Firstly, we show that KPT-185 A2=(ai1i2i3i4i5(2)) is a nonnegative tensor with order 5 and dimension 3, and ai1j2j3j4j5>0ai1j2j3j4j5>0 except for a13333=a21111=a33333=0a13333=a21111=a33333=0. We complete the proof by the following four cases and ai1j2j3j4j5(2)=∑i2,i3=13ai1i2i3ai2j2j3ai3j4j5.Case 1:j2≠j3j2≠j3 and j4≠j5j4≠j5.Then ai2j2j3=ai3j4j5=1ai2j2j3=ai3j4j5=1 for any i2,i3∈ 1,2,3 i2,i3∈ 1,2,3 . Thus ai1j2j3j4j5(2)=∑i2,i3=13ai1i2i3>0.Case 2:j2=j3j2=j3 and j4≠j5j4≠j5.Then ai3j4j5=1ai3j4j5=1 for any i3∈ 1,2,3 i3∈ 1,2,3 . Thusai1j2j3j4j5(2)=∑i2,i3=13ai1i2i3ai2j2j3≥ ai123a211,if j2j3=11;ai113a122+ai131a322,if j2j3=22;ai113a133,if j2j3=33>0.Case 3:j2≠j3j2≠j3 and j4=j5j4=j5.The proof is similar to the proof of Case 2.Case 4:j2=j3j2=j3 and j4=j5j4=j5.In duplication case, we know only a13333=a21111=a33333=0a13333=a21111=a33333=0 by direct computation.Similarly, we can show that A3=(ai1i2…i9(3)) is an almost positive tensor with order 9 and dimension 3 except for a233333333(3)=0 by (4.2), but A4,A5,…A4,A5,… are positive tensors. Thus combining the above cases, we know η(A)=4η(A)=4. □
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