For $T: M_{2,3} \to M_{3,2}$, defined by $T(A)=A^T$.

The $\text{range}(T)$ is the vector space of all $2\times 3$ matrices.

The $\ker(T)$ is the zero vector only since the entire domain is the range.

Therefore the $\text{nullity}(T)$ is $0$.

The $\text{rank}(T)$ is equal to the rank of $A$ which is the number of pivot entries in the rref of $A$. The rank of $A$ is also defined as the $\dim(C(A))$ which is equal to the number of vectors in the basis for $C(A)$. The basis for $C(A)$ can be found by reducing $A$ to rref to see where the pivot entries are and looking at the corresponding columns in $A$.

I apologize but although I am looking at the MathJax tutorial I cannot figure it out.
However, I am hoping for input on whether I am missing anything in my solution.

  • $\begingroup$ Thank you for the edit and edit attempts. $\endgroup$ – EcoMonkey Jul 26 '17 at 21:52
  • 2
    $\begingroup$ The rank of $T$ is the dimension of its range. $\endgroup$ – egreg Jul 26 '17 at 22:42

$C(A)$ only makes sense as a description of a linear transformation when that transformation is encoded with a matrix in the usual way. That is, if $A$ is an $m \times n$ matrix and $T_A$ is the transformation $T_A : \Bbb R^n \to \Bbb R^n$ given by $$ T_A(x) = Ax $$ Then $C(A) = \operatorname{range}(T)$, and $\operatorname{rank}(A) = \operatorname{rank}(T_A) = \dim(C(A)) = \dim(\operatorname{range}(T))$.

More generally, we define the rank of a linear operator $T$ by $\operatorname{rank}(T) = \dim \operatorname{range}(T)$. In our case, $\operatorname{rank}(T) = \dim(M_{3,2}) = 6$.

  • $\begingroup$ Well noted. Thanks for the correction. $\endgroup$ – EcoMonkey Jul 27 '17 at 0:31

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