## Tag Archives: partial derivative

### Partial derivatives

Let $u:\mathbb{R}^n\times\mathbb{R}\to\mathbb{R},(x,t)\mapsto u(x,t)$ be a smooth map, and define the partial derivative as $u_{ij}:=\frac{\partial^2 u}{\partial x_i\partial x_j}$ for each $1\le i,j\le n$. Then $\det(u_{ij}):\mathbb{R}^n\times\mathbb{R}\to\mathbb{R}$ is also a smooth map. The derivative with respect to $t$ is

$\frac{d}{dt}\det(u_{ij})=\frac{d}{dt}(\sum_{k_i}\delta^{k_1\cdots k_n}_{\ 1\cdots\ n}u_{k_1 1}\cdots u_{k_n n})=\sum_{k_i}\delta^{k_1\cdots k_n}_{\ 1\cdots\ n}\frac{d}{dt}(u_{k_1 1}\cdots u_{k_n n})$

$=\sum_{k_i}\delta^{k_1\cdots k_n}_{\ 1\cdots\ n}\sum_{j}u_{k_1 1}\cdots u_{k_{j-1} j-1}\cdot\frac{d}{dt}u_{k_j j}\cdot u_{k_{j+1} j+1}\cdots u_{k_n n}$

$=\sum_{j}\sum_{k_i}\delta^{k_1\cdots k_n}_{\ 1\cdots\ n}u_{k_1 1}\cdots u_{k_{j-1} j-1}\cdot\frac{d}{dt}u_{k_j j}\cdot u_{k_{j+1} j+1}\cdots u_{k_n n}$

$=\sum_{j}\det(u_{i1},\cdots, u_{i, j-1},\frac{d}{dt}u_{i j},u_{i, j+1},\cdots,u_{in})=\sum_{1\le i,j\le n}\frac{d}{dt}u_{i j}\cdot A_{ij},$

where the $A_{ij}$ represents the $i,j$ element of the matrix cofactors. In particular

$\frac{d}{dt}\log\det(u_{ij})=\frac{1}{\det(u_{ij})}\sum_{1\le i,j\le n}\frac{d}{dt}u_{i j}\cdot A_{ij}=\sum_{1\le i,j\le n}\frac{d}{dt}u_{i j}\cdot u^{ij}$.

Now let’s consider the second derivative for a special case that $u$ is affine with respect to $t$, that is, $\frac{d}{dt}u_{ij}$ is independent of $t$:

$\frac{d^2}{dt^2}\log\det(u_{ij})=(\frac{d}{dt}\frac{1}{\det(u_{ij})})\cdot\sum_{1\le i,j\le n}\frac{d}{dt}u_{i j}\cdot A_{ij}$

– – – – – – – – – – – – $+\frac{1}{\det(u_{ij})}\cdot\sum_{1\le i,j\le n}\frac{d}{dt}u_{i j}\cdot (\frac{d}{dt}A_{ij})=I+I\!I$, where

$I=-(\sum_{1\le i,j\le n}\frac{d}{dt}u_{i j}\cdot u^{ij})^2\le0$, and

$I\!I\cdot\det(u_{ij})=\sum_{ij}\frac{d}{dt}u_{i j}\cdot (\sum_{k\neq i,l\neq j}\frac{d}{dt}u_{kl}\cdot B_{ij,kl})$

– – – – – – $=\sum_{k\neq i,l\neq j}\frac{d}{dt}u_{i j} \cdot\frac{d}{dt}u_{kl}\cdot B_{ij,kl}$.

I do not know if $I\!I$ is zero (if it is, then the minus function $-\log\det(u_{ij})$ will be convex).

——

In linear algebra, a symmetric matrix $A\in \mathcal{M}_n(\mathbb{R})$ is said to be positive definite if $v^\tau A v$ is positive, for any non-zero vector $v\in\mathbb{R}^n\backslash\{{\bf 0}\}$.

Proposition. A symmetric $A$, and a symmetric and positive-definite matrix $B$ can be simultaneously diagonalized: that is, there exists $X$ such that $X^\tau AX=\mathrm{diag}(a_1,\cdots,a_n)$ and $X^\tau BX=I$.

Proposition. Suppose $A,B$ are symmetric matrices with $A>0$ and $AB>0$. Then$B>0$.
Proof. Since $B$ is symmetric, there exists an orthogonal matrix $O$ with $O^\tau BO=\mathrm{diag}(b_1,\cdots,b_n)$. Clearly $C=O^\tau AO>0$ and $O^\tau AOO^\tau BO=O^\tau ABO>0$. So the $(ii)$-entries $c_{ii}b_i>0$ and $c_{ii}>0$. Therefore $b_i>0$ and $B>0$.

Proposition. Suppose $A,B>0$. If $A^2-B^2>0$, then $A-B>0$.
Proof. Note that $2(A^2-B^2)=(A+B)(A-B)+(A-B)(A+B)>0$. Then $(A-B)(A+B)>0$. Combining with $(A+B)>0$ we get $(A-B)>0$.