Timeline for Hessian of function of covariance matrices
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 6, 2015 at 6:38 | vote | accept | liubenyuan | ||
| Feb 6, 2015 at 6:38 | |||||
| Jun 24, 2013 at 19:19 | comment | added | loup blanc | $\mathcal{L}''(A)$ is a quadratic form over $\mathbb{R}^{n^2}$. The Hessian of $\mathcal{L}$ in $A$ is the associated matrix $U$. $U$ is defined by $u_{i,j;k,l}=\mathcal{L}''(E_{i,j},E_{k,l})$ where $(E_{i,j})_{i,j}$ is the canonical basis of $\mathcal{M}_n(\mathbb{R})$. | |
| Jun 22, 2013 at 5:22 | comment | added | liubenyuan | I still need to understand your notation of $\mathcal{L}''(A)$ And see if it helped in analyzing the Hessian matrix to decipher whether it is a local minima or not. | |
| Jun 22, 2013 at 5:20 | comment | added | liubenyuan | Thank you blanc. I never thought the second partial derivative of the matrix could be expressed in a so easy way. BTW, if $\mathbf{S}$ is full ranked square matrix, $\mathcal{L}'(A)=\mathbf{0}$ can be solved in a closed form. | |
| Jun 21, 2013 at 16:43 | comment | added | loup blanc | $\mathcal{M}_n( \mathbb{R})$ denotes the $n \times n$ matrices with real entries. The derivative of '$\mathcal{L}_1$' cannot be a real. Yet, using the duality associated to the scalar product $(U,V) \rightarrow tr(U^TV)$, you can say that $\phi'(U)={U^{-1}}^T$, and '$\mathcal{L}_1'(A)={(I+AS)^{-1}}^TS^T$'. To solve $\mathcal{L}'(A)=0$ seems to me difficult. If you know explicitly $S,q$, then you can proceed numerically. | |
| Jun 21, 2013 at 9:14 | comment | added | liubenyuan | and what is $\mathcal{M}_n(\mathbb{R})$ ? I know that $\frac{\partial \mathcal{L}_1(\mathbf{A})}{\partial \mathbf{A}} = tr((I + AS)^{-1}S))$. I am sorry but I am not so familiar with the notation used in your answer. And if $\frac{\partial \mathcal{L}(\mathbf{A})}{\partial \mathbf{A}} = 0$ is the local minima of $\mathcal{L}(\mathbf{A})$, do I still need to prove the Hessian is p.s.d ? | |
| Jun 21, 2013 at 8:17 | comment | added | loup blanc | $H,K \in \mathcal{M}_n (\mathbb{R})$. $\phi' (U)$ is a linear function of $H$ and $\phi'' (U)$ is a symmetric bilinear function of $H,K$. | |
| Jun 21, 2013 at 4:31 | comment | added | liubenyuan | Thanks for your reply. But what is $\mathbf{H}$ and $\mathbf{K}$ ? I used to think that the Hessian of $f(\mathbf{A})$ requires vectorized derivatives, i.e., $H = \frac{\partial^2 \mathcal{L}(\mathrm{vec}\mathbf{A})}{\partial \mathrm{vec}(\mathbf{A})\partial \mathrm{vec}^T(\mathbf{A})}$. | |
| Jun 21, 2013 at 0:56 | history | answered | loup blanc | CC BY-SA 3.0 |