text: Q:
  
  Prove $\frac{\partial ln|X|}{\partial X_{ij}}=tr[X^{-1} \frac{\partial X}{\partial X_{ij}}]$ not using adjoint.
  
  Prove $\frac{\partial ln|X|}{\partial X_{ij}}=tr[X^{-1} \frac{\partial X}{\partial X_{ij}}]$ where $\forall \ p \ \epsilon \ \mathbb{N}, \forall \ X \ \epsilon \ \mathbb{R}^{pxp}, \ $ X is a positive definite matrix.
  
  A:
  
  What I tried based on Th2.14f in Rencher and Schaalje
  Follow the proof (or directly apply) Theorem 2.14f by replacing $A = X$ and $x = x_{ij}$.
  $$
  \frac{\partial ln|X|}{\partial X_{ij}}
  =
  \frac{\partial ln|CDC'|}{\partial X_{ij}}
  $$
  Since positive definite matrices are symmetric, $X$ has a spectral decomposition: $X = CDC'$ where $C$ is an orthogonal matrix and $D$ is a diagonal matrix whose elements are the eigenvalues of X on the main diagonal and zero otherwise.
  \begin{align}
  \frac{\partial ln|CC'D|}{\partial X_{ij}}
  =&\frac{\partial ln|D|}{\partial X_{ij}}\\
  =&\frac{\partial ln(\prod_{i=1}^{n} \lambda_i)}{\partial X_{ij}}\\
  =&\frac{\partial ln(\lambda_1\lambda_2...\lambda_n)}{\partial X_{ij}}\\
  =&\frac{\partial (ln\lambda_1 + ln\lambda_2 + ... + ln\lambda_n)}{\partial X_{ij}}\\
  =&\frac{\partial (\sum_{i=1}^{n} ln\lambda_i)}{\partial X_{ij}}\\
  =&\sum_{i=1}^{n} \frac{1}{\lambda_i}
  \cdot 
  \frac{ ( \partial \lambda_i)}{\partial X_{ij}}\\
  =&tr(D^{-1} \frac{\partial D}{\partial X_{ij}})\\
  \end{align}
  where $D^{-1}$ is a diagonal matrix whose elements are the reciprocals of the corresponding elements in D on the main diagonal and zero otherwise.
  $$
  tr(D^{-1}) = \sum_{i=1}^{n} \frac{1}{\lambda_i} = \sum_{i=1}^{n} D^{-1}_{ii}.
  $$
  $\frac{\partial D}{\partial X_{ij}}$ is a diagonal matrix whose elements are the partial derivatives of the corresponding elements in $D$ with respect to $X_{ij}$.
  $$
  tr(\frac{\partial D}{\partial X_{ij}}) 
  = \sum_{i=1}^{n} \frac{ ( \partial \lambda_i)}{\partial X_{ij}}
  = \sum_{i=1}^{n} \frac{\partial D}{\partial X_{ij}}_{ii}.
  $$
  and $\sum_{i=1}^{n} D^{-1}_{ii} \frac{\partial D}{\partial X_{ij}}_{ii} = tr(D^{-1} \frac{\partial D}{\partial X_{ij}}$) since the diagonal matrix $D^{-1} \frac{\partial D}{\partial X_{ij}}$ is the not only the matrix product but also the entrywise product of $D^{-1}$ and $\frac{\partial D}{\partial X_{ij}}$ since they are diagonal matrices.
  It remains to prove that $tr(D^{-1} \frac{\partial D}{\partial X_{ij}}) = tr(X^{-1} \frac{\partial X}{\partial X_{ij}}).$
  \begin{align}
  tr(D^{-1} \frac{\partial D}{\partial X_{ij}})
  &=
  tr(D^{-1} \frac{\partial D}{\partial X_{ij}} + \frac{\partial CC'}{\partial X_{ij}})
  \\
  &=
  tr(D^{-1} \frac{\partial D}{\partial X_{ij}} + C\frac{\partial C'}{\partial X_{ij}} + C'\frac{\partial C}{\partial X_{ij}})
  \\
  &=tr(CD^{-1} \frac{\partial D}{\partial X_{ij}}C' + C\frac{\partial C'}{\partial X_{ij}} + CD^{-1}C'\frac{\partial C}{\partial X_{ij}}DC')
  \\
  &=tr((CD^{-1}C')(C \frac{\partial D}{\partial X_{ij}}C' + CD\frac{\partial C'}{\partial X_{ij}} + CD^{-1}C'\frac{\partial C}{\partial X_{ij}}DC'))
  \\
  &=tr(CD^{-1}C' [\frac{C\partial (DC')}{\partial X_{ij}} + \frac{\partial C}{\partial X_{ij}}(DC')])\\
  &=tr(CD^{-1}C' \frac{\partial (CDC')}{\partial X_{ij}})\\
  &=tr(X^{-1} \frac{\partial X}{\partial X_{ij}})\\
  &\hspace{10cm}{\it QED}\\
  \end{align}
  Rencher, Alvin, and G. Bruce Schaalje. "Matrix Algebra." Linear Models in Statistics. Second ed. New Jersey: John Wiley and Sons, Inc., 2008. 5-68. Print.
  
  