In my analysis class, my professor (and the textbook) reference the fact that $\mathbf{D}T(\bar{x}) = T$ for any $\bar{x} \in \mathbb{R}^n$ where $T \in \mathcal{L}(\mathbb{R}^n,\mathbb{R}^n) $. I'm wondering about the precise meaning of this statement. Is it that the differential of any linear transformation is also linear? Or is that the differential of any linear transformation is itself (and therefore also linear)? The second seems confusing to me because the derivative of a function is often not itself. For example, $f(x) = 2x$ is a linear transformation but $f'(x) = 2 \ne 2x$ for all $x \in \mathbb{R}$.


In this context the notation is slightly different than what you are used to.

Consider $f(x)=x^2$. In this notation the derivative of $f$ at the point $x$ acting on a vector $h$ is $(f')_x h = 2x h$. The derivative is the linear function $``$ Multiply by $2x$".

For $f(x) = 2x$ the derivative at the point $x$ acting on the vector $h$ is $(f')_x h = 2h$. Now the linear transformation is just $``$ Multiply by $2$".


If you consider $T = 2$ as a $1\times 1$ matrix, then $f' = T$ as claimed. (the map is "take $x$ and multiply by $T$")

In general, if you consider $$ f(x) = Tx $$

then $Df = T$, literally. You could check this by writing out the formula for $Tx$ in coordinates, then taking various partial derivatives of the various components to compute $Df$. In particular, the $i$th component of $Tx$ is

$$ \sum_j t_{ij} x_j $$

and so $$(Df)_{ij} = \partial_j (Tx)_i = t_{ij}$$

as claimed.


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