Tensor Rank of 2X2 Matrix: Is It Always 2?

[siyacar]

Should not the definition of "Rank" agree in the two cases below? :

1)rank of a 2X2 matrix and

2) "tensor rank" of the same 2X2 matrix

Here is my particular example?

|1 1|
|0 1|

This matrix has rank 2. What is its tensor rank? Still 2?

Thnk you

 

[dextercioby]

You should start from the definitions. What is the <rank> of a tensor defined to be ?

 

[siyacar]

Thank you for your response.
Background: I am just learning the notion of tensor rank. My real effort was understanding a certain article where
2x2x2 and 4x4x4 etc, arrays are involved. I was warming up, so to speak, and ran into some problems.

I do know the definition of tensor rank I believe. It is the minimum number of "diads" in the decomposition (in the case of a matrix), and the minimum number of "triads"
used in the decomposition for a 2x2x2 array.

So if @ denotes the tensor rank, the matrix I have posted shoud have tensor rank 2 since it can be written as:

(1,0) @ (1,1) + (0,1) @ (0,1)

Here is my question:
I will write a 2x2x2 array below; by first writing the front face as a matrix and the back face as a matrix.

Front |-1 0| Back |0 1|
|0 1| |1 0|


In this article it is asserted that the rank if this 2x2x2 array is 3, BUT it 2 (not 3) if the entries are considered to be Complex numbers instead of Real numbers.
I was trying to verify this (I was not able to).
If the assertion is correct, then, while the rank of a matrix does not depend on the base field of the entries, it does depend on the filed for a 2x2x2 array.

Any help is appreciated in verifying that the tensor rank is 2 for this array when considered as Complex entries. I thank you for your time