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_{1}f

_{1}(x) + c

_{2}f

_{2}(x) +...+ c

_{n}f

_{n}(x)=0, then would the linear independence be preserved if we differentiate the equation with respect to x?

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- Thread starter AdrianZ
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Fredrik

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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).

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Deveno

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(by this, of course, i mean that the answer involves the colonel).

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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.

(by this, of course, i mean that the answer involves the colonel).

Would you be more precise?

- #5

Fredrik

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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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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).

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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 {Tv

To show that It's sufficient, let's take an arbitrary linear combination of Tv

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?

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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People in the USA pronounce "Colonel" and "kernel" the same. The KFC franchise, originally named Kentucky Fried Chicken, was founded by Colonel Sanders.

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Thanks for the help guys.

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