I have 3 proof questions from my book that I have tried and I would like to see if my solutions are valid and/or there is a simpler way to prove them.
Firstly, the notation $ker(L)$ means all $f$ such that $L(f)=0$. The kernel is called the zero kernel if $f=0$ is the only solution, otherwise it is called nontrivial.
- Prove that the linear operator $L$ has zero kernel iff $L$ is injective.
Attempt: Allow that $L$ has zero kernel. Then, $L(f)=0 \implies f=0$.
Rewrite as: $L(f) = 0*L(g) \implies L(f) = L(0*g) \implies L(f) = L(0)$ (done by linearity), and we know that $f=0$, so $L(f) = L(0) \implies f=0$, which is the definition of injective. The converse is pretty trivial. (ie Let $L(f)=L(0)$, since $L$ is injective, $f=0$.)
- Prove that $\lambda$ is an eigenvalue for $L$ iff $L-\lambda I$ has nontrivial kernel. Moreover, if $\lambda$ is an eigenvalue, then $ker(L-\lambda I)$ is the eigenspace belonging to $\lambda$.
Attempt: Assume $\lambda \neq 0$, and, that $ker(L)$ is not only the zero kernel. Then, $L(f) = \lambda f \implies (L-\lambda I)(f)=0$, where $I$ denotes identity function. From here, if we had assumed that only the zero kernel exists, then this also implies that $ker(L)$ is the zero kernel as well. In that case, then, we have a contradiction, since we assumed that $ker(L)$ was nontrivial. Therefore, the kernel must be nontrivial (This is where I'm not sure if my logic is sound). The converse is again trivial.
For the second part, I want to show that this is an eigenspace, that is, this kernel is the set of all eigenfuctions. So, I take some function $f$ and allow that $f \in ker(L-\lambda I)$, then, $f$ satisfies $L- \lambda I)(f) = 0 \implies L(f) = \lambda f$, which is the definition of the eigenfunction, so this does form an eigenspace.
- Let $L$ be a linear operator on $R^3$. Prove $\lambda$ is an eigenvalue for $L$ iff $det(L- \lambda I) = 0$.
In $R^3$ we have 3-tuples, so I assume $X \neq 0$, where $X$ denotes a 3x1 vector. $L(X)$ would then yield 3 different equations, $L(x_i) = \lambda x_i$, where $i = 1,2,3$ Then, $(L - \lambda I)(x_i) = 0$, and not all $x_i = 0$. At this point, I think that this should yield a 3x3 set of equations since $L$ is assumed to be a second order differential operator. Then, since we have assumed a nontrivial zero must exist, this set of equations is linearly dependent which immediately implies that $det(L- \lambda I) = 0$.
I tried to make this as concise as I could, sorry if it's a bit of a read!
Suppose $L$ is a Linear Operator.
Ques 1.
Remark: There is something wrong.
The definition of injective is given by $L(f)=L(g)$ iff. $f=g$ . But you didn't show.
Proof : If $kerL = {0}$ , still assume $f \not =g $ and $ L(f)=L(g)$ to get $L(f-g)\in kerL \Rightarrow a-b=0$
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Ques 2.
Remark: You have Misunderstanded the definition of eigenvalue.
Definition: $\lambda$ is a eigenvalue of L ,if for some vector $f\not = 0,L(f)=\lambda f$
Proof: It's easy to show $f\in ker(L-\lambda I)$ , hence $ker(L-\lambda I) \not = {0}$
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Ques 3.
Remark: Good!.In fact , it can be induced from Ques 2.
since $det(L)=0 $ is equaivalent to $kerL \not = {0}$ (A basic theorme in Linear Algebra)
The information of "second differential operator" is useless here.
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