3
$\begingroup$

$M$ 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.

$\endgroup$
1
  • 2
    $\begingroup$ If you have difficulty manipulating $n \times n$ matrix, try some easy and concrete cases, like $n=2,3,4$. $\endgroup$
    – xbh
    Aug 29 '18 at 12:33
3
$\begingroup$

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 $n-1$.

$\endgroup$
2
$\begingroup$

By row operations first add all rows to the first one (what do you see?) and then add all the rows to the second one ...

EDIT

You can first multiply all rows with $n\not=0$

The idea is to bring $M$ to an echelon form.

$\endgroup$

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.