em12
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Hope this is the right section. I'm having trouble ironing out an apparent inconsistency in matrix trace derivative rules.
Two particular rules for matrix trace derivatives are
\frac{\partial}{\partial\mathbf{X}} Tr(\mathbf{X}^2\mathbf{A})=(\mathbf{X} \mathbf{A}+\mathbf{A} \mathbf{X})^T
and
\frac{\partial}{\partial\mathbf{X}} Tr(\mathbf{X}\mathbf{A}\mathbf{X}^T)=\mathbf{X} \mathbf{A}^T+\mathbf{X}\mathbf{A}
Now assume that \mathbf{A} is diagonal (or maybe even just symmetric) and \mathbf{X} is anti-symmetric. Then by the cyclic property of the trace, -Tr(\mathbf{X}^2\mathbf{A})=Tr(\mathbf{X}\mathbf{A}\mathbf{X}^T). So the two derivatives should be equal up to a minus sign, no?
However, the first rule returns the derivative
- (\mathbf{X}\mathbf{A}+\mathbf{A}\mathbf{X})
and the second returns
2\mathbf{X}\mathbf{A}.
Am I missing something?
Two particular rules for matrix trace derivatives are
\frac{\partial}{\partial\mathbf{X}} Tr(\mathbf{X}^2\mathbf{A})=(\mathbf{X} \mathbf{A}+\mathbf{A} \mathbf{X})^T
and
\frac{\partial}{\partial\mathbf{X}} Tr(\mathbf{X}\mathbf{A}\mathbf{X}^T)=\mathbf{X} \mathbf{A}^T+\mathbf{X}\mathbf{A}
Now assume that \mathbf{A} is diagonal (or maybe even just symmetric) and \mathbf{X} is anti-symmetric. Then by the cyclic property of the trace, -Tr(\mathbf{X}^2\mathbf{A})=Tr(\mathbf{X}\mathbf{A}\mathbf{X}^T). So the two derivatives should be equal up to a minus sign, no?
However, the first rule returns the derivative
- (\mathbf{X}\mathbf{A}+\mathbf{A}\mathbf{X})
and the second returns
2\mathbf{X}\mathbf{A}.
Am I missing something?