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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.

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Lion
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    The rank is the dimension of the range. Since your transformation is onto, the rank is the same as the dimension of $W$. Also, the nullity is the dimension of the kernel. – J126 Jul 12 '14 at 01:11
  • @JoeJohnson126 OK, now I get it. Thank you for your answer! – Lion Jul 12 '14 at 01:21
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    Oh. I didn't see that. It should be $\dim(V)$, not $\dim(T)$. Linear transformations don't have dimensions. – J126 Jul 12 '14 at 01:21
  • @JoeJohnson126 OK, thank you again. By the way, I also confuse that what is the definition of dimension of matrix (or vector)? Is that the number of components included by matrix or the rank of the matrix? – Lion Jul 12 '14 at 01:26
  • Yes. So an $n\times m$ matrix is of dimension $nm$. An $n\times 1$ column vector has dimension $n$. – J126 Jul 12 '14 at 12:09

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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 $\{v\in V : T(v) = \vec{0}\}$.

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