A question on Rank and trace of a special matrix

Question

I want to share the following question which was asked in a competitive exam:

For a fixed positive integer $n\geq 3$, let $A$ be the $n\times n$ matrix defined by $A=I-\dfrac{1}{n}J$, where $J$ is the $n\times n$ matrix with all entries equal to 1. Which of the following statements is/are not true?

1. $A^k=A$ for every positive integer $k$.
2. $\operatorname{Trace}(A)=n-1$
3. $\operatorname{Rank}(A)+\operatorname{Rank}(I-A)=n$
4. $A$ is invertible.

Well I see that my question is on hold. But I have posted this in Q & A style and have given quite a clear explanation of my way of approach in solving it.

Answer 1

1.

It's not hard to show that $A^{2}=A$, using the fact that $J^{2}$ is the $nxn$ matrix with all entries equal to $n$:

$A^{2}=(I-\frac{1}{n}J)(I-\frac{1}{n}J)=I-\frac{2}{n}+\frac{1}{n^{2}}J^{2}=I-\frac{1}{n}J=A$

Now suppose $A^{k}=A$ holds for some positive integer $k$. Then

$A^{k+1}=AA^{k}=AA=A^{2}=A$

Hence, by induction $A^{k}=A$ holds for every positive integer $k$.

Answer 2

Any suggestion will be appreciated.

Simple matrix multiplication reveals that $A^2=A$ and hence it can be shown that $A^k=A$ for every positive integer $k$, so that $(1)$ is true.

The simplest thing that we can check is $$Trace(A)=Trace(I)-Trace\left(\dfrac{1}{n}J\right)=n-1$$ Hence $(2)$ is true.

Now, since $A=I-\dfrac{1}{n}J$, so $I-A=\dfrac{1}{n}J$ and so $rank(I-A)=1$. Also, performing the elementary operations $R_n\rightarrow R_n+R_n-1+\ldots+R_1$ and then $R_i\rightarrow R_i-R_1\forall i=2,3,\ldots,n-1$, it is clear that $Rank(A)=n-1$ and so $(3)$ is true.

Now, it can be seen that $x=(1,1,\ldots,1)_{1\times n}^T$ satisfies $Ax=0$, so that $0$ is an eigenvalue of $A$. Hence $A$ is singular and so $A$ is not invertible. So, $(4)$ is not true.