Does differentiation preserve linear independence?

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If we take the derivative of n functions that are linearly independent to each other and we write it down like c1f1(x) + c2f2(x) +...+ cnfn(x)=0, then would the linear independence be preserved if we differentiate the equation with respect to x?
 

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  • #2
Fredrik
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Is this the statement you're trying to prove or disprove? If [itex]\{f_1,\dots,f_n\}[/itex] is a linearly independent set, then so is [itex]\{f_1',\dots,f_n'\}[/itex].

If yes, then try to solve this problem first: Suppose that V is a vector space, that T:V→V is linear, and that [itex]\{v_1,\dots,v_n\}\subset V[/itex] is a linearly independent set. Do we need to assume anything else to ensure that [itex]\{Tv_1,\dots,Tv_n\}[/itex] is linearly independent?

(I can't just tell you the answer. The question is too much like a homework problem).
 
  • #3
Deveno
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oh, i know this one. it has to do with that guy that invented kentucky fried chicken, right?

(by this, of course, i mean that the answer involves the colonel).
 
  • #4
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Is this the statement you're trying to prove or disprove? If [itex]\{f_1,\dots,f_n\}[/itex] is a linearly independent set, then so is [itex]\{f_1',\dots,f_n'\}[/itex].

If yes, then try to solve this problem first: Suppose that V is a vector space, that T:V→V is linear, and that [itex]\{v_1,\dots,v_n\}\subset V[/itex] is a linearly independent set. Do we need to assume anything else to ensure that [itex]\{Tv_1,\dots,Tv_n\}[/itex] is linearly independent?

(I can't just tell you the answer. The question is too much like a homework problem).
Actually I neither want to prove nor to disprove it. It was a guess that I came up with. like a conjecture. I'm asking if anyone knows the answer, because I've seen that method frequently be used in many areas of mathematics.
If It's true, then a proof that requires no more advanced math than linear algebra will be highly appreciated.
About your question, it depends on T. I haven't thought about it yet but I guess if T is an invertible mapping, then the answer would be yes.
oh, i know this one. it has to do with that guy that invented kentucky fried chicken, right?

(by this, of course, i mean that the answer involves the colonel).
Would you be more precise?
 
  • #5
Fredrik
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Yes, I know the answer. :smile:

The problem I came up with doesn't require anything advanced. But we need to assume something that ensures that Tx=0 implies that x=0. So we assume that T is injective, or equivalently, that ker T={0}. (Hence the chicken reference).

Now, is the derivative operator [itex]f\mapsto f'[/itex] injective? (I like to denote it by D. So Df=f').
 
  • #6
Deveno
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define T:V→V (we'll have to assume that all functions in V are differentiable, at least) by:

T(f) = f' = Df

show that T preserves linear independence iff ker(T) = {0}.

(ker(T) is the null space of T. it might help to think about constant functions, first).
 
  • #7
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Yes, I know the answer. :smile:

The problem I came up with doesn't require anything advanced. But we need to assume something that ensures that Tx=0 implies that x=0. So we assume that T is injective, or equivalently, that ker T={0}. (Hence the chicken reference).

Now, is the derivative operator [itex]f\mapsto f'[/itex] injective? (I like to denote it by D. So Df=f').
Well, It's obviously necessary that Tx=0 implies x=0, because if not, the set {Tv1,Tv2,...,Tvn} contains the zero vector and will not be linearly independent. so KerT={0} is indeed a necessary condition.
To show that It's sufficient, let's take an arbitrary linear combination of Tvi's (i=1,...,n). If c1Tv1+c2Tv2+...+cnTvn=0 then we can write T(c1v1+c2v2+...+cnvn)=0. Hence, c1v1+c2v2+...+cnvn is in KerT and since KerT={0} we conclude that c1v1+c2v2+...+cnvn=0. but we had assumed that vi's were linearly independent, so that implies all ci's are zero.
according to a theorem in calculus and real analysis, if Df=0 then f is a constant function. we conclude that the kernel of D consists of all constant functions.
so D as a linear operator is not generally injective, but for all non-constant functions, can we say that D preserves linear independence?

so if T is an invertible operator, we surely can say that it preserves linear independence. 'cuz it's kernel is trivial. right?

define T:V→V (we'll have to assume that all functions in V are differentiable, at least) by:

T(f) = f' = Df

show that T preserves linear independence iff ker(T) = {0}.

(ker(T) is the null space of T. it might help to think about constant functions, first).
Yes. Thanks for the help. By the way, what's the story of the colonel and fried chicken in linear algebra? haven't figured that out yet. lol.
 
  • #8
Fredrik
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Looks like you have understood what I was going for. The set {f,g} where f is a constant function and g is a first-degree polynomial, is a counterexample to the original hypothesis (because f' and g' are both constants). But if we start with a set without any constant members, then as you observed, linear independence will be preserved.

People in the USA pronounce "Colonel" and "kernel" the same. The KFC franchise, originally named Kentucky Fried Chicken, was founded by Colonel Sanders.
 
  • #9
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Thanks for the help guys.
 

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