Why Do Arithmetic-Progression Columns Give a Zero Determinant?

If the entries in each column form an arithmetic progression as you move down the rows, then every column has a very restricted form. For a square matrix with at least three rows, that restriction forces the determinant to be zero.

The key is to write each column using its first term and common difference. This exposes a shared two-dimensional space containing all the columns.

Write a column as a vector

Suppose a column has $n$ entries and is an arithmetic progression with first term $a$ and common difference $d$. Its entries are

$$a,\ a+d,\ a+2d,\ \ldots,\ a+(n-1)d.$$

As a column vector, this is

$$\begin{pmatrix} a\\a+d\\a+2d\\\vdots\\a+(n-1)d \end{pmatrix} = a\begin{pmatrix}1\\1\\1\\\vdots\\1 \end{pmatrix} +d\begin{pmatrix}0\\1\\2\\\vdots\\n-1 \end{pmatrix}.$$

The two vectors on the right depend only on the number of rows, not on the particular column. The values $a$ and $d$ can differ from column to column, but every column is still a linear combination of those same two vectors.

Why the determinant vanishes

A determinant of a square matrix is zero if its columns are linearly dependent. If $n\ge 3$, the matrix has at least three columns, and all of them lie in the span of just two vectors. That span has dimension at most two, so three or more columns in it must be linearly dependent. Therefore,

$$\boxed{\det A=0\quad\text{for }n\ge 3.}$$

This argument does not require the columns to have the same first term or the same common difference. Each column may use its own pair of values $(a_j,d_j)$.

Worked example

Consider the $4\times4$ matrix whose columns are four different arithmetic progressions:

$$A=\begin{pmatrix} 1&2&0&-1\\ 3&1&3&3\\ 5&0&6&7\\ 7&-1&9&11 \end{pmatrix}.$$

For example, the first column is $1,3,5,7$ and the third is $0,3,6,9$. Define

$$u=\begin{pmatrix}1\\1\\1\\1\end{pmatrix}, \qquad v=\begin{pmatrix}0\\1\\2\\3\end{pmatrix}.$$

The columns of $A$ can be written as

$$c_1=1u+2v,\qquad c_2=2u-v,\qquad c_3=0u+3v,\qquad c_4=-u+4v.$$

Thus all four columns belong to the span of $u$ and $v$, which has dimension at most two. Four vectors in that space are linearly dependent, so $\det A=0$. There is no need to expand the determinant or perform row operations.

What happens for a 2×2 matrix?

The zero-determinant result needs at least three rows. In a $2\times2$ matrix, a column with first term $a_j$ and difference $d_j$ is simply $(a_j,a_j+d_j)^T$. For two such columns,

$$A=\begin{pmatrix} a_1&a_2\\ a_1+d_1&a_2+d_2 \end{pmatrix}, \qquad \det A=a_1d_2-a_2d_1.$$

This can be nonzero. For instance, columns $(1,2)^T$ and $(1,3)^T$ are both arithmetic progressions of length two, and their determinant is $1\cdot3-1\cdot2=1$.

A single-entry column is also a progression in the vacuous sense, but the one-by-one determinant is just its entry; it is not necessarily zero.

Common mistakes

• Assuming the progressions need identical differences. They do not. The argument works when every column has its own first term and common difference.
• Applying the result to only some columns. If only two columns are arithmetic progressions, the remaining columns may lie outside the two-dimensional span. The determinant is not forced to vanish.
• Confusing columns with rows. If the rows, rather than the columns, are arithmetic progressions across the columns, the same reasoning applies to the rows. A matrix and its transpose have the same determinant, so the determinant is zero for square matrices of size at least three in that case too.

A broader pattern

The same idea works for columns whose entries follow a polynomial pattern in the row index. If every column has the form

$$c_j=\bigl(p_j(0),p_j(1),\ldots,p_j(n-1)\bigr)^T,$$

where each $p_j$ is a polynomial of degree at most $k$, then each column is a combination of the fixed vectors $(1,\ldots,1)^T$ and $(i^r)_{i=0}^{n-1}$ for $r=1,\ldots,k$. All columns therefore lie in a space of dimension at most $k+1$. A square determinant must be zero when $n>k+1$.

Arithmetic progressions are the degree-at-most-one case, which explains the two-vector span and the vanishing determinant for $n\ge3$.