I Tangent Spaces and Subspaces ... McInerney Theorem 3.3.13 ...

Summary
Thread involves finding the dimension of a tangent subspace ...
I am reading Andrew McInerney's book: First Steps in Differential Geometry: Riemannian, Contact, Symplectic ... and I am focused on Chapter 3: Advanced Calculus ... and in particular on Section 3.3: Geometric Sets and Subspaces of ##T_p ( \mathbb{R}^n )## ... ...

I need help with an aspect of the proof of Theorem 3.3.13 ... ...

Theorem 3.3.13 (together with a relevant definition) reads as follows:



McInerney - Defn 3.3.12 & Theorem 3.3.13 ... ... Page 1 ... .png

McInerney - 2 - Defn 3.3.12 & Theorem 3.3.13 ... ... Page 2 ... .png




In the above text from McInerney we read the following:

" ... ... The fact that ##\phi## has rank ##n -1## follows by computing the Jacobian matrix at any point in ##U## ... ... "


Can someone please demonstrate rigorously, formally and explicitly that ##\phi## has rank ##n -1## ... ...


My computations with respect to the Jacobian ##[D \phi(p) ]## were as follows:

We have ##\phi ( x_1, \ ... \ ... \ x_{n-1} ) = ( x_1, \ ... \ ... \ x_{n-1}, f( x_1, \ ... \ ... \ x_{n-1}) )##

Now put ...

##f_1( x_1, \ ... \ ... \ x_{n-1} ) = x_1##

##f_2( x_1, \ ... \ ... \ x_{n-1} ) = x_2##

... ... ...

... ... ...

##f_{n-1}( x_1, \ ... \ ... \ x_{n-1} ) = x_{n-1}##

##f_n( x_1, \ ... \ ... \ x_{n-1} ) = f( x_1, \ ... \ ... \ x_{n-1} )##


Then ... the Jacobian ...


##[D \phi(p) ] = \begin{bmatrix} \frac{ \partial f_1 }{ \partial x_1} & ... & ... & \frac{ \partial f_1 }{ \partial x_{n-1} } \\ ... & ... & ... & ... \\ ... & ... & ... & ... \\ \frac{ \partial f_{n-1} }{ \partial x_1} & ... & ... & \frac{ \partial f_{n-1} }{ \partial x_{n-1} } \\ \frac{ \partial f }{ \partial x_1} & ... & ... & \frac{ \partial f }{ \partial x_{n-1} } \end{bmatrix}##


##= \begin{bmatrix} 1 & 0 & 0 & ... & ... & 0 \\ 0 & 1 & 0 & ... & ... & 0 \\ ... & ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... & ... \\ \frac{ \partial f }{ \partial x_1} & ... & ... & ... & ... & \frac{ \partial f }{ \partial x_{n-1} } \end{bmatrix}##

... now ... how do we show that the rank of ##[D \phi(p) ]## is ##n-1## ...?




Hope someone can help ...

Peter
 

fresh_42

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Hi Peter,

if you write down a bit more of the Jacobi matrix, then it will be
##[D \phi(p) ] = \begin{bmatrix} 1 & 0 & 0 & ... & ... & 0 \\ 0 & 1 & 0 & ... & ... & 0 \\ ... & ... & ... & ... & ... & ... \\ ... & ... & ... & ... & 1 & 0 \\ ... & ... & ... & ... & 0 & 1 \\ \frac{ \partial f }{ \partial x_1} & ... & ... & ... & ... & \frac{ \partial f }{ \partial x_{n-1} } \end{bmatrix}##

We have a matrix of dimensions ##n## times ##n-1##, so the rank can at most be ##n-1## because we cannot have ##n## independent columns, if there are only ##n-1##. The rank is also at least ##n-1## since the first ##n-1## rows are linear independent.
 
Thanks fresh_42 ...

Appreciate your help ...

Peter
 

mathwonk

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note also the matrix is that of a linear map from R^n-1 to R^n, hence the rank cannot be more than n-1.
 
Thanks mathwonk ...

Peter
 

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