An isotropic tensor must remain identical under any rotation:
t' = R.t.R-1 = t
For three dimensions, for a rotation around the first axis this leads to:
<br />
\text{t'}(1,1)=t(1,1) \\<br />
\text{t'}(1,2)=t(1,2) \cos (\alpha )-t(1,3) \sin (\alpha ) \\<br />
\text{t'}(1,3)=t(1,2) \sin (\alpha )+t(1,3) \cos (\alpha ) \\<br />
\text{t'}(2,1)=t(2,1) \cos (\alpha )-t(3,1) \sin (\alpha ) \\<br />
\text{t'}(2,2)=t(3,3) \sin ^2(\alpha )+t(2,2) \cos ^2(\alpha )-(t(2,3)+t(3,2)) \sin (\alpha ) \cos (\alpha ) \\<br />
\text{t'}(2,3)=-t(3,2) \sin ^2(\alpha )+t(2,3) \cos ^2(\alpha )+(t(2,2)-t(3,3)) \sin (\alpha ) \cos (\alpha ) \\<br />
\text{t'}(3,1)=t(2,1) \sin (\alpha )+t(3,1) \cos (\alpha ) \\<br />
\text{t'}(3,2)=-t(2,3) \sin ^2(\alpha )+t(3,2) \cos ^2(\alpha )+(t(2,2)-t(3,3)) \sin (\alpha ) \cos (\alpha ) \\<br />
\text{t'}(3,3)=t(2,2) \sin ^2(\alpha )+t(3,3) \cos ^2(\alpha )+(t(2,3)+t(3,2)) \sin (\alpha ) \cos (\alpha ) \\
The isotropy condition then leads you to a bunch of relations:
t(1,2) = 0
t(1,3) = 0
t(3,1) = 0
t(2,1) = 0
t(2,2) = t(3,3)
t(2,3) = 0
t(3,2) = 0
The rotations around another axis yields only one additional conclusion:
t(1,1) = t(2,2)