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    Suppose the square matrix A of dimension n is of the form
    (BC)
    (DE)

    Where B is a square matrix of dimension k and E is a square matrix of size n-k.

    C has dimension n x (n-k) and D has dimension (n-k) x n.

    If D=0 show that detA = detB detE.
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    (Original post by alexmufc1995)
    Suppose the square matrix A of dimension n is of the form
    (BC)
    (DE)

    Where B is a square matrix of dimension k and E is a square matrix of size n-k.

    C has dimension n x (n-k) and D has dimension (n-k) x n.

    If D=0 show that detA = detB detE.
    It follows from just using Cramer's rule and expanding around the first column. Alternatively, can you take linear combinations so that C becomes 0? (Then it becomes obvious, because we've decomposed V into two subspaces on which A acts independently, so the volume scale on V is the same as the product of the volume scales on the two subspaces.) If not, when can't you do it?
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    (Original post by Smaug123)
    It follows from just using Cramer's rule and expanding around the first column. Alternatively, can you take linear combinations so that C becomes 0? (Then it becomes obvious, because we've decomposed V into two subspaces on which A acts independently, so the volume scale on V is the same as the product of the volume scales on the two subspaces.) If not, when can't you do it?
    I'd thought of eliminating C, but was unsure how I could easily generalise this.

    Please could you elaborate on the use of Cramer's rule? Thanks
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    (Original post by alexmufc1995)
    I'd thought of eliminating C, but was unsure how I could easily generalise this.

    Please could you elaborate on the use of Cramer's rule? Thanks
    Just write out the matrix in full, and expand around the first column using the standard rule for finding determinants in terms of sub-determinants. The subdeterminants can be found inductively (strictly, perform induction on the size of A). You end up with a disgusting mess that turns out to simplify to the correct thing.

    "Generalise"? It's already as general as it needs to be, unless I've misunderstood you :P
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    (Original post by Smaug123)
    Just write out the matrix in full, and expand around the first column using the standard rule for finding determinants in terms of sub-determinants. The subdeterminants can be found inductively (strictly, perform induction on the size of A). You end up with a disgusting mess that turns out to simplify to the correct thing.

    "Generalise"? It's already as general as it needs to be, unless I've misunderstood you :P
    I've had a go with the expansion down the first column but, as you said it's a disgusting mess and I'm sure this can't be what the question is looking for.

    I meant, how can you show that C can, in general, be reduced to 0 using EROs and ECOs?
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    (Original post by alexmufc1995)
    I've had a go with the expansion down the first column but, as you said it's a disgusting mess and I'm sure this can't be what the question is looking for.

    I meant, how can you show that C can, in general, be reduced to 0 using EROs and ECOs?
    Well, it can't always. Under what circumstances can it not?
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    Have to say, I would just use the \sum_\rho \text{sgn}(\rho) a_{1\rho{1}}...a_{n\rho(n)} definition and observe that for a_{1\rho(1)}...a_{k\rho(k)} to be non-zero we must have \rho(i) \in \{1, 2, ..., k\} for 1<=i<=k. Which means the only terms that contribute are where \rho permutes {1, 2, ... , k}. But it then follows that \rho also permutes {k+1,..., n}, and so...
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    (Original post by DFranklin)
    Have to say, I would just use the \sum_\rho \text{sgn}(\rho) a_{1\rho{1}}...a_{n\rho(n)} definition and observe that for a_{1\rho(1)}...a_{k\rho(k)} to be non-zero we must have \rho(i) \in \{1, 2, ..., k\} for 1<=i<=k. Which means the only terms that contribute are where \rho permutes {1, 2, ... , k}. But it then follows that \rho also permutes {k+1,..., n}, and so...
    Ah, that's neater - I haven't thought about determinants for some time, so the only thing about them left in my mind is as a volume form.
 
 
 
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