Given two linear transformations L and K, show ##K = \lambda L## holds

[JD_PM]

TL;DR

Exercise on linear transformations and dimensional theorem.

Let $V$ be a real vectorspace of finite dimension $n$. Let $L, K:V \rightarrow \Re$ be linear transformations so that $ker(L) \subset ker(K)$. Then there's a parameter $\lambda \in \Re$ so that $K=\lambda L$

a) Show that $K=\lambda L$ holds when $K=0$.

b) Suppose that $K \neq 0$. Compute $dim(kerK)$ and show that $dim(kerK)=dim(kerL)$.

c) Show that $K=\lambda L$ holds when $K \neq 0$.a) Here I think we just have to set $\lambda=0$and then the equation holds. However I am not that convinced (it seems too easy).

b) To compute $dim(kerK)$ I used the dimension theorem for the linear transformation $K:V \rightarrow \Re$

$$dim(V) = dim(kerK) + dim(Imk)$$

We are given that $dim(V)=n$ and the co-domain of both linear transformations is $\Re$. Thus I get

$$dim(kerK)=n-1$$

To compute $dim(kerL)$ I used the dimension theorem for the linear transformation $L:V \rightarrow \Re$

$$dim(V) = dim(kerL) + dim(ImL)$$

We are given that $dim(V)=n$ and the co-domain of both linear transformations is $\Re$. Thus I get

$$dim(kerL)=n-1$$

Then indeed we get $$dim(kerL)=dim(kerK)$$ Is this OK?

c) I do not really know how to approach this section. Could you please give a hint?
Any help is appreciated.

Thanks.

 

[Infrared]

Let $v$ be a vector not in $\text{Ker}(K)$ and choose $\lambda$ so that $K(v)=\lambda L(v).$ You can do this because $\text{Ker}(L)\subset\text{Kerl}(K)$, so $L(v)\neq 0$. What can you say about the kernel of $K-\lambda L$?

 

[wrobel]

Let $X,Y,Z$ be vector spaces perhaps infinite dimensional. And let $A:X\to Y,\quad B:X\to Z$ be linear operators; and $A(X)=Y$.

THEOREM. Assume that $\ker A\subseteq \ker B$. Then there exists a unique linear operator $\Lambda:Y\to Z$ such that $B=\Lambda A$.

Indeed, For any element $y\in Y$ there is an element $x\in X$ such that $Ax=y$. By definition put
$\Lambda y:=Bx$. This definition is correct: if there exists another $x'\in X$ such that $Ax'=y$ then $x-x'\in\ker A\subseteq\ker B$ thus $Bx=Bx'$. qed

 

[WWGD]

Let $X,Y,Z$ be vector spaces perhaps infinite dimensional. And let $A:X\to Y,\quad B:X\to Z$ be linear operators; and $A(X)=Y$.

THEOREM. Assume that $\ker A\subseteq \ker B$. Then there exists a unique linear operator $\Lambda:Y\to Z$ such that $B=\Lambda A$.

Indeed, For any element $y\in Y$ there is an element $x\in X$ such that $Ax=y$. By definition put
$\Lambda y:=Bx$. This definition is correct: if there exists another $x'\in X$ such that $Ax'=y$ then $x-x'\in\ker A\subseteq\ker B$ thus $Bx=Bx'$. qed

I don't see where we assumed A is surjective. Besides, please don't provide a full solution, we want to lead the OP to find it by themselves.

 

[wrobel]

I don't see where we assumed A is surjective

this assumption does not shrink generality: you can always take $A(X)$ on a role $Y$ and then extend $\Lambda$ from $A(X)$ to the whole space. But such an extension is not unique.

Besides, please don't provide a full solution, we want to lead the OP to find it by themselves.

It is a very special situation. There is a classical theorem and I believe that it would be better if the OP would know this classical theorem and its regular proof instead of inventing strange proofs of special cases of this classical theorem.

 

[WWGD]

this assumption does not shrink generality: you can always take $A(X)$ on a role $Y$ and then extend $\Lambda$ from $A(X)$ to the whole space. But such an extension is not unique.It is a very special situation. There is a classical theorem and I believe that it would be better if the OP would know this classical theorem and its regular proof instead of inventing strange proofs of special cases of this classical theorem.

Never mind, I realize in this case the map must be onto because the image is a vector space and R is 1-dimensional over itself.

 

[JD_PM]

What can you say about the kernel of $K-\lambda L$?

Based on the definition and assuming that $v$ is not in $ker(K-\lambda L)$ I get

$$K(v) \neq \lambda L(v)$$

But could you please explain why this is relevant?

 

[Infrared]

I didn't assume $v$ is not in the kernel of $K-\lambda L$. I assumed $v$ was not in the kernel of $K$.

 

[wrobel]

By the way, the standard generalization of the fact from the top is as follows. Let $f,f_1,\ldots,f_n:X\to\mathbb{R}$ be linear functions such that $\bigcap_{i=1}^n\ker f_i\subseteq\ker f.$ Then there exist constants $\lambda_1,\ldots,\lambda_n$ such that $f=\sum_{i=1}^n\lambda_i f_i$.

This fact follows from the above theorem. Indeed, let
$$Y=A(X)\subseteq\mathbb{R}^n,\quad Z=\mathbb{R},\quad B=f,\quad Ax=(f_1(x),\ldots,f_n(x))^T.$$
By the proved above theorem there exist a linear function $\Lambda: A(X)\to\mathbb{R}$ such that $B=\Lambda A$. If needed, $\Lambda$ can be extended to the whole $\mathbb{R}^n$ and thus $\Lambda A=(\lambda_1,\ldots,\lambda_n)(f_1,\ldots,f_n)^T$.

 

[mathwonk]

work on V/ker(L). then you are reduced to showing two linear maps R-->R are scalar multiples of each other, which is trivial.