# Calculating the nullity and rank of a linear map between complex matricies

A question on an assignment asks to find the rank and nullity of the linear map B, from the set of all matrices of size $2\times 3$ with complex entries, given by;

$$B\begin{pmatrix} a & b & c\\ d & e & f \end{pmatrix} =\begin{pmatrix} a & 0 & 2c\\ 0 & e-d & f \end{pmatrix}$$

Am I right in thinking that:

$$\mathrm{ker}(B) = \begin{pmatrix} 0 & x & 0\\ y & y & 0 \end{pmatrix}$$

And that therefore nullity $=$ dim(ker($B$)) $= 2$

And using the Rank-Nullity Theorem,

$$\mathrm{rank}(B) = \dim( \begin{pmatrix} a & b & c\\ d & e & f \end{pmatrix} )-2 = 6-2 = 4$$

• size of $B$ is $2\times 2$ so your rank computation is a little too much... Dimension is not the sum of length and width. Good thinking about the kernel though.
– user13838
Nov 8, 2011 at 1:11
• @percusse the map is not left multiplication by a $2 \times 2$ matrix $B$, but an "abstract" linear transformation from the space of $2\times 3$ matrices to themselves. The dimension of both the domain and codomain is 6. Nov 8, 2011 at 1:15
• @BillCook Bah, the parentheses of B got me there. Sorry for the ignorance.
– user13838
Nov 8, 2011 at 1:19
• @percusse that's ok. I've made several mistakes today that are at least that silly. :) Nov 8, 2011 at 1:23

$$B\begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \qquad \Longrightarrow \qquad \begin{pmatrix} a & 0 & 2c \\ 0 & e-d & f \end{pmatrix}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

Therefore, $a=c=f=0$ and $e=d$. Thus $b,d$ are free to take on any real value. So, yes.

$$\mathrm{ker}(B) = \left\{ b\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} + d\begin{pmatrix} 0 & 0 & 0 \\ 1 & 1 & 0 \end{pmatrix} \;{\Huge |}\; b,d \in \mathbb{R} \right\}$$

Clearly the kernel is 2-dimensional. The domain being 6-dimensional means that the rank must be $6-2=4$. You are correct.

You can also see that the rank is 4 directly. Notice that

$$\left\{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right\}$$

are all elements in the range of $B$ [set $a,\dots,f=0$ with the exception of (1) $a=1$, (2) $c=1/2$, (3) $e=1$, and (4) $f=1$.]

Clearly these elements span the range and are independent. Thus the dimension of the range is 4 (so rank=4).

• Thanks for the confirmation Bill! Also, your extra comment on the rank is very helpful.
– Sam
Nov 8, 2011 at 1:15