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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?

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    $\begingroup$ About your Edit ; what make you say that column-rank is $n+1$ ? $\endgroup$
    – Jean Marie
    Commented Jun 13, 2019 at 16:34
  • $\begingroup$ Because we have $n$ columns, that are not zero and yeah you're right, its actually $n$ (or are these all linear dependent of the first two columns?) $\endgroup$
    – Doesbaddel
    Commented Jun 13, 2019 at 16:40
  • $\begingroup$ A good way to show that $A=[C_1|C_2|\cdots|C_n]$ has column-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 $C_3=2C_2-C_1$, etc. $\endgroup$
    – Jean Marie
    Commented Jun 13, 2019 at 17:02
  • $\begingroup$ Oh, thank you that makes sense! Now, I'll just need to add the special case for $A\in\mathbb{K}^{1\times 1}$ where the rank is 1. $\endgroup$
    – Doesbaddel
    Commented Jun 13, 2019 at 17:11

1 Answer 1

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(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$.

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  • $\begingroup$ Thank you for your detailed answer, I will work my way through! $\endgroup$
    – Doesbaddel
    Commented Jun 13, 2019 at 11:18
  • $\begingroup$ Couldn't we just substract $a_{1,1}$ from all $a_{i,j}$ and get the same result? $\endgroup$
    – Doesbaddel
    Commented Jun 13, 2019 at 11:49
  • $\begingroup$ Doing this, you wouldn't work on the same matrix. Besides, I don't see the interest of subtracting $2$ to all entries. $\endgroup$
    – Jean Marie
    Commented Jun 13, 2019 at 12:01
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    $\begingroup$ Subtracting this row to all other rows would amount to subtract the product $\begin{pmatrix}1\\1\\ \vdots \\1\end{pmatrix}(a_{11},...a_{1n})$ giving a rather simple matrix looking like $A$. This could be a track... $\endgroup$
    – Jean Marie
    Commented Jun 13, 2019 at 12:12
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    $\begingroup$ Indeed the column rank is always equal to the row rank. May be you could write a sketch of your computation as an Edit to your question ; I will have a look to it in a few hours. $\endgroup$
    – Jean Marie
    Commented Jun 13, 2019 at 12:48

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