Loring Tu's “An Introduction to Manifolds”, P95, Example 8.19
My problem is: How to verify that the final sentence " $\frac{d}{dt}|_{t=0}gc(t)=gc^\prime(0) $ by $\Bbb R-$linearity and Proposition 8.15 " ? here, $g$ is a matrix, $c^\prime(t)$ is not the calculus derivative(derivative of a real-valued function), but the velocity vector(linear map of tangent spaces)
I really appreciate any help you can provide
Proposition 8.15 as follows
Let $g$ have entries $(g)_{i,j}$, and similarly for each $t$ let the value of the curve $c(t)$ have entries $(c(t))_{i,j}$. Then the formula for matrix multiplcation says $$ (g c(t))_{i,j} = \sum_{k = 1}^n (g)_{i, k} (c(t))_{k, j}. $$ Differentiating this we obtain (we can differentiate a matrix by differentiating it entrywise) $$ \frac{d}{d t} (g c(t))_{i,j} = \frac{d}{d t}\sum_{k = 1}^n (g)_{i, k} (c(t))_{k, j} = \sum_{k = 1}^n (g)_{i, k} \frac{d}{d t}(c(t))_{k, j}, $$ since each of the entries $(g)_{i, k}$ are just constant. But the entries $\frac{d}{d t}(c(t))_{k, j}$ are exactly the entries of the matrix $c'(t)$, so we obtain $$ \frac{d}{d t} (g c(t))_{i,j} = \sum_{k = 1}^n (g)_{i, k} (c'(t))_{k, j} = (g c'(t))_{i,j}. $$ Hence the derivative at $0$ the entries are $(g c'(0))_{i,j}$, i.e. $\frac{d}{d t}\lvert_{t = 0} g c(t) = g c'(0)$, as desired.