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"Prove that if T is a unitary operator on a finite dimensional inner product space,then there exists a unitary operator U such that T^2=U" My question is,is it possible to solve the question even if the space is considered over the real field? Or is it possible only if the field is complex?

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I assume you meant to say there exists a unitary $U$ such that $U^2 = T$.

If this is the case, then the statement is not true over the real numbers (or in general, if the base field fails to be algebraically closed). For example, there is no real unitary operator $U$ such that $$ U^2 = \pmatrix{1&0\\0&-1} $$

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    Yes you are right,i meant U^2=T.could you please tell me how you found that the matrix you cited fails to satisfy the condition? – 2017-01-30
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    Consider the eigenvalues of $T$. What eigenvalues could $U$ have? – 2017-01-30
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    I guess you are trying to imply that √-1 doesnt exist in R and so we cannot find a diagonal matrix U such that U^2=T.but what is the guarantee that we cannot find a non diagonal matrix? – 2017-01-30
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    That's not it. Any *real* matrix with an eigenvalue of $\sqrt{-1}$ must have its other eigenvalue equal to $-\sqrt{-1}$. – 2017-01-30
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    True,since i and -i are complex conjugates,right? – 2017-01-30
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    Exactly.${}{}{}{}$ – 2017-01-30
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    I still couldnt figure out this : "Consider the eigenvalues of T. What eigenvalues could U have? " – 2017-01-30
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    @AnwitaBhowmik what's the relationship between the eigenvalues of $U$ and the eigenvalues of $U^2$? – 2017-01-30
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    Umm,i dont know... – 2017-01-30
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    Alright, then the fact you're missing for the question is this: if $\lambda$ is an eigenvalue of $U$, then $\lambda^2$ must be an eigenvalue of $U^2$ – 2017-01-30