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I have column vectors $u_1,...,u_n$ forming the matrix $U$. Now these vectors are independent for the sake we can also assume that that they are mutually orthogonal. I.e. $U$ has full rank. If I square each entry of U, will the resulting matrix be still full rank any ideas or counter examples? thanks

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Counterexample:

$$ \left(\begin{array}{cc} 1& 1\\ 1 &-1 \end{array}\right) $$

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    There should be similar counterexamples for all dimensions $n$. – 2017-01-04
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    There are. In fact, $n \times n$ binary (1, -1) matrices of full rank can be constructed for any $n$ (e.g., cyclic permutations of the vector $[-1, 1, 1, \ldots, 1]$ work for any $n>2$). Their element-wise square is always rank-one. – 2017-01-04