# Block multiplication of matrices with a matrix having determinant 1

Suppose we have two matrices, A∈Mm×n(F)A \in M_{m\times n} (\mathbb {F}) and B∈Mm×p(F)B\in M_{m\times p} (\mathbb {F}), where C(B)⊆C(A)C(B) \subseteq C(A). How do you show that there exists a matrix, P∈Mn+p×n+p(F)P \in M_{n+p\times n+p} (\mathbb {F}) such that |P|=1|P|=1 and that block multiplication, [AB]P=[A0m,p]\begin{bmatrix}A&B\end{bmatrix}P = \begin {bmatrix} A&0_{m,p}\end{bmatrix} is well-defined?

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What are C(B)C(B) and C(A)C(A)?
– arctic tern
2 days ago

C(B)C(B) is the column space of BB and C(A)C(A) is the column space of AA.
– Gregory Braun
2 days ago

I was able to narrow it down to this. Let PP be an upper triangular block matrix, such that P=[CD0F]P = \begin{bmatrix}C & D\\0 & F\end{bmatrix}. Then [AB]P=[ACAD+BF]=[A0]\begin{bmatrix}A&B\end{bmatrix}P = \begin{bmatrix}AC & AD + BF\end{bmatrix} = \begin{bmatrix}A & 0\end{bmatrix}. Then, the matrix block, CC must be InI_{n}. Now, determinant, |P|=1=|C||F||P| = 1 = |C||F| where it can be concluded that |D|=1|D| = 1. Also, AD+BF=0AD + BF =0. The only use I have for the clue on the column spaces is that p≤rp\le r.
– Gregory Braun
yesterday

OK, maybe it’s not true that p≤np\le n.
– Gregory Braun
yesterday

@arctictern, C(B)C(B) and C(A)C(A) are the column spaces of BB and AA respectively.
– Gregory Braun
yesterday

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