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I found the following identity:

$$ \frac{\partial( \det (X^T A X ))}{\partial X} = 2\det(X^TAX)AX(X^TAX)^{-1} $$

on the matrix cookbook. It is equation 47 on page 8. Note that $X$ is an $n \times m$ matrix and $A$ is a symmetric $n \times n$ matrix.

I could not find the identity in their cited references. Does anyone know of a textbook or paper that has this identity?


NOTE: I asked this question on Mathoverflow and was advised to repost it here.

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    This may help: http://www.psi.toronto.edu/matrix/calculus.html, also this one http://www.psi.toronto.edu/matrix/calculus.html2012-08-29
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    Thanks, I was looking for a textbook or paper so I can cite it, but I E-mailed Mike Brooks -- the author of the page you recommended -- for a (citable) reference2012-08-29
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    You can also cite online links.2012-08-29
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    True, but I'd prefer to have a more standard reference that has been through a review process, one that may even include a derivation.2012-08-29
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    This identity is easy to obtain, you just need to use the very definition of the derivative. To cite references is just as you will be citing references for the identity $x'=1$.2012-08-29
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    @Godot. This is my first post here, and I've used MathOverflow and StackExchange several times and have never received a rude remark as yours...the above identity is hardly the the same as $x' = 1$.2012-08-29
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    @dblazevski My remark is not intended to make an offense to you. I just wanted to stress that maybe giving references would not be appropriate, as giving references to all trivialities would be just confusing to the reader.2012-08-29
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    @dblazevski Ok I understand. I just would like to encourage you to prove it yourself. Good luck :)2012-08-29
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    @chaohuang the second link is the same as the first one :-)2012-08-29
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    I looked in vain for a similar result back in 1997, and ended up [writing it up](http://www.math.ntnu.no/~hanche/notes/diffdet/) for myself. (But this was for teaching, not research.) My formula can be written as $(d/dt)\ln\det\Phi(t)=\operatorname{tr}(\Phi(t)^{-1}\dot\Phi(t)$, where $\Phi$ is a matrix valued function of a single variable and the dot denotes a derivative. Your result should be derivable from that, with $\Phi=X^TAX$ and $t$ being the various components of $X$ in turn.2012-08-29

2 Answers 2

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Of course we must assume $X^T A X$ is invertible for this equation to make sense.

Take $\tilde{X} = X + t Y$. Then $$\eqalign{\tilde{X}^T A \tilde{X} &= X^T A X + t (Y^T A X + X^T A Y) + O(t^2)\cr &= \left(I + t (Y^T A X + X^T A Y)(X^T A X)^{-1}\right) X^T A X + O(t^2)\cr}$$ so $ \det(\tilde{X}^T A \tilde{X}) = \det(I + tM) \det(X^T A X) + O(t^2)$ where $M = (Y^T A X + X^T A Y)(X^T A X)^{-1}$. Now $\det(I+tM) = \det(\exp(tM)) + O(t^2) = \exp(\text{tr}(tM)) + O(t^2) = 1 + t\; \text{tr}(M) + O(t^2)$. We have $$ \eqalign{\text{tr}(M) &= \text{tr}(Y^T A X (X^T A X)^{-1} + X^T A Y (X^T A X)^{-1})\cr &= \text{tr}(Y^T A X (X^T A X)^{-1}) + \text{tr}((X^T A X)^{-1} X^T A Y)\cr &= 2 \text{tr}(Y^T A X (X^T A X)^{-1})}$$ Taking $Y = E_{ij}$, the matrix with $1$ in entry $(i,j)$ and $0$ everywhere else, this says $$ \frac{\partial}{\partial X_{ij}} \det(X^T A X) = 2 \text{tr}(E_{ji} A X (X^TAX)^{-1}) \det(X^T A X) = 2 \det(X^T A X) (A X (X^T A X)^{-1})_{ij} $$ which is the meaning of $$ \frac{\partial}{\partial X} \det(X^T A X) = 2 \det(X^T A X) A X (X^T A X)^{-1} $$

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    Thank you very much! I tried using the chain rule (applied to $D_x (det(x)) = det(x) x^{-T}$ and index notation, but was left with headache.2012-08-29
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The requested formula is absolutely incorrect. Let $\phi:X\rightarrow X^TAX=U\rightarrow \det(X^TAX)$. The hypothesis $X^TAX$ invertible is useless. $U'_X:H\rightarrow 2X^TAH$ and $\phi'(X):H\rightarrow tr(adj(U)U'(H))$ (where $adj(U)$ is the classical adjoint of $U$). Finally $\phi'(X)(H)=2tr(adj(X^TAX)X^TAH)$. In particular $\phi'(X)=0$ iff $adj(X^TAX)X^TA=0$.