0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ About your Edit ; what make you say that column-rank is $n+1$ ? $\endgroup$ – Jean Marie Jun 13 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 Jun 13 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 Jun 13 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 Jun 13 at 17:11
2
$\begingroup$

(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$.

$\endgroup$
  • $\begingroup$ Thank you for your detailed answer, I will work my way through! $\endgroup$ – Doesbaddel Jun 13 at 11:18
  • $\begingroup$ Couldn't we just substract $a_{1,1}$ from all $a_{i,j}$ and get the same result? $\endgroup$ – Doesbaddel Jun 13 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 Jun 13 at 12:01
  • 1
    $\begingroup$ 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 Jun 13 at 12:12
  • 1
    $\begingroup$ 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 Jun 13 at 12:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.