# Differentiation of the null space: I would like to know the value of d(v)/dt.

#### Raja88

##### New member
A 3 dimensional vector [FONT=MathJax_Math-italic]v[FONT=MathJax_Main]=[FONT=MathJax_Main][[FONT=MathJax_Math-italic]w[FONT=MathJax_Math-italic]x[FONT=MathJax_Main],[FONT=MathJax_Math-italic]w[FONT=MathJax_Math-italic]y[/FONT][/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math-italic]w[FONT=MathJax_Math-italic]z[/FONT][/FONT][FONT=MathJax_Main]][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT], The product of the vector's null space and the vector itself is given as [FONT=MathJax_Math-italic]J[FONT=MathJax_Main]([FONT=MathJax_Math-italic]v[FONT=MathJax_Math-italic]T[FONT=MathJax_Main])[FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Math-italic]v[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]..(1) , where [FONT=MathJax_Math-italic]J[FONT=MathJax_Main]([FONT=MathJax_Math-italic]v[FONT=MathJax_Main])[FONT=MathJax_Main]=[FONT=MathJax_Math-italic]n[/FONT][FONT=MathJax_Math-italic]u[/FONT][FONT=MathJax_Math-italic]l[/FONT][FONT=MathJax_Math-italic]l[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]v[FONT=MathJax_Math-italic]T[/FONT][/FONT][FONT=MathJax_Main])[/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]and assumed as orthonormal. Now, the vector can be written as [FONT=MathJax_Math-italic]v[FONT=MathJax_Main]=[FONT=MathJax_Main]([FONT=MathJax_Math-italic]J[FONT=MathJax_Main]([FONT=MathJax_Math-italic]v[FONT=MathJax_Math-italic]T[/FONT][/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main])-1[/FONT][/FONT][/FONT][/FONT][/FONT]
[/FONT][/FONT]

I would like to know the value of [FONT=MathJax_Math-italic]d[FONT=MathJax_Main]([FONT=MathJax_Math-italic]v[FONT=MathJax_Main])[FONT=MathJax_Main]/[FONT=MathJax_Math-italic]d[FONT=MathJax_Math-italic]t.[/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]
[/FONT]

Whether [FONT=MathJax_Math-italic]d[FONT=MathJax_Main]([FONT=MathJax_Math-italic]v[FONT=MathJax_Main])[FONT=MathJax_Main]/[FONT=MathJax_Math-italic]d[FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main]=([/FONT][FONT=MathJax_Math-italic]n[/FONT][FONT=MathJax_Math-italic]u[/FONT][FONT=MathJax_Math-italic]l[/FONT][FONT=MathJax_Math-italic]l[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]v[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]t[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main])T)-1[/FONT][/FONT][/FONT][/FONT][/FONT][/FONT][/FONT]

#### HallsofIvy

##### Elite Member
??? There is nothing at all in the definition of the vector that says it is a function of t!

(I know the "null space" of a linear operator but not of a vector. I presume you mean what I would call the "orthogonal complement" of the vector v- the set of all vectors u such that the dot product of u and v is 0.)

#### Raja88

##### New member
Thanks for the reply..

It is missed v(t)=[wx,wy,wz]. (Actually v(t) is 3D camera coordinates. I need the time derivative of the coordinates)

Yeah you are right. J(v) is orthogonal to v. Since orthogonal, the authors took the null space values of v as J.

hence [FONT=MathJax_Math-italic]v[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]J[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]v[FONT=MathJax_Math-italic]T[/FONT][/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main])-1 =([/FONT][FONT=MathJax_Math-italic]n[/FONT][FONT=MathJax_Math-italic]u[/FONT][FONT=MathJax_Math-italic]l[/FONT][FONT=MathJax_Math-italic]l[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]v[FONT=MathJax_Math-italic]T[/FONT][/FONT][FONT=MathJax_Main]))-1[/FONT][FONT=MathJax_Main]...(1)[/FONT]

To find the time derivatives , I am differentiating the eqn (1). dv/dt

Now i have a doubt that whether dv/dt=(null(dv/dt))-1

#### HallsofIvy

##### Elite Member
If v is a vector depending on t, then dv/dt is also a vector. In particular, if v is a three dimensional vector then dv/dt is a three dimensional vector and null(dv/dt) is a two dimensional space. Now what does the "-1" in "$$\displaystyle (null(dv/dt))^{-1}$$ mean?

#### HallsofIvy

##### Elite Member
If v is a vector depending on t, then dv/dt is also a vector. In particular, if v is a three dimensional vector then dv/dt is a three dimensional vector and null(dv/dt) is a two dimensional space. Now what does the "-1" in "$$\displaystyle (null(dv/dt))^{-1}$$ mean? How do you take a -1 power or inverse of a vector space?

#### Raja88

##### New member
"-1" here refers to inverse. Since i told J(vT)=null[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]v[FONT=MathJax_Math-italic]T[/FONT][/FONT][FONT=MathJax_Main]), is orthonormal, , Inverse of [/FONT]null[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]v[FONT=MathJax_Math-italic]T[/FONT][/FONT][FONT=MathJax_Main]) is ([/FONT]null[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]v[FONT=MathJax_Math-italic]T[/FONT][/FONT][FONT=MathJax_Main]))T, T represents transpose, i.e. J(vT)-1=(J[/FONT][FONT=MathJax_Main](v[/FONT]T[FONT=MathJax_Main]))[/FONT][FONT=MathJax_Main]T

For example v=[/FONT]
[0.1919, 0.1789, 0.6180]T

J(VT)= [ -0.2665 -0.9205; 0.9448 -0.1908;
-0.1908 0.3410]

Hence, [FONT=MathJax_Main]Inverse of [/FONT]null[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]v[FONT=MathJax_Math-italic]T[/FONT][/FONT][FONT=MathJax_Main]) =[/FONT][-0.2665 0.9448 -0.1908;
-0.9205 -0.1908 0.3410]

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