I have attached the question above in picture. It says the answer is $D$.
Now, I do know that Rank-Nullity theorem is being applied here. Let me explain how I have worked it out and finally leading to my doubt, so please bear with me..(sorry for my poor English)
So, the first thing they said that there is a matrix $T$ which is of order $m\times n$, Now they have already defined the set $V$ in this. Let us suppose there exists a linear transformation such that
$\{S: M_{n\times p}\to M_{m\times p}; S(x) = TX\}$
Now, in the question, they said $V$ is a set such that $TX= O$ so, meaning The set $V=\ker(S)$
so therefore $\dim V = N(S)$ (where $N$ is nullity of set $S$)
NOW, this is where I am stuck. Where on Earth are they getting $P$ multiplied?
Yes! I can see the $P$, column, but I have referred 2-3 reference books in linear Algebra, particularly the one by Kenneth Hoffman & Ray Kunze. Aside from this, I also referred to Introduction to Algebra by Serge Lang and I scan through the Rank Nullity theorem and every time they talk about Rank Nullity theorem they always refer to Linear Transformation such as $T: \Bbb R^n\to\Bbb R^m$ where the formula is $\operatorname{Rank}(T) + \operatorname{Nullity}(T) = n$, where on earth is another number being Multiplied to this whole thing?
Best Answer
Here is a start: think of $T \in M_{m,n}(\mathbb R)$ as a linear transformation $T: \mathbb R^n \rightarrow \mathbb R^m$. By the rank-nullity theorem, you know the nullity of $T$ is $n - \text{rank}(T) = d$. This means that the nullspace of $T$, which is the subspace of vectors $Y \in \mathbb R^n$ such that $TY = 0$, is spanned by $d$ linearly independent vectors $Y_1, \cdots, Y_d$.
Coming back to your problem: what type of matrices $X \in M_{n,p}(\mathbb R)$ satisfy $TX = 0$? You can show $X = [C_1 | \cdots | C_p]$ is a matrix whose columns $C_i$ are elements of the nullspace of $T$. You might want to exhibit an explicit basis for $V$ in terms of the basis $\{Y_1, \cdots, Y_d\}$ of $T$, but the idea is that each column "is" $d$-dimensional and there are $p$ columns, hence the dimension of $V$ is $pd = p(n - \text{rank}(T))$.