Let $A$ be an $m\times n$ matrix and $B$ an $n\times m$ matrix. Show that $$ |I_m-AB|=|I_n-BA|. $$
I don't know where to start.
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1Hint: The determinant is fixed by conjugation by any invertible matrix. – Travis Willse Oct 27 '14 at 01:39
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@Travis Thanks! What does conjugation mean? :P – Dia McThrees Oct 27 '14 at 01:40
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@DiaMcThrees : $M\mapsto N^{-1}MN$ is conjugation of $M$ by $N$. – Michael Hardy Oct 27 '14 at 01:41
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1@Travis : Wouldn't your idea require the matrices to have just as many rows as columns? – Michael Hardy Oct 27 '14 at 01:42
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Conjugation by a matrix $P$ is the operation $S \mapsto P^{-1} S P$; it used, for example, when changing bases. In particular, $\det (P^{-1} S P) = \det(P^{-1}) \det(S) \det(P) = (\det P)^{-1} \det S \det P = \det S$. – Travis Willse Oct 27 '14 at 01:42
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Ah, I didn't see that the matrices aren't square. In that case you might try to modify this approach by suitably padding $A, B$. – Travis Willse Oct 27 '14 at 01:43
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Also, $A$ and $B$ don't necessarily have full rank. – aschepler Oct 27 '14 at 01:46
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It would help to know what the vertical bars on either end of each side are supposed to mean. – Charlie Frohman Oct 27 '14 at 01:46
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I'm wondering if thinking about the singular value decomposition of $AB$ might help. I've seen this proved, but at this moment I don't remember how it was done. – Michael Hardy Oct 27 '14 at 01:48
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@CharlieFrohman the vertical bars mean determinant – Dia McThrees Oct 27 '14 at 01:48
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Bars are ambiguous, they can mean many things. I made it more explicit. – Patrick Da Silva Oct 27 '14 at 01:49
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1@MichaelHardy I found this – Dia McThrees Oct 27 '14 at 01:49
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So far only I have up-voted the question, although many have commented within only a few minutes. – Michael Hardy Oct 27 '14 at 01:53
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Hint.
Compute the product of the block matrices $$ \pmatrix{I_m & A \\ 0 & I_n} \pmatrix{I_m - AB & 0 \\ B & I_n} \pmatrix{I_m & -A \\ 0 & I_n}.$$
This somewhat mystifying calculation can be thought of as performing different "block" row and column operations on the matrix $\pmatrix{I_m & A \\ B & I_n}$, which don't change the determinant.
That is, $$\pmatrix{I_m & -A \\ 0 & I_n} \pmatrix{I_m & A \\ B & I_n}$$ subtracts $A$ times the bottom row from the top row, while $$ \pmatrix{I_m & A \\ B & I_n} \pmatrix{I_m & -A \\ 0 & I_n}$$ subtracts the first column times $A$ from the second column. The two different ways of eliminating the $A$ lead to two different triangular block matrices with the same determinant.
user187373
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+1, Is this the sylvester's determinant criterion?. Is there any other proof other than the block determinant argument? – dineshdileep Oct 27 '14 at 04:25
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Another way of approaching this: define $$ C = \pmatrix{I&A\\B&I}, \quad D = \pmatrix{I&-A\\0&I} $$ and note that $\det(CD) = \det(DC)$.
Ben Grossmann
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+1, Is this the sylvester's determinant criterion?. Is there any other proof other than the block determinant argument? – dineshdileep Oct 27 '14 at 04:25
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@dineshdileep if you mean this, then yes. I would think that there must be such proofs, but it's hard to beat block-matrix arguments for brevity. – Ben Grossmann Oct 27 '14 at 04:27