A particular manipulation of vectors

Question

I have a velocity vector defined as, $$V^i=frac{dX^i}{dt}$$
Then, is the following manipulation correct?
$$frac{mathbf{V}boldsymbol{cdot}mathbf{V}}{V^i}=frac{V_kdelta^k_{,,,,a}V^a}{V^i}=V_kdelta^k_{,,,,a}frac{dX^a/dt}{dX^i/dt}=V_kdelta^k_{,,,,a}frac{dX^a}{dX^i}=V_kdelta^k_{,,,,a}delta^a_{,,,,i}= V_i$$
If not where am I making a mistake? Please help.
Edit: Actually I have an expression like,
$$A_imathbf{e}^i=frac{mathbf{V}boldsymbol{cdot}mathbf{V}}{V^i}mathbf{e}^iequivmathbf{A}=frac{mathbf{V}boldsymbol{cdot}mathbf{V}}{mathbf{V}}implies A_i=frac{V_kV^k}{V^i}$$
And the question is whether,
$$A_i=V_i$$

Thiago · Accepted Answer

Your derivation fails in the third equality, when you assume that $frac{dX^a/dt}{dX^i/dt}=frac{dX^a}{dX^i}$. For that to be possible, you would need $X^a(t)$ to be an invertible function, which is not possible for a curve in general. That is, you cannot write $t=t(X^a)$.
Consider for example a 2D circular motion where $vec{X}=(cos t,sin t)$. Given an instant of time, you can find the position uniquely. But given the position, you cannot find $t$ uniquely.
Also, note that $mathbf{A}=frac{mathbf{V}cdotmathbf{V}}{mathbf{V}}$ makes no sense (at least not in the usual sense of vector algebra), because the inverse of a vector is not a vector. In other words, vector division is not defined (again, not in the usual sense of vector algebra.)

Quillo · Answer

Example in 2D, but it's the same in any other number of dimensions.
Represent $mathbf{v}$ in some basis as $(v_x,v_y)$, then
$$
A_x = frac{v_x^2+v_y^2}{v_x} = {bigg{|}} frac{dx}{dt} {bigg{|}} + {left(
frac{dy}{dt}
right)}^2 
{left(
frac{dx}{dt}
right)}^{-1}  
$$
and similarly for $A_y$. So, the answer is no, $A_x neq v_x$.

A particular manipulation of vectors

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