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:
- Substracting the first row from all other rows
- Swap two rows
- $-2\cdot(row_1)+(row_2), \dots, (-(m-1))\cdot(row_1)+(row_2)$
$$\begin{pmatrix}2 & 3&\dots & 1+n\\ 3&4 &\dots & 2+n\\ \vdots & \vdots &\ddots &\vdots\\ m+1 & m+2 &\dots &m+n \end{pmatrix} \overbrace{\iff}^{1} \begin{pmatrix}2 & 3&\dots & 1+n\\ 1& 1 &\dots & 1\\ 2 & 2 & \dots & 2\\ \vdots & \vdots &\ddots &\vdots\\ m-1 & m-1 &\dots & m-1 \end{pmatrix}$$
$$\overbrace{\iff}^{2} \begin{pmatrix}1& 1 &\dots & 1\\ 2 & 3&\dots & 1+n\\ 2 & 2 & \dots & 2\\ \vdots & \vdots &\ddots &\vdots\\ m-1 & m-1 &\dots & m-1 \end{pmatrix} \overbrace{\iff}^{3} \begin{pmatrix}1& 1 &\dots & 1\\ 0 & 1&\dots & n-1\\ 0 & 0 & \dots & 0\\ \vdots & \vdots &\ddots &\vdots\\ 0 & 0 &\dots & 0 \end{pmatrix}$$
$\implies \operatorname{rank}_{\text{Row}}(A)=2\neq\operatorname{rank}_{\text{Column}}(A)=n+1$, but Row and Column rank should be equal, what have I done wrong?
Best Answer
(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
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$.