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$$\mathfrak{L}(0)=0$$

Image of 0 under $$\mathfrak{L}(.)$$ is zero "0" if $$\mathfrak{L}(.)$$ is linear

Example of linear map for $$\R^n$$ (in a n dimension case)

$$A:\R^n \to \R^n$$

$$x \mapsto y=Ax$$

$$A \in R^{nxn} $$ nxn matrix

Insert Figure



In the case above:

$$\mathfrak{D}(Domain)=\R^n$$

$$\mathfrak{R}(Range)=\R^n$$

Analyzing a more general where m≠n:

$$A:\R^n \to \R^n$$

$$x \mapsto y=Ax$$

$$A \in R^{nxn}$$

where: n=row and m=columns

y=Ax

Now consider:

y=Ax

where: y=nx1 matrix

Ax=nx1

b=nx1

$$M: \R^m \mapsto \R^n$$

$$x \mapsto y=Ax+b$$

Clearly:

$$M(0)=b\neq 0\Rightarrow M(.)$$

The $$\neq$$ signal means that $$M(.)$$ is not homogenic

$$M(.)$$ is not a linear map, affine map

Example: Rotation followed by translation

$$y=Rx+b$$

m=n=2

$$\begin{Bmatrix} y_1\\y_2

\end{Bmatrix}=\begin{bmatrix} R_{11} &R_{12} \\ R_{21} &R_{22} \end{bmatrix}\begin{Bmatrix} x_1\\x_2

\end{Bmatrix}+\begin{Bmatrix} b_1\\b_2

\end{Bmatrix}$$

Note: Equivalency $$A\equiv B$$

"if and only of" $$\equiv$$ "equivalent to" $$\equiv$$ "necessary and sufficient condition

$$A\Rightarrow B$$ A is a sufficient condition for B

$$A\Leftarrow B$$ B is a sufficient condition for A

A is a necessary condition for B

$$\begin{bmatrix}

A\Rightarrow B\end{bmatrix}\Leftrightarrow \begin{bmatrix}

\mathbf{B}\Rightarrow \mathbf{A}\end{bmatrix}$$

Where: B is non B or negation of B

and A is non A or negation of A

1) $$(\Rightarrow) $$ (sufficient condition) 2) $$(\Leftarrow) $$ (necessary condition)

Homework

Question: Is div(.) an x of $$M:\R^m\to \R^n $$ and $$M$$ linear?

$$\R^m$$,$$\R^n$$ are species of vectors (tensors,column matrix). Div maps, vector fields

vector fields (vector-value function) into a scalar function.

In other words, domain and range of div(.) are function spaces

=Transformation of coordinates=

Second order linear PDE`s
Coordinates:

$$x=\phi (\bar{x},\bar{y})$$

$$y=\psi (\bar{x},\bar{y})$$

Linear coordinate transformation

Let consider:

$$u=(x,y)= u(\phi (\bar{x},\bar{y}),y(\psi (\bar{x},\bar{y}))$$

$$u=(x,y)= u(\bar{x},\bar{y})$$ * abuse of notation by using "u"

$$u=(x,y)= \bar{u}(\bar{x},\bar{y})$$ * this is more rigorous notation

Example:

Let: $$u(x)=ax+b$$

Consider: $$x=x(\bar{x})=sin\bar{x}$$ where x $$\to $$ general function or $$(\phi (\bar{x}))$$

$$u(x)=u(\phi (\bar{x}))=a sin \bar{x}+b$$

$$u(x)=u(\bar{x})=\bar{u}$$

where:

$$u(\bar{x})$$ is an abuse of notation

and $$\bar{u}(\bar{x})$$ is a more rigorous notation

$$u_x(x,y)=\frac{\partial u}{\partial x\,}(x,y)=u_x(\phi (\bar{x},\bar{y}),y(\psi (\bar{x},\bar{y}))$$

$$=\frac{\partial \bar{u}}{\partial x\,}(\bar{x},\bar{y})=\frac{\partial \bar{u}}{\partial \bar{x}\,}\frac{\partial \bar{x}}{\partial x\,}+\frac{\partial \bar{u}}{\partial \bar{y}\,}\frac{\partial \bar{y}}{\partial x\,}$$

where:

$$\frac{\partial \bar{u}}{\partial \bar{x}\,}\frac{\partial \bar{x}}{\partial x\,}=\bar{u}_\bar{x}$$

$$\frac{\partial \bar{u}}{\partial \bar{y}\,}\frac{\partial \bar{y}}{\partial x\,}=\bar{u}_\bar{y}$$

Define:

$$\bar{x}=\bar{x}(x,y)=\phi (x,y)$$

$$\bar{y}=\bar{y}(x,y)=\psi (x,y)$$

$$u_y(x,y)=\frac{\partial \bar{u}}{\partial y\,}(\bar{x},\bar{y})=\bar{u}_{\bar{x}}\frac{\partial \bar{x}}{\partial y\,}+\bar{u}_{\bar{y}}\frac{\partial \bar{y}}{\partial y\,}$$

In a matrix form:

$$=\partial \bar{x}(u)=\begin{bmatrix}\frac{\partial \bar{x}}{\partial x\,} & \frac{\partial \bar{y}}{\partial x\,}\end{bmatrix} \begin{bmatrix}\partial \bar{x} \\ \partial \bar{y}\end{bmatrix}(\bar{u})$$

Homework

$$\begin{Bmatrix}\partial x \\\partial y\end{Bmatrix}(u)=\begin{bmatrix}\frac{\partial \bar{x}}{\partial x} & \frac{\partial \bar{y}}{\partial x}\\ \frac{\partial \bar{x}}{\partial y} & \frac{\partial \bar{y}}{\partial y}\end{bmatrix}\begin{Bmatrix}\partial \bar{x} \\\partial \bar{y}\end{Bmatrix}(\bar {u})$$

$$\begin{bmatrix}\frac{\partial \bar{x}}{\partial x} & \frac{\partial \bar{y}}{\partial x}\\ \frac{\partial \bar{x}}{\partial y} & \frac{\partial \bar{y}}{\partial y}\end{bmatrix}$$ This matrix is known as the Jacobian matrix

Note: Easier and more general

$$(x_1,...,x_n)\to (\bar{x}_1,...,\bar{x}_n)$$

Ind. notation: $$\bar{x}_i=\bar{x}_i(x_1,...,x_n)$$

$$\mathbf{J}_{nxn}=\begin{bmatrix}\partial \bar{x}_i\\ \partial \bar{x}_j\end{bmatrix}_{nxn}$$

where:

i= row index

j=column index