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After reading this part of Rudin's book i have one question: $A'(\mathbf{x})=A$ seems to me little bit weird because:

1) $A'(x)$ - it's derivative of operator $A$ at point $\mathbf{x}\in \mathbb{R^n}$ and it's number but RHS is function.

2) Let $A(x)=2x$ then it's linear operator on $\mathbb{R}^1$. Let $x_0=1$ then $A'(x_0)=2$ but $2\neq 2x$ and $A'(x)=A$ is meaningless.

I searched the same questions on MSE but I can't understand their meaning. I would be very grateful if anyone explain it to me in detail.

RFZ
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  • The first sentence: "We have defined derivatives ... to be linear transformations". At any fixed point $x$, the derivative of a multivariate function is a linear transformation, in this example denoted by $A'(x)$ – user159517 Jan 22 '16 at 17:25
  • In 2), you're misinterpreting $A'(x_0)$. This is NOT a number but rather a $1 \times 1$ matrix whose only entry is $2$, with the point being that $A'(x_0)$ is meant to be a linear transformation. The linear transformation determined by the $1 \times 1$ matrix $A'(x_0)$ is the same as that determined by $A$. – Santiago Canez Jan 22 '16 at 17:30
  • @SantiagoCanez, can you explain please how you get this matrix? – RFZ Jan 22 '16 at 18:08

3 Answers3

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Note that the definition of a derivative in this context is slightly different from what you are used to.

Note that $f(x) = x$ is a map from $\Bbb R \to \Bbb R$. So, its derivative at $x$ should be a linear map from $\Bbb R \to \Bbb R$. Indeed, we end up with $$ [f'(x)](y) = 1\cdot y $$ which can be neatly represented by the number $1$. Similarly, the derivative of $f(x) = Ax$ is given by $$ [f'(x)](y) = A y $$ which can be neatly represented by the matrix $A$. Let's try a another function, and maybe you'll start getting the idea. For $f(x) = x^3$ (from $\Bbb R$ to $\Bbb R$), we have $$ [f'(x)](y) = 3x^2 y $$ which gives us a linear function of $y$ at any value of $x$.

Hopefully this clarifies things a bit.

Note that the derivative of $f(x)$ is defined to be the (unique) linear function $[f'(x)](y)$ such that $$ \lim_{h \to 0}\frac{f(x+h) - f(x) - [f'(x)](h)}{\|h\|} = 0 $$ If we start with $f(x) = Ax$, then we can show that $[f'(x)](y) = Ay$ is the derivative of $f$ by this definition. When we plug in, we have $$ \lim_{h \to 0}\frac{f(x+h) - f(x) - [f'(x)](h)}{\|h\|} = \\ \lim_{h \to 0}\frac{A(x+h) - A(x) - Ah}{\|h\|} = \\ \lim_{h \to 0}\frac{0}{\|h\|} $$ so the necessary condition holds.

Ben Grossmann
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It would be nice if mathematicians were better about saying what the types of things were.

Write $Df$ for $f'$. Then $D : (\mathbb{R}^n \to \mathbb{R}^m) \to (\mathbb{R}^n \to L(\mathbb{R^n}, \mathbb{R^m}))$. You don't notice this at first because $L(\mathbb{R},\mathbb{R}) \cong \mathbb{R}$, i.e. in the 1D case from high school, the linear transformation that results is represented by a single number.

So what's happening is $A : L(\mathbb{R}^n, \mathbb{R}^m)$ which is just a special type of function $\mathbb{R}^n \to \mathbb{R}^m$ so $DA : \mathbb{R}^n \to L(\mathbb{R}^n, \mathbb{R}^m)$ so it's still expecting the point about which to make the linear approximation. So $DA(x) : L(\mathbb{R}^n, \mathbb{R}^m)$ just like $A$. It will serve you well to think about what the types of things are.

Derek Elkins left SE
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The differential of a function defined on $\mathbb{R}^n\rightarrow \mathbb{R}^m$ at $x$ is a LINEAR function $L_x:\mathbb{R}^n\rightarrow \mathbb{R}^m$ such that $\lim\limits_{h\rightarrow 0} (f(x+h)-f(x)-L_x(h))=0$. If $n=1$, the differential of $f:\mathbb{R}\rightarrow \mathbb{R}$ at $x$ is identified with a number since a linear map of $\mathbb{R}$ is of the form $y\rightarrow ay$ and $a$ determines the map. So if $A$ is a linear map and $x\in \mathbb{R}$ $\lim\limits_{h\rightarrow 0}(A(x+h)-A(x)-A(h))=0$ so the differential of $A$ is $A$.

RFZ
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Tsemo Aristide
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