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Given three matrix $X,Y,Z$ satisfying the followings:

$$XZ=ZX$$ $$YZ=ZY$$ $$rank(XY-YX+I)=1$$

Prove that $Z=aI$ for sum real number $a$.

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    One dirty but certainly feasible approach is to make $Z$ a Jordan normal form over $\mathbb C$. If you are familiar with centralisers of Jordan forms, the rest is in principle relatively straightforward (although somewhat tedious in practice). If this is a homework question, there are perhaps some neat tricks to finish it off. – 2017-02-27
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    Can you be more specific please? – 2017-03-02
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    Let $Z$ be in Jordan form (over $\mathbb C$). Then the three given conditions imply that $Z$ actually has no non-trivial Jordan block, i.e. it is a scalar multiple of the identity matrix. Since $Z$ is real, this multiple must be real too. – 2017-03-02

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$XY=YX$ implies $XY-YX=0$ we deduce $XY-YX+I=I$ so $X,Y,Z$ are one dimensional matrices since the rank of $I$ is $1$ and $Z=aI$.

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    Sorry, my mistake, I've just fixed it – 2017-02-27
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    @Tsemo Aristide: How do you find that $XY = YX$? – 2017-02-27
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    @Tsemo Aristide: as a counterexample to show that $I$ does not need to have rank 1: Consider $Z$ to be the 2 by 2 identity matrix, $X = \begin{pmatrix} 1 & 1\\0 & 1\end{pmatrix}$ and $Y = \begin{pmatrix} 0 & 1\\1 & 0 \end{pmatrix}$. Then we have that $XZ = ZX, YZ = YZ$ and the rank of $XY - YX + I$ is 1, but none of these matrices are one dimensional. – 2017-02-27
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    @Student Tsemo Aristide answered the question as originally posed. The OP has made an edit that completely changes the problem. – 2017-02-27
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    Oh okey, this explaines everything! I was really confused why we could find that $XY = YX$. – 2017-02-27
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    @callus: I am not sure what to do with my comment now, should I remove it? – 2017-02-27
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    @Student No, in fact, I think it's important to leave it. If someone else takes a look at this page, there's a good chance they will be in the same boat you were in. – 2017-02-27