2

Let $A:\mathbb R^3\to \mathbb R^3$ be a linear map. I need the following technical result to solve a problem I'm solving:

$$A(\alpha(t))'=A(\alpha'(t))$$

Where $\alpha:I\to \mathbb R^3$ is a parametric curve.

Intuitively, I know this is true, but I would like to prove it formally.

user42912
  • 23,582

1 Answers1

3

Let $T$ be the derivative of $A \circ \alpha$, evaluated at a given $t \in I$. Then $T$ is the unique linear map $\mathbb{R} \rightarrow \mathbb{R}^3$ such that

$$\frac{||A \circ \alpha(t + h) - A \circ \alpha(t) - T(h)||}{|h|}$$

goes to zero as $h$ goes to $0$. Let $S: \mathbb{R} \rightarrow \mathbb{R}^3$ be the linear map $S(h) = A(\alpha'(t)h)$. Then

$$\frac{||A \circ \alpha(t + h) - A \circ \alpha(t) - S(h)||}{|h|} = \frac{||A[ \alpha(t + h) - \alpha(t) - \alpha'(t)h]||}{|h|}$$

where $|| \cdot ||$ is a fixed norm on $\mathbb{R}^3$.

The numerator is bounded by $||A|| \cdot ||\alpha(t+h) - \alpha(t) - \alpha'(t)h||$, where $||A||$ is the norm of the linear transformation $A$, and by definition of $\alpha'(t)$,

$$\frac{||\alpha(t+h)-\alpha(t) -\alpha'(t)h||}{|h|}$$

goes to zero as $h$ goes to $0$. So $S$ satisfies the same property as $T$. By uniqueness, $S = T$.

Another way using the chain rule:

The chain rule says that the derivative of $A \circ \alpha$, evaluated at $t$, is the composition of the linear transformation $\alpha'(t): \mathbb{R} \rightarrow \mathbb{R}^3$ (the derivative of $\alpha$ evaluated at $t$), followed by the linear transformation $A'(\alpha(t)): \mathbb{R}^3 \rightarrow \mathbb{R}^3$ (the derivative of $A$ evaluated at $\alpha(t)$). So

$$(A \circ \alpha)'(t) = A'(\alpha(t)) \circ \alpha'(t)$$

Since $A$ is a linear transformation, its derivative at every point of $\mathbb{R}^3$ is constant and equal to $A$. Thus

$$(A \circ \alpha)(t) = A \circ \alpha'(t)$$

as linear maps $\mathbb{R} \rightarrow \mathbb{R}^3$.

D_S
  • 33,891
  • So you proved the derivative of the composition $A(\alpha(t))'$ is $A(\alpha'(t)\alpha(t))$, but I need that the derivative of this composition were $A(\alpha'(t))$, can we say $A(\alpha'(t)\alpha(t))=A(\alpha'(t))$? – user42912 Oct 15 '17 at 16:02
  • I wrote it wrong, it should be correct now. If by $A(\alpha'(t))$, you mean the linear transformation $A \circ \alpha'(t): \mathbb{R} \rightarrow \mathbb{R} \rightarrow \mathbb{R}^3$, then that's correct. – D_S Oct 15 '17 at 16:29
  • Thank you. I'm sorry, I still don't get it. Isn't the derivative of a linear transformation the linear trasnformation itself, i.e., $A'=A$? so shouldn't we have $A(\alpha(t))'=A(\alpha(t))$? – user42912 Oct 16 '17 at 01:50
  • The derivative of $A$ at any point is $A$ itself, yes. For the derivative of $A \circ \alpha$, you need to use the chain rule, which is basically what I did. – D_S Oct 16 '17 at 02:00
  • Using the chain rule as you said: $(A\circ \alpha)'(t)=A'(\alpha(t))\circ\alpha'(t)=A(\alpha(t))\circ\alpha'(t)$ I still don't get what I want. What am I missing? – user42912 Oct 16 '17 at 04:26
  • If $f: \mathbb{R}^3 \rightarrow \mathbb{R}^3$, then the derivative $ f': \mathbb{R}^3 \rightarrow \textrm{Hom}_{\mathbb{R}}(\mathbb{R}^3)$ is a function which associates to each point $p \in \mathbb{R}^3$ a certain linear map $f'(p): \mathbb{R}^3 \rightarrow \mathbb{R}^3$. If $f$ is already a linear map, then $f'(p) = f$ for all $p \in \mathbb{R}^3$. So $A'(\alpha(t)) = A$. – D_S Oct 16 '17 at 04:39
  • So what you wrote: $A(\alpha(t)) = A'(\alpha(t))$ does not make sense; the right side is a linear map, the left side is an element of $\mathbb{R}^3$. It should be $A = A'(\alpha(t))$. See my edit. – D_S Oct 16 '17 at 04:48
  • Haaaaaaaaa, I got it!!! Thank you very very much!!!!!!!!! – user42912 Oct 16 '17 at 05:20