Let $M\lt\mathbb R^{10},\;\dim M=4$. Find the dimension of $\mathcal L=\{T\in L\left(\mathbb R^{10},\mathbb R^8\right):M\subseteq Ker T\}$ 
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!
 A: 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 $48$.
A: I think maybe some of your confusion is notation. When it says $$\mathcal L=\{T\in L\left(\mathbb R^{10},\mathbb R^8\right):M\subseteq Ker T\}$$ that colon should be read as “such that”. In other words, $\mathcal{L}$ is the set of $8 \times 10$ matrices $T$ for which $M \subseteq \ker T$.
First, to show $\mathcal{L}$ is a vector space, let $S, T \in \mathcal{L}$. Then for any $m \in M$, we know $m \in \ker S, \ker T$, i.e. $Sm = Tm = \mathbf{0}$. Then $(S + T) m = \mathbf{0}$, so $M \subseteq \ker (S + T)$. Thus $S + T \in \mathcal{L}$. A similar argument will show that $\mathcal{L}$ is closed under scalar multiplication.
Now to calculate $\dim \mathcal{L}$, maybe one of the strongest methods is rank-nullify. To use this, we’d wanna find a linear operator $A : L(\mathbb{R}^{10}, \mathbb{R}^8) \to V$ such that $\ker A = \mathcal{L}$. One option I see is this. We essentially want an operator $A$ which tells us how a matrix $T \in L( \mathbb{R}^{10} , \mathbb{R}^8)$ acts on $M$. Let $f_1, f_2, f_3, f_4$ be a basis for $M$, and let $B$ be the matrix in $L( \mathbb{4} , \mathbb{R}^{10})$ whose columns are $f_1, f_2. f_3, f_4$. Define $A: L( \mathbb{R}^{10}, \mathbb{R}^8) \to L( \mathbb{R}^4, \mathbb{R}^8)$ by
$$AT = TB .$$
Then you can check that $\ker A = \mathcal{L}$, and from here use rank-nullity to find $\dim \mathcal{L}$.
A: First of all
$$
\begin{aligned}
\mathcal L
&=\{\ T\in L\left(\mathbb R^{10},\mathbb R^8\right)
\ :\ M\subseteq \operatorname{Ker} T\ \}
\\
&=\{\ T\in L\left(\mathbb R^{10},\mathbb R^8\right)
\ :\ T(M)=0\ \}
\\
\end{aligned}
$$
is a vector space. (Take for this $S,T$ to be two elements in this set, and let $a,b$ be two real scalars. Then $aS+bT$ is a linear map, and in each $v$ in $M$ we have $(aS+bT)(v)=aS(v)+bT(v)=a\cdot 0+b\cdot 0=0+0=0$, so $(aS+bT)(M)$ is the zero vector space $0$.
Now let $m_1,m_2,m_3,m_4$ be a basis of $M$. Extend it using vectors $n_1,n_2,n_3,n_4,n_5,n_6$ to a basis of $\Bbb R^{10}$ and denote by $N$ the vectorspace generated by these six added vectros. It has dimension $6$. Note that each $T$ in $\mathcal L$ is determined by its restriction on $N$. Conversely, each $T_1:N\to \Bbb R^8$ can be uniquely extended to a linear map in $L(\Bbb R^{10},\Bbb R^8)$ which vanishes on $M$. This implies
$$
\mathcal L\cong L(N,\Bbb R^8)\cong L(\Bbb R^6,\Bbb R^8)\ ,
$$
a space of dimension $6\cdot 8$.
