# Rank of an $n\times n$ matrix

$M$ is a square matrix $n\times n$, where the elements $M_{i,i} = 1 - 1/n$ and $M_{i,j} = - 1/n\ (i\neq j)$.

I want to show the rank of this matrix is $(n-1)$. How can I do that?

I am stuck on how to operate with a $n\times n$ matrix. I tried using rank of sum $≤$ sum of ranks.

Letting $M = [-1/n]_{n\times n} + I_{n\times n}$, then the rank of $[-1/n]_{n\times n}$ equals to $1$ and the rank of $I_{nxn}$ equals to $n$. So that doesn't narrow it down.

• If you have difficulty manipulating $n \times n$ matrix, try some easy and concrete cases, like $n=2,3,4$.
– xbh
Aug 29 '18 at 12:33

Let $x_1,x_2,\ldots,x_n$ be the columns of $M$. Then consider any linear combination $$0=a_1x_1+a_2x_2+\cdots+a_nx_n$$ The $i$th component of this vector equation, inserting the $i$th entry of each of the $x_k$, may be manipulated to say $$a_i=\frac{a_1+\cdots+a_n}{n}$$ where the right-hand side is the same for any $i$. Thus we must have $a_1=a_2=\cdots=a_n$. Therefore $\ker M=\operatorname{Span}([1,1,\ldots,1]^T)$, and by the nullity-rank theorem the rank of $M$ is $n-1$.
You can first multiply all rows with $n\not=0$
The idea is to bring $M$ to an echelon form.