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Exact(46)
A ̃ has full column rank.
with equality holding when matrix φ has full column rank.
If the CFOs are different, A has full column rank.
An important observation is that when A ̃ has full column rank, A - has full column rank too.
The φ matrix has full column rank, i.e., rank=N r N t KL=8.
The matrix φ has full column rank with rank=N r N t KL=8.
Similar(14)
Lemma 2. If G ≥ (L + 1), the matrix A ̄ d ( n ) has full column-rank N T (L + 1)(Q + 1).
We observe that the NRK × NRK matrix R-1[j v ] lies in the noise subspace of B[j v ], i.e., R-1[j v ]B[j v ] = 0. Suppose the NRK × NT (K - G(P + 1)) matrix B[j v ] has full column-rank NT (K - G(P + 1)).
this matrix does not have full column rank.
It is straightforward to show that a necessary condition for C.3 is for each (mathbf {T}_{mathsf {V}_{j}}phantom {dot {i}!}) to have full column rank.
Since does not have full column rank and thus is not invertible, (14) can be solved not by eigen decomposition but instead by a generalized eigenvector problem.
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com