The problem is:
Let $V$ be the real vector space of all real $2×3$ matrices and let $W$ be the real vector space of all real $4×1$ column vectors. If T is a linear transformation from $V$ onto $W$. What is the dimension of the subspace $\{v\in V:T(v)=0\}$?
I see this problem in GRE practice book. I search this problem in Mathematics exchange and there is an answer.
@André Nicolas gives an answer that since $\dim(V)=6$ and $\dim(W)=4$, then $\ker(T)=6-4=2$. I am confused that why $\ker(T)=\dim(V)-\dim(W)$? I just know the Rank plus nullity theorem that $nullity(T)+rank(T)=\dim(T)$. So how his formula is true? Thanks in advance.
The key is that the linear transformation is "onto" $W$. So the dimension of the image is the dimension of $W$, namely 4. Thus by the rank-nullity theorem, dim(kernel) + dim(image) = dim(domain), we get dim(kernel) + 4 = 6. So the dimension of the kernel is 2. Of course the kernel is precisely the subspace $\{v\in V : T(v) = \vec{0}\}$.