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I've seen two different results on the derivatives of determinants of matrices:

$$\frac{\partial |X|}{\partial X_{ij}}=X_{ij}.\tag1$$

$$\frac{\partial\det(X)}{\partial X}=|X|(X^{-1})^{T}.\tag2$$

These seem to imply different things. Which is right and why? Can't find it proven anywhere.

Hakim
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conjectures
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  • Maybe this can suggest a hint ?http://math.stackexchange.com/questions/38701/how-to-calculate-the-gradient-of-log-det-matrix-inverse. – shyamupa May 30 '14 at 17:15
  • The first one is manifestly wrong, because implies $|X|=\frac{1}{2}X_{ij}^2+(\text{terms-not-depending-on-}X_{ij})$ – Vincenzo Tibullo May 30 '14 at 17:18

1 Answers1

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Given, by Laplace Theorem, and for each choice of $i$ $$ |X|=\sum_{j=1}^n (-1)^{i+j}X_{ij}M_{ij} $$ where $M_{ij}$ is the cofactor of $X_{ij}$ and does not contain $X_{ij}$, we have $$ \frac{\partial |X|}{\partial X_{ij}}=(-1)^{i+j}M_{ij} $$ If the matrix is invertible, we have $$ X^{-1}_{ij}=\frac{1}{|X|}(-1)^{i+j}M_{ji} $$ and so $$ \frac{\partial |X|}{\partial X_{ij}}=|X|X^{-1}_{ji} $$

Vincenzo Tibullo
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  • Cheers for this. I was just about clawing my way there. Name of the Laplace theorem is really useful. Gives a handle other than 'that thing you do to get det()'. – conjectures May 30 '14 at 17:40