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Let $A$ be real symmetric $n\times n$ matrix whose only eigenvalues are $0$ and $1$. Let the dimension of the null space of $A-I$ be $m$. Pick out the true statements.

  1. The characteristic polynomial of $A$ is $(\lambda-1)^m(\lambda)^{m-n}$.

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

  3. The rank of $A$ is $m$.

This is what I did: I found geometric multiplicity corresponding to eigenvalue value $0$ to be $n-m$(using symmetric matrix is diagonalizable ) while geometric multiplicity of eigenvalue value 1 is $m$(that is given) .so the characteristic polynomial of $A$ should be $(\lambda-1)^m(\lambda)^{n-m}$. While I am not sure about other two statements. Any kind of help is highly appreciated. Thanks.

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I think you are right. – Un Chien Andalou May 12 '12 at 13:38
@Makausi thank you – srijan May 12 '12 at 15:04
welcome and all the best for tomorrow – Un Chien Andalou May 12 '12 at 15:19
up vote 3 down vote accepted

As you noticed, since $A$ is a real symmetric matrix, it is diagonalizable, and the properties satisfied (or not) by $A$ will be satisfied also by $D$, where $D$ is a diagonal matrix which consists of the eigenvalues of $A$. Indeed, $\dim\ker(A-I)$ and characteristic polynomial are invariant up a change of basis. To see that, write $A=P^{-1}DP$, and $\ker(A-I)=\ker(D-I)$, as it can be seen by a double inclusion. For the characteristic polynomial, use determinant for example.

  • The first is true if we switch $m$ and $n$ in the power of $\lambda$, namely $p_A(\lambda)=(\lambda-1)^m\lambda^{n-m}$, otherwise the degree would be $2m-n$, which is not $m$, except in the case $m=n$.
  • The second is false if $k=0$ unless $A=I$, but true for $k\geq 1$.
  • The third is also true. Indeed, we look at the rank of $A$ is the same as the rank of $D$, and the rank of a diagonal matrix is easy to determine.
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Sorry sir i didn't get $dimker(A−I)$ and characteristic polynomial are invariant up a change of basis. – srijan May 12 '12 at 13:17
And why third is true? Can you explain please? – srijan May 12 '12 at 13:18
How second statement is true for all $k\geq 1$ ? Can you explain sir? – srijan May 12 '12 at 13:24
First show that it's enough to show it for $D$. – Davide Giraudo May 12 '12 at 13:25
Dear sir i got all the points you made except for the first statement. My observation is characteristic polynomial of $A$ is $(\lambda-1)^m(\lambda)^{n-m}$. Does it make no difference ? – srijan May 12 '12 at 13:57

To answer 2 and 3, consider diagonalizing $A$. And note that it is enough to show 2 for $k=1$.

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Dear sir what i have observed is correct or not? – srijan May 12 '12 at 13:19

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