Let $M\leqslant\mathbb R^{10},\dim M=4$ and
$$\mathcal L=\{T\in L\left(\mathbb R^{10},\mathbb R^8\right):M\subseteq Ker T\}$$Prove $\mathcal L$ is a subspace of $L\left(\mathbb R^{10},\mathbb R^8\right)$ and find its dimension.
My attempt:
We have already discussed why the statement $M\subset Ker(T)$ is false because of the following case:
Let $e=\{e_1,\ldots,e_{10}\}$ be some basis for $\mathbb R^{10}$ and let $A, B\in\mathcal L$ s.t. $Ker A\ne Ker B$, i.e., $$Ae_i=0\;\&\; Be_i\ne 0$$
$\implies (\alpha A+\beta B)(e_i)=\underbrace{\alpha Ae_i}_{=0}+\underbrace{\beta Be_i}_{\ne 0}\notin\mathcal L,\alpha,\beta\in\mathbb R$
If $\mathcal L$ isn't closed under addition and scalar multiplication, it can't be a vector space, and, therefore, a subspace.
$(*)$ Note: At this point, I'm facing a problem similar to one in my prior post.
Since $M$ doesn't have to be a proper subset of $Ker T,\;\dim M=Ker T=4$.
Then I concluded $Ker A= Ker B\implies\operatorname{range}A=\operatorname{range}B=6,\forall A,B\in\mathcal L\lt L$ (in order to: $\alpha A+\beta B\in\mathcal L$).
My first thought of finding $\dim\mathcal L$ was comparing it with $\dim L\left(\mathbb R^{10},Im T\right),$ but it's inefficient, but I know fore sure $\dim L\left(\mathbb R^{10},\mathbb R^8\right)=80$
May I ask for advice on solving this task? Thank you in advance!
Best Answer
Write $\Bbb R^{10}=M\oplus M^\prime$ for some choice of complementary space $M^\prime$ of dimension $6$. Then any linear function $T:\Bbb R^{10}\rightarrow\Bbb R^{8}$ decomposes uniquely as $$ T_1\oplus T_2:M\oplus M^\prime\rightarrow\Bbb R^{8} $$ where $T_1$ and $T_2$ are nothing but the restriction of $T$ to $M$ and $M^\prime$ respectively (this follows from the elementary fact that a linear map between vector spaces is completely determined by its values on a basis of the domain).
It should be clear now that your space $\cal L$ consists of the linear maps $T$ such that $T_1=0$.
Thus $\cal L$ is isomorphic to ${\rm Hom}(M^\prime,\Bbb R^8)$ and thus its dimension is $48$.