Rank of a Hankel-like rectangular matrix

Question

Suppose $A \in \mathbb{R}^{m \times n}$ with $a_{i,j}=i+j$ for all $(i,j) \in [m] \times [n]$.

$$A=\begin{pmatrix}a_{1,1} &\dots & a_{1,n}\\ \vdots & \ddots & \vdots\\ a_{m,1} & \dots & a_{m,n} \end{pmatrix}$$

Determine the rank of $A$ with $m,n>0$

$a_{i,j}=i+j$ means, that the matrix looks like this: $$A=\begin{pmatrix}2_{1,1} & 3_{1,2}&\dots & 1+n_{1,n}\\ 3_{2,1}&4_{2,2} &\dots & 2+n_{2,n}\\ \vdots & \vdots &\ddots &\vdots\\ m+1_{m,1} & m+2_{m,2} &\dots &m+n_{m,n} \end{pmatrix}$$

In order to determine the rank, we need to put $A$ in row echelon form:

$$A=\begin{pmatrix}2 & 3 & \dots & 1 + n \\ 0 & -\frac{1}{2} & \dots & \frac{1 - n}{2}\\ \vdots & \vdots & \ddots & \vdots \\ 0 & \frac{7 - m}{2} & \dots & \frac{-m + 1 - nm + n}{2} \end{pmatrix}$$

That's what I did so far. How to go on?

EDIT:

1. Substracting the first row from all other rows
2. Swap two rows
3. $-2\cdot(row_1)+(row_2), \dots, (-(m-1))\cdot(row_1)+(row_2)$

$$\begin{pmatrix}2 & 3&\dots & 1+n\\ 3&4 &\dots & 2+n\\ \vdots & \vdots &\ddots &\vdots\\ m+1 & m+2 &\dots &m+n \end{pmatrix} \overbrace{\iff}^{1} \begin{pmatrix}2 & 3&\dots & 1+n\\ 1& 1 &\dots & 1\\ 2 & 2 & \dots & 2\\ \vdots & \vdots &\ddots &\vdots\\ m-1 & m-1 &\dots & m-1 \end{pmatrix}$$

$$\overbrace{\iff}^{2} \begin{pmatrix}1& 1 &\dots & 1\\ 2 & 3&\dots & 1+n\\ 2 & 2 & \dots & 2\\ \vdots & \vdots &\ddots &\vdots\\ m-1 & m-1 &\dots & m-1 \end{pmatrix} \overbrace{\iff}^{3} \begin{pmatrix}1& 1 &\dots & 1\\ 0 & 1&\dots & n-1\\ 0 & 0 & \dots & 0\\ \vdots & \vdots &\ddots &\vdots\\ 0 & 0 &\dots & 0 \end{pmatrix}$$ $\implies \operatorname{rank}_{\text{Row}}(A)=2\neq\operatorname{rank}_{\text{Column}}(A)=n+1$, but Row and Column rank should be equal, what have I done wrong?

Answer 1

(I consider here the cases where $m,n \geq 2$ ; if $m=1$ or $n=1$, make a special immediate treatment giving rank$(A)=1$).

We are going to establish that in general

rank(A)=$2$.

Instead of considering echelon form of $A$, let us express $A$ under the form :

$$A=\begin{pmatrix} 1\\2\\ \vdots \\m\end{pmatrix}\begin{pmatrix} 1&1&\cdots \ 1\end{pmatrix}+\begin{pmatrix} 1\\1\\ \vdots \\ 1\end{pmatrix}\begin{pmatrix} 1&2&\cdots \ n\end{pmatrix}\tag{1}$$

Said otherwise :

$$A=\begin{pmatrix} 1&1\\2&1\\ \vdots&\vdots \\ m&1\end{pmatrix}\begin{pmatrix} 1&1&\cdots &1\\1&2&\cdots&n\end{pmatrix}\tag{2}$$

which is a rank-$2$ matrix whatever the values of $m,n \geq 2$.

Proof of the fact that, in general, $A$ is rank-2 :

It amounts, by one of the definitions of the rank of a matrix, to prove that the range of $A$ is 2-dimensional.

The range of $A$ is the subspace of all $AV$ for any $V$ with entries $a_k$, i.e., using (2), the set of all vectors of the form :

$$\begin{pmatrix} 1&1\\2&1\\ \vdots&\vdots \\ m&1\end{pmatrix}\begin{pmatrix} \sum a_k\\ \sum ka_k\end{pmatrix}$$ i.e., as the set of linear combinations :

$$(\sum a_k) \begin{pmatrix} 1\\2\\ \vdots \\m\end{pmatrix}+(\sum ka_k)\begin{pmatrix} 1\\1\\ \vdots \\ 1\end{pmatrix}$$

which generates a dimension-$2$ subspace because the two column vectors are independent. Therefore, the dimension of the range of $A$ is 2.

Edit : An alternative way to show that $A=[C_1|C_2|\cdots|C_n]$ has rank equal to $2$, is by using the fact that

$$C_2-C_1=C_3-C_2=...=C_{n}-C_{n-1}$$

(equal to the vector with all its entries equal to $1$) giving $n-1$ relationships permitting to express all columns $C_i$, $(i>2)$ as linear combinations of $C_1$ and $C_2$, for example $C_3=2C_2-C_1$, $C_4=C_3+C_2-C_1=3C_2-2C_1$, and, in the general case :

$$C_k=(k-1)C_2-(k-2)C_1$$

As all columns can be expressed as linear combinations of the same $2$ columns, which are independent (being non proportional). Thus rank$(A)$=2$.

Answer 2

Another way to show the same:

Here, for the Hankel-like matrix $A$ we have $A_{ij}=f(i+j-1)=i+j=\underset{u_i^T}{\begin{bmatrix} i & 1 \end{bmatrix}}\underset{v_j}{\begin{bmatrix} 1 \\ j \end{bmatrix}}$, with $u_i=\begin{bmatrix} 1 \\ i \end{bmatrix}$ and $v_j=\begin{bmatrix} 1 \\ j \end{bmatrix}$.

Define the matrices $U\in \mathbb R^{n\times 2}$ and $V\in \mathbb R^{2\times n}$ as:

$U=\begin{bmatrix} u_1^ T\\ u_2^ T\\ \vdots \\ u_ n^ T \end{bmatrix}$ and $V=\begin{bmatrix} v_1& v_2 & \dots & v_ n \end{bmatrix}$,

so that

$UV=\begin{bmatrix} 1 & 1 \\ 2 & 1 \\ \ldots & \ldots \\ n & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 & \ldots & 1 \\ 1 & 2 & \ldots & n\end{bmatrix}=A \in \mathbb R^{n\times n}$, where $A$ is written as a product of two low-rank (rank-2) martices $U$ and $V$.

Now, $\mathrm{rank}(A)= \mathrm{rank}(UV) \leq \min(\mathrm{rank}(U), \mathrm{rank}(V)) = 2$ (from here), i.e., $\mathrm{rank}(A) \leq 2 \quad \ldots \quad (1)$

Also, The rank of a matrix is the size (order) of the largest square submatrix with a non-zero determinant. Here we have a $2 \times 2$ submatrix of $A$ as $\begin{bmatrix} 2 & 3 \\ 3 & 4\end{bmatrix}$ with determinant $8-9=-1 \neq 0$, hence we conclude $\mathrm{rank}(A) \geq 2\quad \ldots \quad (2)$

Combining (1) and (2), we have $rank(A)=2$.