The inverse, rank and product of tensors

Tensors product and hyperdeterminant of boundary formats product

“…If Det(C i ) = 0, then by Theorem 3.6, we know (I m 0 +1 , x (1) , · · · , x (p+q−1) ) · (C i ) T = 0 has solution X = (x (1) , · · · , x (p+q−1) ) with all x (i) = 0, ∀i ∈ [p + q − 1]. Then, by Lemma 2.4 and Theorem 2.5, (I m 0 +1 , x (1) , · · · , x (p+q−1) ) · (C i ) T = (I m 0 +1 , x (1) , · · · , x (p+q−1) )…”

“…If Det(B i ) = 0, by Theorem 3.6, we know (I m i +1 , y (1) , · · · , y (q) ) · (B i ) T = 0 just has a trivial solution. Since vectors x (i) , · · · , x (i+q−1) are all nonzero, we know…”

“…By Definition 2.1, equation (3) can be rewritten as the following form: (1) , · · · , x (p) ) · A T . .…”

“…(e i m 0 +1 , x (1) , · · · , x (p) ) · A T ⎞ ⎟ ⎠ = (I m 0 +1 , x (1) , · · · , x (p) ) · A T . Now, we give a reformulation of Theorem 3.1 of Chapter 14 in [9].…”

“…Noting that the definition of hyperdeterminant used here is different from the one used in Section 3. In 2014, Bu and others generalized Definition 1.2 as following: [1].) Let A ∈ C n 1 ×n 2 ×···×n 2 and B ∈ C n 2 ×···×n k+1 be order m 2 and k 1 tensors, respectively.…”