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If $A_{m\times n}$ is a matrix such that $\sum_{j=1}^n a_{ij}=0$ for each $i=1,2,…,m,$ then why the columns of $A$ are linearly dependent set, and hence $\operatorname {rank}(A)<n$?

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  • $\begingroup$ It's hard to understand what you wrote. LaTeXize it. $\endgroup$ – DonAntonio Sep 22 '13 at 12:57
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    $\begingroup$ She tried, @DonAntonio, but was missing braces around subscripts and bounds of summation, making it difficult to make out. Hopefully fixed (there was also some copy and paste inserted in dollar signs, so I'm hoping I got the summation correctly). $\endgroup$ – amWhy Sep 22 '13 at 13:00
  • $\begingroup$ @amWhy, thank you for fixing my question, it's exactly what I tried to wrote. $\endgroup$ – Diane Vanderwaif Sep 22 '13 at 13:06
  • $\begingroup$ You're welcome, @Diane. Note that when there is more than one character in a subscript, you need to enclose the subscript in braces. E.g. $A_{m \times n}$ = A_{m \times n} $\endgroup$ – amWhy Sep 22 '13 at 13:14
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If $m < n$, then

$$\operatorname{rank} A \le \min \{m,n\} = m < n.$$

Let us now assume that $m \ge n$. Note that

$$A \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} = 0_{m \times 1},$$

Aiming for a contradiction, we assume that $A$ if of a full rank (i.e., $\operatorname{rank} A = n$). That means that $A$ has a submatrix $B$ of order $n$ such that $B$ is of a full rank (made of the linearly independent rows in $A$), i.e., $B$ is invertible. We know that

$$B \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} = 0_{n \times 1},$$

which is a contradiction with the invertibility of $B$, so $A$ is not of a full rank.

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Hint: the equation that is given actually gives an explicit linear dependence relation between the columns.

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  • $\begingroup$ Since I am a pedantic kind of mathematician, I should add: provided that $n>0$, i.e., provided there are any columns at all. For $n=0$ that statement to prove is wrong: no columns makes an (empty) linearly independent set. But you may forget I said that. $\endgroup$ – Marc van Leeuwen Sep 22 '13 at 13:05
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I assume $m=n$ just to make it look good .

we have $\sum_{j=1}^m a_{ij}=0$ for all $1\leq i\leq m$

i..e, $\sum_{j=1}^m a_{1j}+\sum_{j=1}^m a_{2j}+\dots+\sum_{j=1}^m a_{mj}=0$

i.e., $(a_{11}+a_{12}+\dots+a_{1m})+(a_{21}+a_{22}+\dots+a_{2m})+\dots +(a_{m1}+a_{m2}+\dots+a_{mm})=0$

i.e., $(a_{11}+a_{21}+\dots+a_{m1})+(a_{12}+a_{22}+\dots+a_{m2})+\dots +(a_{1m}+a_{2m}+\dots+a_{mm})=0$

Do you see "the next step" would conclude that columns are linearly dependent. (??)

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