Rank of matrix whose row sums are $0$

linear algebramatricesmatrix-rank

I have a matrix

$$\begin{bmatrix} 1-c_1&-c_2&-c_3 & \dots & -c_n\\-c_1&1-c_2&-c_3 & \dots & -c_n \\ -c_1&-c_2&1-c_3 & \dots & -c_n \\ \vdots & \vdots & \vdots & \ddots & \vdots\\-c_1&-c_2&-c_3 & \dots & 1-c_n\end{bmatrix}$$

And $c_1 + \dots + c_n = 1$. Every $c_i \in (0,1)$. Then what is the rank of this matrix?


My attempt: I add column $2, 3, \dots, n$ to the first column. Then the first column changes to zero column. Thus the rank is at most $n-1$. I guess the rank is just $n-1$. But I don't know how to justify my guess.

Best Answer

Let $M$ be the matrix. Write $\vec c = (c_1, \cdots, c_n)$. Then for all $v = (v_1, \cdots, v_n)^t$,

$$M v = v - (\vec c\cdot v) \vec 1,$$ where $\vec 1 = (1, \cdots, 1)^t$.

In particular, $Mv = v$ for all $v$ so that $\vec c \cdot v = 0$. Hence the rank is at least $n-1$. Together with your observation, the rank is exactly $n-1$.

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