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.



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, $n=\text{number of columns of }A, =3$).

  • $\begingroup$ What is the point of this question then? They could have just asked me to get these values for A. It just confused me instead. $\endgroup$ – jem Aug 25 '14 at 9:51
  • $\begingroup$ @jem You're right. It's common to talk about the kernel, image etc. of a transformation, but every linear transformation $T(\vec x)$ can be represented $A \vec x $ for some suitable matrix $A$, so it doesn't really matter. $\endgroup$ – beep-boop Aug 25 '14 at 11:07

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 $Ax=\mathbf{0}$.


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.

  • $\begingroup$ Different matrices give rise to different multiplication transformations. Indeed, $A=B$ if and only if $T_A=T_B$ where $T_A(x)=Ax$ and similarly for $T_B$. However it is true that a transformation can have different matrix representations with respect to different bases, but that's a slightly different story. $\endgroup$ – EPS Aug 24 '14 at 22:11
  • $\begingroup$ @Sam: You are right, it takes a matrix and a base to determine a transformation. I thought jem just wanted to know the difference between transformations and matrices, but was able to do the caculations. $\endgroup$ – Lehs Aug 25 '14 at 5:42

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