0
$\begingroup$

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,

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

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – user13838
    Nov 8, 2011 at 1:11
  • $\begingroup$ @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. $\endgroup$
    – Bill Cook
    Nov 8, 2011 at 1:15
  • $\begingroup$ @BillCook Bah, the parentheses of B got me there. Sorry for the ignorance. $\endgroup$
    – user13838
    Nov 8, 2011 at 1:19
  • $\begingroup$ @percusse that's ok. I've made several mistakes today that are at least that silly. :) $\endgroup$
    – Bill Cook
    Nov 8, 2011 at 1:23

1 Answer 1

1
$\begingroup$

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

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).

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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .