show the mapping is a patch

[logistic_guy]

here is the question

$\displaystyle \bold{Z}(x,y) = (x,xy,y)$.
Show the mapping $\displaystyle \bold{Z}: \bold{R^2} \rightarrow \bold{R^3}$ is a patch.

Definition

Patch -- from Wolfram MathWorld

A patch (also called a local surface) is a differentiable mapping x:U->R^n, where U is an open subset of R^2. More generally, if A is any subset of R^2, then a map x:A->R^n is a patch provided that x can be extended to a differentiable map from U into R^n, where U is an open set containing A...

mathworld.wolfram.com

the definition don't say anything about regularity

to solve this i check first if $\displaystyle \bold{Z}$ is injective then to check the regularity condition that is the Jacobian matrix have full rank. if i solve the Jacobian matrix how to know it have full rank?

 

[Steven G]

I'm not really up on this material. Having said that, the definition of Patch states that you need to show that U = R² is an open subset of R². You also need to also show that the mapping Z is a differentiable mapping (yes, you show that using the Jacobian matrix)

Also, please try not to putts under arithmetic--arithmetic means pre-algebra.

 

[logistic_guy]

I'm not really up on this material. Having said that, the definition of Patch states that you need to show that U = R² is an open subset of R². You also need to also show that the mapping Z is a differentiable mapping (yes, you show that using the Jacobian matrix)

Also, please try not to putts under arithmetic--arithmetic means pre-algebra.

$\displaystyle J = \begin{bmatrix}\frac{\partial x}{\partial x} & \frac{\partial y}{\partial x} & \frac{\partial f}{\partial x} \\ \frac{\partial x}{\partial y} & \frac{\partial y}{\partial y} & \frac{\partial f}{\partial y} \end{bmatrix} = \begin{bmatrix} 1 & y & 0 \\ 0 & x & 1 \end{bmatrix}$

how to know if this matrix have full rank or don't?

 

[blamocur]

how to know if this matrix have full rank or don't?

It does have a full rank because columns 1 and 3 form 2x2 identity matrix.

 

[Steven G]

$\displaystyle J = \begin{bmatrix}\frac{\partial x}{\partial x} & \frac{\partial y}{\partial x} & \frac{\partial f}{\partial x} \\ \frac{\partial x}{\partial y} & \frac{\partial y}{\partial y} & \frac{\partial f}{\partial y} \end{bmatrix} = \begin{bmatrix} 1 & y & 0 \\ 0 & x & 1 \end{bmatrix}$

how to know if this matrix have full rank or don't?

...by showing that it has two independent columns.

 

[logistic_guy]

It does have a full rank because columns 1 and 3 form 2x2 identity matrix.

thank blamocur

...by showing that it has two independent columns.

thank Steven