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I know that rank of square matrix is equal to transpose of that but I am interested to know is that fact is also true for square of that matrix.
Is it $\text{rank}(A)=\text{rank}(A^2)$?

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    For some kind of matrices it is true, for example for orthogonal ones or more generally for all with full rank. – 2017-01-30
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    @widawensen thanks – 2017-01-31

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Hint: $\begin{bmatrix}0&1\\0&0\end{bmatrix}^2=\begin{bmatrix}0&0\\0&0\end{bmatrix}$

What is the rank of $\begin{bmatrix}0&1\\0&0\end{bmatrix}$? What is the rank of $\begin{bmatrix}0&1\\0&0\end{bmatrix}^2?$

Something that is true however, is that $\text{rank}(AB)\leq \text{min}(\text{rank}(A),\text{rank}(B))$ so in this case $\text{rank}(A^2)\leq \text{rank}(A)$