Let \(\mathbb{F}\) be \(\mathbb{R}\) or \(\mathbb{C}\), and let \(\mathbf{A}\in\mathbb{F}^{m\times n}\). The rank of \(\mathbf{A}\), written \(\operatorname{rank}\mathbf{A}\), is the dimension of its column space \(\operatorname{Col}\mathbf{A}\), the subspace spanned by its columns. A square matrix \(\mathbf{Q}\in\mathbb{F}^{k\times k}\) is unitary if \(\mathbf{Q}^{*}\mathbf{Q}=\mathbf{I}_k\). In the real case, this condition is \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_k\), and \(\mathbf{Q}\) is called orthogonal. The square-shape condition matters: a rectangular matrix with orthonormal columns need not be unitary.

For a rectangular \(\boldsymbol{\Sigma}\), the main diagonal has \(k=\min(m,n)\) positions. If \(m<n\), columns after column \(m\) are structurally zero; if \(m>n\), rows after row \(n\) are structurally zero. These extra positions are not additional singular values under the convention above. Whether or not a convention appends extra zeros to a singular-value list, the number of nonzero singular values is unchanged.

Why the SVD Diagonal Gives the Rank

Proof. Write a full SVD \(\mathbf{A}=\mathbf{U}\boldsymbol{\Sigma}\mathbf{V}^{*}\), with \(\mathbf{U}\) and \(\mathbf{V}\) square and unitary. Since \(\mathbf{V}^{*}\mathbf{V}=\mathbf{I}_n\), the map \(\mathbf{V}^{*}\) is onto: for any \(\mathbf{x}\in\mathbb{F}^n\), \(\mathbf{x}=\mathbf{V}^{*}(\mathbf{V}\mathbf{x})\). Therefore \(\operatorname{Col}(\boldsymbol{\Sigma}\mathbf{V}^{*})=\operatorname{Col}\boldsymbol{\Sigma}\): every output of \(\boldsymbol{\Sigma}\mathbf{V}^{*}\) is an output of \(\boldsymbol{\Sigma}\), and, by surjectivity, every output of \(\boldsymbol{\Sigma}\) is attained. Multiplication on the left by \(\mathbf{U}\) maps \(\operatorname{Col}\boldsymbol{\Sigma}\) bijectively onto \(\operatorname{Col}\mathbf{A}\), since \(\mathbf{U}\) is invertible with inverse \(\mathbf{U}^{*}\). A bijective linear map sends a basis to a basis, so the two column spaces have the same dimension. Hence \(\operatorname{rank}\mathbf{A}=\dim\operatorname{Col}\boldsymbol{\Sigma}\). Let \(J=\{i: \sigma_i\ne0\}\). For any \(\mathbf{x}\in\mathbb{F}^n\), the \(i\)-th coordinate of \(\boldsymbol{\Sigma}\mathbf{x}\), for \(1\le i\le k\), is \(\sigma_i x_i\), and any remaining output coordinates are zero. Thus every output belongs to the span of the standard coordinate vectors \(\mathbf{e}_i\) for \(i\in J\). Conversely, for each \(i\in J\), \(\boldsymbol{\Sigma}(\mathbf{e}_i/\sigma_i)=\mathbf{e}_i\), using the input coordinate vector in \(\mathbb{F}^n\) and the output coordinate vector in \(\mathbb{F}^m\). Therefore the column space of \(\boldsymbol{\Sigma}\) is exactly the span of those \(\mathbf{e}_i\). They are distinct standard coordinate vectors and hence linearly independent, so \(\dim\operatorname{Col}\boldsymbol{\Sigma}=|J|\), the number of nonzero singular values. This proves the claim. \(\square\)

The proof separates the argument into two parts. The unitary factors preserve the dimension of the column space because they make invertible coordinate changes. The rectangular diagonal factor contributes one independent output direction for each nonzero diagonal entry, and no direction for a zero entry.

Reading Rank from Rectangular SVDs

Common Mistake: Counting Nonzero Columns

A nonzero column does not necessarily contribute a new independent direction. In the last example, both columns of \(\mathbf{A}\) are nonzero. The incorrect calculation “there are two nonzero columns, so the rank is \(2\)” counts columns without checking whether they are independent. In fact, the second column is twice the first:

Thus the column space is spanned by just one nonzero vector, in agreement with the single nonzero singular value \(5\). Count the nonzero diagonal singular values, not the nonzero entries or nonzero columns of the original matrix. Also keep the factor’s shape in view: a zero column forced by a wide rectangular shape is not an extra singular value under the convention that lists \(\min(m,n)\) values.

This rank test is useful because it reads the number of independent output directions directly from an SVD, even when a matrix is rectangular. The diagonal factor also indicates which singular-vector directions are active: a positive diagonal entry maps its corresponding direction to a nonzero output, whereas a zero entry contributes nothing. The rank conclusion itself depends only on how many such diagonal entries are nonzero, not on the particular singular vectors chosen.