why is $(dX)_{q}: \mathbb{R}^2 \rightarrow \mathbb{R}^3 $ injective equivalent to saying that $X_{u}(q) $ and $X_{v}(q) $ are linearly independent?

Question

I'm studying surfaces and in the definition I found a note that refers to an equivalence for one of the conditions in the definition, but I don't understand why those two things are equivalent. That is, why is $(dX)_{q}: \mathbb{R}^2 \rightarrow \mathbb{R}^3 $ injective equivalent to saying that $X_{u}(q) = \frac{\partial X}{\partial u}(q)$ and $X_{v}(q) = \frac{\partial X}{\partial v}(q)$ are linearly independent ? I'm referring to definition 2.2 and remark 2.5. I have added images:

Answer 1

Just try to write down the matrix associated to the linear mapping $(dX)_q$. Then remember that an injective linear mapping has trival kernel.

Answer 2

$(dX)_q$ is a linear map represented by a matrix having 3 rows and 2 columns. the first column is $\frac{\partial X}{\partial u}(q)$ and the second one $\frac{\partial X}{\partial v}(q)$.

Saying that $(dX)_q$ is injective is equivalent to say that its kernel is reduced to the zero vector which is equivalent to say that the two column vectors are linearly independent.