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

#### Peter

##### Well-known member
MHB Site Helper
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 $$\displaystyle T_p ( \mathbb{R}^n )$$ ... ...

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:

In the above text from McInerney we read the following:

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

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

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

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

Now put ...

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

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

... ... ...

... ... ...

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

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

Then ... the Jacobian ...

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

= \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 $$\displaystyle [D \phi(p) ]$$ is $$\displaystyle n-1$$ ...?

Hope someone can help ...

Peter

#### Opalg

##### MHB Oldtimer
Staff member
$[D \phi(p) ] = \begin{bmatrix} 1 & 0 & \ldots & \ldots & 0 \\ 0 & 1 & \ldots & \ldots & 0 \\ \ldots & \ldots & \ddots & \ldots & 0 \\ \ldots & \ldots & \ldots & \ddots & 0 \\ 0&0&\ldots&\ldots&1 \\ \frac{ \partial f_n }{ \partial x_1} & \ldots & \ldots & \ldots & \frac{ \partial f }{ \partial x_{n-1} } \end{bmatrix}$ is an $n\times(n-1)$ matrix, so it cannot have rank greater than $n-1$. On the other hand, its first $n-1$ rows form the identity $(n-1)\times(n-1)$ matrix, which has rank $n-1$.

#### Peter

##### Well-known member
MHB Site Helper
$[D \phi(p) ] = \begin{bmatrix} 1 & 0 & \ldots & \ldots & 0 \\ 0 & 1 & \ldots & \ldots & 0 \\ \ldots & \ldots & \ddots & \ldots & 0 \\ \ldots & \ldots & \ldots & \ddots & 0 \\ 0&0&\ldots&\ldots&1 \\ \frac{ \partial f_n }{ \partial x_1} & \ldots & \ldots & \ldots & \frac{ \partial f }{ \partial x_{n-1} } \end{bmatrix}$ is an $n\times(n-1)$ matrix, so it cannot have rank greater than $n-1$. On the other hand, its first $n-1$ rows form the identity $(n-1)\times(n-1)$ matrix, which has rank $n-1$.

Thanks Opalg ...