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The question is :

Suppose a real matrix $A$ satisfies $A^3=A$, $A \neq I$, $A \neq 0$.If $Rank(A) = r$ and $Trace(A) = t$, then

(A) $r \geq t$ and $r+t$ is odd.

(B) $r \geq t$ and $r+t$ is even.

(C) $r < t$ and $r+t$ is odd.

(D) $r

If I take the matrix $\pmatrix {1&0\\0&0}$ then $(A)$,$(C)$ and $(D)$ are violated and so indirectly it indicates that the correct option is $(B)$.But I fail to prove it directly.Please anybody help me by telling the actual way of proceeding.

Thank you in advance.

EDIT :

Since the polynomial $x^3-x$ annihilates $A$ so it follows that the minimal polynomial of $A$ has distinct linear factors.Hence $A$ is diagonalisazable.So $A$ is similar to a diagonal matrix $D$.Since two similar matrices have the same rank it follows that $Rank(A)$ is same as the number of non-zero eigen values of $A$.Now if $A$ has $r$ non-zero eigen values then $t \leq r$ since all the eigen values of $A$ are $\leq$ $1$.Hence the result follows.

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    What is $a$ in your EDIT? Is it $A$? – 2017-02-19
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    Your edit is correct, but you still haven't proved that $r + t$ is even. – 2017-02-19
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    Yeah @user49640 I have overlooked that thing.Sorry for this.But I am trying to solve it.Any help will be appreciated. – 2017-02-19
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    You've done well so far. With the ideas you used to prove the first part, the second part shouldn't be hard. – 2017-02-19
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    Yeah I got it.Suppose we have $M$ 1's and $N$ -1's.If $M=N$ then $r=2M$ and $t=0$.Which implies $r+t = 2M$, an even number.If $M>N$ then $r=M+N$ and $t=M-N$.Which implies $r+t = 2M$,an even number.Similarly when $N>M$ we have $r+t = 2M$,again an even number and we are done. – 2017-02-19
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    i.e. in general if we have $M$ $1$'s and $N$ $-1$'s we always have $r+t = 2M$, an even number. – 2017-02-19
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    @ArnabChatterjee another way to see this is that if all the nonzero eigenvalues are 1, then r=t (so they have the same parity) and if you change one of the 1 eigenvalues to -1 it changes the trace by 2, preserving the parity or t. So r and t are either both even or both odd. – 2017-02-19
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    When you're happy with your work, you should post it as an answer, rather than as an edit to the question. – 2017-02-19
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    @ArnabChatterjee. Wow! You have done a great job. – 2017-02-19

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Since the polynomial $x^3-x$ annihilates $A$ so it follows that the minimal polynomial of $A$ has distinct linear factors.Hence $A$ is diagonalisazable.So $A$ is similar to a diagonal matrix $D$.Since two similar matrices have the same rank it follows that $Rank(A)$ is same as the number of non-zero eigen values of $A$.Now if $A$ has $r$ non-zero eigen values then $t \leq r$ since all the eigen values of $A$ are $\leq$ $1$.Hence the result follows.

Yeah I got it.Suppose we have M $1$'s and N $-1$'s.If $M=N$ then $r=2M$ and $t=0$.Which implies $r+t=2M$, an even number.If $M>N$ then $r=M+N$ and $t=M−N$.Which implies $r+t=2M$,an even number.Similarly when $N>M$ we have $r+t=2M$,again an even number and we are done.

i.e. in general if we have $M$ $1$'s and $N$ $−1$'s we always have $r+t=2M$, an even number.Thus $(B)$ is the only correct option.