# Derivative is surjective if each row vector in the jacobian are linearly independent?

I am reading Differential topology by Guillermin, on one of the page the author made a statement. (I am paraphrasing)

suppose that $g_1,....g_l$ are smooth function from a manifold $X \rightarrow R$ The set $Z$ of common zero is a submanifold if $(0,.....0)$ is a regular value of $g = g_1 \times g_2.... \times g_l$ You can quickly verify that $dg_x: T_x(X) \rightarrow R^l$ is surjective if and only if the l functionals $d(g_1)_x.....d(g_l)_x$ are linearly independent on $T_x(X)$.

Seems to me the last statement is implying the derivative, jacobian is surjective if and only if the row vectors are linearly independent? So first question is, how does the linearly indpendent of the $d(g)i)_x$ translates to linearly independence of the row vectors?

Second question: If so, what's the proof of that (from linear algebra)? I know that column vectors of a linear map being linearly independent implies the map is injective, so I'm guessing the dual notion must be true (since for square matrices, column rank = row rank and a linear map is injective if and only if it is surjective).

• my "intuitive proof" for second question would be , suppose the simplest case, that $v2$ = $a \times v1$ where $v1$ and $v2$ are the first and second row vectors and $a \in R$, then I can never have an output vector with the 1st component being 0 and the second component being non zero. – Ecotistician May 16 '18 at 6:00
• "intuitive proof" for 1st question would be the apply a contradiction for the case with a vector (1,1,1,1,1..) – Ecotistician May 16 '18 at 6:05
• On the other hand, if the row vectors are linearly independent, their rank = number of rows => they span the R^m where m= number of rows, and so by Gaussian elimination we can reduce the rows to m rows of $e_1....e_m$ vectors which gives any solution in $R^m$ – Ecotistician May 16 '18 at 11:18
• A linear transformation is surjective if it is full rank. Full rank means either all the rows or columns are linearly independent. If you are mapping from $R^n$ to $R^m$ , with $m<n$, then the number of rows is $m$, and columns $n$. Surjectivity occurs if your matrix has rank $m$. Moreover, if you are going from $n$ to $m$ with $n$ larger than $m$, then you cannot have surjectivity. – rubikscube09 May 16 '18 at 20:08