For a real valued $m$ by $n$ matrix $M$ with $m<n$, why is the following true?
$$\det(I+MM^T) = \det(I+M^TM)\;?$$
To me this seems counter-intuitive as $\det(M^TM) = 0$.
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Why is det$(M^TM) = 0$? – Michael Harrison Mar 18 '16 at 15:43
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1because $MM^T$ and $M^TM$ have the same nonzero eigenvalues ... – user251257 Mar 18 '16 at 15:44
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@MichaelHarrison My mistake. Thank you. – Mar 18 '16 at 15:44
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@user251257 That's very interesting. How do we know that? – Mar 18 '16 at 15:46
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@Lembik: singular value decomposition – user251257 Mar 18 '16 at 15:47
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@Théophile Yes it does. – Mar 18 '16 at 15:49
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@MichaelHarrison No I was actually right. Notice that $m< n$ so $\det(M^TM) = 0$. – Mar 18 '16 at 16:14
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This is an example of Sylvester's determinant law which states that for an $m\times n$ matrix $A$ and an $n\times m$ matrix $B$, $\operatorname{det}(I_m + AB) = \operatorname{det}(I_n + BA)$. In your case, $A = M$ and $B = M^T$.
Michael Albanese
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Thank you. Could you explain why adding the identity makes such a difference? It certainly isn't true without this addition. – Mar 18 '16 at 16:17
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1@Lembik One explanation is this. It's almost true. Recall that if $m=n$, it's certainly true that $\det(MM^T)=\det(M^TM)$. With $m<n$, it is still true if $M$ is rank deficient, since both $MM^T$ and $M^TM$ are singular. It fails for $m<n$ only when $M$ is full-rank, as $M^TM$ is necessarily singular while $MM^T$ is not. However as user251257 pointed out, they still have the same non-zero eigenvalues. Now, recall that adding an identity matrix increments each eigenvalue by 1, and that the determinant is the product of all eigenvalues. It follows that both products equal. – syeh_106 Mar 19 '16 at 04:46
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using the SVD decomposition : $M = U S V^*, M^* = V S^* U^*$,
$$I + M M^* = U(SS^* +I)U^*, \qquad I + M^* M = V (S S^* +I) V^*$$
and $det(V A) = det(A) = det(A V)$ for any unitary matrix $V$, hence :
$$det(I+MM^* ) = det(I+M^* M) = det(I+S S^* )$$
if $M$ is real, then $M^* = M^T$ and you get what you asked.
reuns
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If you use a skinny SVD, then you can't assume $UU^=I$. I you use a full SVD, then one of the $SS^$ should be $S^*S$. – user251257 Mar 18 '16 at 16:23
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ok, with $M_{z(k)}$ denoting the zero padding of $M$ such that it becomes a $k \times k$ matrix, $det(I_m + M M^T) = det(I_n +M_{z(n)} M_{z(n)}^T)$, this is quite obvious since $M_{z(n)} M_{z(n)}^T = (M M^T)_{z(n)}$ – reuns Mar 18 '16 at 16:28
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