A linear map that is multiplication by a matrix The problem statement, all given variables and data
Let $T$ be multiplication by matrix $A$:
$$A=
        \begin{bmatrix}
        1 & -1 & 3 \\
        5 & 6 & -4 \\
        7 & 4 & 2 \\
        \end{bmatrix}
$$
Find the range, kernel, rank and nullity of T.
Attempt at a solution
I can find all of these values except for A. What does the question mean when T is a multiplication by matrix A? How do I get range/kernel etc. of T.. I know how to get them for A.
Thanks.
 A: Any matrix $A$ represents a linear function $T$, or more precisely $T_A$, defined by multiplication, that is, by $T(x)=Ax$. Now it makes sense to talk about the range and kernel of $T$. Rank and nullity are simply the dimension of range and kernel of $T$, respectively.
Regarding your question, note that, by definition or a little argument, the rank of any matrix $A$, agrees with the rank of the linear map $T_A$, and the nullity of $A$ is the same as the nullity of $T_A$.
For instance,
 $$\ker(T)=\{x\in \mathbb{R}^3\mid T(x)=\mathbf{0}\}=\{x\in \mathbb{R}^3\mid Ax=\mathbf{0}\}.$$
Thus, to find the kernel of $T$, you need to solve the homogeneous system $Ax=\mathbf{0}$.
A: The set of linear maps $T:\mathbb{R^n} \to \mathbb{R^m}$ is in 1-1 correspondence with $\mathbb{M}_{m,n}$, the set of $m \times n$ matrices, so it makes sense to talk about the kernel, range etc. of a transformation $T$ by the kernel, range etc. of its corresponding matrix.
In short, everything you have to find for the transformation $T$, is just what you have to find for the matrix $A$.


*

*$$\ker(T)=\ker(A)=\{ \vec x \in \mathbb{R}^3\ : A \vec x=\vec 0\}$$

*$$\text{null}(T)=\text{null}(A)=\dim[\ker(A)]$$

*The range of $T$ is spanned by the columns of $A$.

*$\text{rank}(T)=\text{rank}(A)=\dim[\text{range}(A)]=\text{number of linearly-independent columns of }A$

Note: the nullity of $A$ (or nullity of $T$, if you like) is the number of free variables in a row-echelon form of $A$.

Shortcut: if you have the nullity and you can't be bothered to find the rank (or vice-versa), you can use the rank-nullity theorem, which states that $$\text{rank}(A)+\text{null}(A)=n$$ (in this case, $n=\text{number of columns of }A, =3$).
A: It is a question of abstraction: $A$ is the matrix, the scheme, while $T$ is the transformation that multiplication with the matrix results in. $A$ and $T$ is not the same, specially since different matrices with respect to other basis can result in the same transformation.
