Show that $dF_x$ is surjective for all $x$

Question

I am trying to tackle question 2.3.8 on GP, but I haven't figure out the following question yet.

Suppose $m > 1$. Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a smooth map. Consider $f + Ax$ for $A \in \mathrm{Mat}_{m\times n}$. Define $F: \mathbb{R}^n \times \mathrm{Mat}_{m\times n} \rightarrow \mathrm{Mat}_{m\times n}$ by $F(x,A) = df_x + A$. How can I show that $dF_x$ is surjective for all $x$?

I am really grateful to @Ross B. 's answer on the question The derivative of a linear transformation, $DF$ would be a rank-3 tensor with elements

$$ (DF)_{i,j,k} = \frac{\partial^2 f_i}{\partial x_j \partial x_k} $$

Some authors also define matrix-by-vector and matrix-by-matrix derivatives differently be considering $m \times n$ matricies as vectors in $\mathbb{R}^{mn}$ and "stacking" the resulting partial derivatives.

Then I got really lost trying to show that $dF$ is surjective.

Some thoughts I had so far:

$f: \mathbb{R}^n \rightarrow \mathbb{R}^m$, and its tangent plane has dimension $m \times n$, therefore $dF$ maps from $m \times n \times n$ to $m \times n$. Then I am thinking of to prove that $\forall m \in \mathrm{Mat}_{m\times n}, \exists x \in \mathbb{R}^n, A \in \mathrm{Mat}_{m\times n}$ such that $ d f_x + A = m$. I am not sure if this is correct, nor how to prove this.

Thank you very much.

To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. — Zev Chonoles, Jun 10 '13 at 20:36
Hi @ZevChonoles, thanks a lot for your instruction. I added where the problem comes from and why I got stuck. Thank you very much. — 1LiterTears, Jun 10 '13 at 22:12
No, if you look at the function $g\colon M(n)\to M(n)$, $g(A)=A$, what's the derivative? — Ted Shifrin, Jun 10 '13 at 22:24
@Matt I mean, GP says m=n, but the problem I am dealing with is $\mathbb{R}^n \rightarrow \mathbb{R}^m$ - thanks! — 1LiterTears, Jun 10 '13 at 22:27
@TedShifrin Oh, I guess I got confused again - the derivative of linear transformation is itself.... right? Thank you -) — 1LiterTears, Jun 10 '13 at 22:28
I guess I was really hungry earlier..I think it should be $dF(x,A) = d(df_x + A) = df_x + A$ — 1LiterTears, Jun 10 '13 at 22:59
@Matt Hope that won't happen? I assumed $n > m, A \neq 0$... — 1LiterTears, Jun 10 '13 at 23:36
No, it's not $df_x + A$. Don't write down this formal formulas without understanding what you're doing :) If you want to show $dF_{(x,A)}\colon \mathbb R^n \times M(n) \to M(n)$ is surjective, think about an appropriate directional derivative that will give you what you want. That is, you need to think about $dF_{(x,A)}(v,B)$ and what you know and what you don't. — Ted Shifrin, Jun 10 '13 at 23:47
Hi @TedShifrin, thank you so much for helping me again and again.. Though, I am curious that if it is not $df_x +A$, what it could be? $df_x$ is linear, and $A$ is a matrix, I don't understand where could be wrong.. :) — 1LiterTears, Jun 10 '13 at 23:56
@BenjaLim In the problem, it is given that $f$ is a smooth map (I just added this in the problem description.) But to my understanding, $df_x$ is a linear map, since it works on the tangent plane - or this is not true..? — 1LiterTears, Jun 11 '13 at 00:32
@MathSnail Yes that is true. The derivative is a linear map between two finite dimensional vector spaces. Perhaps you should have said more explicitly in the beginning of the question that $f$ is *any* smooth map. — , Jun 11 '13 at 00:36
Hint: $f$ is actually irrelevant for this proposition/proof. Follow my advice and do directional derivatives, not in coordinates. — Ted Shifrin, Jun 11 '13 at 00:47
Hi @TedShifrin, I have been thinking hard on the directional derivatives and made some progress as edited in the question. But I still could not draw the conclusion. Would you mind kindly giving me some help? Thank you very much! — 1LiterTears, Jun 11 '13 at 17:10
What is $dF_{(x,A)}(0,B)$ for any $m\times n$ matrix $B$? BTW, GP 2.3.8 specifically has $n\times n$ matrices. To get from $F$ a submersion to $F_A$ never $0$ for generic $A$ you will need $m\times n$ with $m>1$. — Ted Shifrin, Jun 12 '13 at 00:01
Hi Ted, I guess I am dealing with a slight derivation of 2.3.8.. — 1LiterTears, Jun 12 '13 at 00:08
I guess $dF_{x,A}(0,B) = d_{x,A}(df_x+A) = d^2f_x + dA_A = d^2f_x + I_A = d^2f_x + A$..? — 1LiterTears, Jun 12 '13 at 00:13
No. Please use the definition of the derivative of a map between vector spaces: $dg_a(v) = \lim\limits_{t\to 0}\dfrac{g(a+tv)-g(a)}t$. — Ted Shifrin, Jun 12 '13 at 01:01
Thank you Ted. $\displaystyle dF_{x,A}(0,B) = \lim_{t \rightarrow 0}\frac{F(x,A+tB)-F(x,A)}{t}= \lim_{t \rightarrow 0}\frac{df_x + A+ tB-df_x - A}{t}= B$ — 1LiterTears, Jun 12 '13 at 04:07

score 0 · Accepted Answer · edited Apr 13 '17 at 12:20

This question arises many very interesting discussions: like The dimension of the derivative and The derivative of a linear transformation. These discussions are really very helpful and inspiring for me. Thanks all the wonderful teachers!

However, the solution to show that $dF_x$ is surjective for all $x$ is rather straight forward, thanks to Professor @Ted Shifirin's patient guidance. I hope I carried out Ted's beautiful proof correctly:

Consider $\displaystyle dF_{x,A}(0,B) = \lim_{t \rightarrow 0}\frac{F(x,A+tB)-F(x,A)}{t}= \lim_{t \rightarrow 0}\frac{df_x + A+ tB-df_x - A}{t}= B$. Therefore, $\forall m \in \mathrm{Mat}_{m\times n}, \exists B \in \mathrm{Mat}_{m\times n}$, such that $dF_{x,A}(0,B) = m$. In particular, $B=m$.

Show that $dF_x$ is surjective for all $x$

1 Answers1