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Consider $$A=I-\frac{1}{n}J$$ where $I$ is an $n\times n $ idenity matrix, $J$ is an $n\times n $ matrix with all entries as $1$.

Which of the following is NOT true?

  • $A^k=A$ for every positive integer.
  • Trace of $A$ is $n-1$
  • $\text{Rank $(A)$+Rank $(I-A)$}=n$
  • $A$ is invertible.

My first concern was to conclude fourth option is false and it looks more natural for me to conclude fourth is false.

Reason is $I-\frac{1}{n}J$ is invertible iff $\frac{1}{n}J$ is nilpotent i.e., $J$ is nilpotent and I am sure just by looking at it that $J$ would never be nilpotent.

So, $A$ would never be invertible.

Trace of $A$ would be just $\text{ Trace of $A$ - $\frac{1}{n}$Trace of $J$}=n-\frac{1}{n}(n)=n-1$

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

I see $-\frac{1}{n}J+\frac{1}{n^2}J^2=0$ but I am not sure if this implies $A^k=A$ for all $k$.

I can not say anything about third option.

Could some one help me to clear this.

Thank you.

Jean-Claude Arbaut
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  • For the fourth, just try to sum each row. Now what is the result of $A \cdot (1,1,\dots,1)^T$ ? – Jean-Claude Arbaut Dec 23 '13 at 13:42
  • Result would be $(n-1,n-1,\dots,n-1)^T$... I am unable to relate this to my question.. more help would be necessary I guess... :) –  Dec 23 '13 at 13:48
  • No, result would be $(0,0,\dots,0)^T$. – Jean-Claude Arbaut Dec 23 '13 at 13:54
  • @arbautjc : really? I would check it once again... your answer sound to be more natural because in that case we can directly say that $A$ not invertible :D –  Dec 23 '13 at 13:58

2 Answers2

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Notice that $$J^2=nJ$$ hence $$A^2=\left(I-\frac 1 nJ\right)^2=I-\frac2nJ+\frac{1}{ n^2}J^2=A$$ hence we can answer the first point. The second point is also easy since $\mathrm{tr}$ is linear. Now since the polynomial $x^2-x=x(x-1)$ annihilates $A$ and $A\ne \alpha I_n$ then $0$ and $1$ are eigenvalues of $A$ and then $A$ isn't invertible, but since $A$ is diagonalizable then $$\dim E=n=\dim\ker A+\dim\ker(A-I)=\mathrm{Rank} A+\mathrm{Rank }(I-A)$$ by the rank-nullity theorem and this answers the third point.

  • This makes a lot of sense to me.... I do not immediately understand the logic after you write since $A$ is diagonalizable.... I would need some time... Thank you :) –  Dec 23 '13 at 14:03
  • Since the eigenvalues of $A$ are $0$ and $1$ and $A$ is diagonalizable then $E=\ker A\oplus \ker(A-I)$. –  Dec 23 '13 at 14:05
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    :-) ${}{}{}{}{}{}{}$ – Mikasa Dec 23 '13 at 14:31
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A is the orthogonal projector along the direction of all ones, $e=(1,1,...,1)$, $A=I-\frac{e\,e^T}{e^Te}$. The usual identities for projectors apply.

The column-row product $e\cdot e^T$ is just the matrix where every entry is $1$, so it is $J$. $e^Te=n$ should be obvious. The general form of any corank-1-projector is $P=I-\frac{u\,v^T}{v^Tu}$. In the case $u=v$ this gives an orthogonal projector, the case $u=v=e$ gives the matrix $A$.

Now either you know the properties of projectors, esp. $P^2=P$, and that orthogonal projectors are additionally symmetric $P^T=P$, or you have to prove them in general or only for this example.

Lutz Lehmann
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  • I am sorry, I could not understand your idea... I do not understand anything except that $A=I-\frac{e,e^T}{e^Te}$... Please take some more pain to teach me more clearly... –  Dec 23 '13 at 13:51
  • @PraphullaKoushik For the first: a projector $P$ has always $P^2=P$. The second and third are often shown in a linear algebra course, look at your favourite reference. – Jean-Claude Arbaut Dec 23 '13 at 13:55
  • Yes Yes.... I got some vague idea... I would work on this.... This is really helpful.. Thank you :) –  Dec 23 '13 at 14:04
  • @arbautjc : Yes Yes... I got it :) Thanks Man.. –  Dec 23 '13 at 14:05