2
$\begingroup$

I have a question regarding vector subspaces: show $U=\{A\in M_{22} \mid A^2=A\}$ is not a subspace of $M_{22}$.

I have said: let $A={(a_1, a_2, a_3, a_4) = \begin{pmatrix} a_1 & a_2 \\ a_3 & a_4\\ \end{pmatrix} \in R}$

Is this the correct way to proceed? If I check it under scalar multiplication I end up with a $2\times 2$ matrix with elements $ca_1$, $ca_2$, $ca_3$ and $ca_4$. (I can't figure out how to embed a matrix within a line sorry).

I am unsure how to relate the caveat $A^2=A$

$\endgroup$
6
  • $\begingroup$ If you want to check whether it is a subspace, you don't need to start from that. You can just use matrix operations. $\endgroup$ – KittyL Aug 2 '15 at 9:37
  • $\begingroup$ So you want to prove that the set $U$ is a vector subspace of $M_{22}$? $\endgroup$ – Prahlad Vaidyanathan Aug 2 '15 at 9:38
  • $\begingroup$ @Prahlad: Probably he's being asked to determine whether $U$ is a subspace and justify his answer ... $\endgroup$ – hmakholm left over Monica Aug 2 '15 at 9:40
  • $\begingroup$ @KittyL can you explain in more detail please? $\endgroup$ – Angie01 Aug 2 '15 at 9:42
  • $\begingroup$ @PrahladVaidyanathan I have edited the question. $\endgroup$ – Angie01 Aug 2 '15 at 9:42
1
$\begingroup$

For example, to check whether it is close under scalar multiplication, you should check whether the matrix $cA$ also satisfies the definition of $U$, i.e., whether

$$(cA)^2=cA$$ given $c$ a scalar.

Matrix operation gives $(cA)^2=c^2A^2=c^2A$ since $A^2=A$. Does this satisfy the definition of $U$?

Similarly, to check whether it is close under addition, you need to check whether $$(A_1+A_2)^2=(A_1+A_2)$$ given $A_1,A_2\in M_{22}$.

$\endgroup$
1
$\begingroup$

Try the identity matrix: $I^2=I$, so $I\in U$. What can you say about $2I$?

$\endgroup$
1
$\begingroup$

There is an easy example of a matrix with this property: the identity matrix $$\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$$

You can check that scalar multiplication fails by using any scalar that is not $\pm 1$ and the identity matrix. (Where do the scalars come from?)

But I want to address another part of your question: how to come up with other matrices with this property.

To come up with other examples, square the matrix $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$:

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix}= \begin{pmatrix} a^2+bc & ab+bd \\ ac+cd & bc+d^2 \end{pmatrix} $$

Since you want $A^2=A$, you want the following equations to be satisfied:

$$ a=a^2+bc \\ b=ab+bd \\ c=ac+cd \\ d=bc+d^2 $$

The easiest way to generate such a matrix is to use $1$s and $0$s:

For example, if we let $a=0$ and $b=0$, then we just need $$c=cd \\ d=d^2$$

And for these to be true, we can take $c=-1, d=1$ (we could have also taken $c=1$)

Let's check: $$B^2=\begin{pmatrix} 0 & 0 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} 0 & 0 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ -1 & 1 \end{pmatrix}=B$$

You can check that even though $B, I \in U, (B+I)^2 \neq B+I$, so $B+I \not\in U$. (So $U$ is not closed with respect to addition; but checking scalar multiplication is much easier.)

For some other matrices of $U$, take $a=1, b=c=d=0$, or $a=b=1, c=d=0$. There are many more.

$\endgroup$
2
  • $\begingroup$ Thanks @coldnumber. From the second part of your answer, it looks to me like you are showing that $B^2=B$. Are you just showing that there are cases that scalar multiplication holds true? It just confused me a bit is all. $\endgroup$ – Angie01 Aug 3 '15 at 0:33
  • $\begingroup$ Showing that $B^2=B$ is equivalent to showing that $B \in U$. In that part I'm not showing anything about scalar multiplication or addition, just giving some examples of matrices of $U$. $\endgroup$ – coldnumber Aug 3 '15 at 0:35
0
$\begingroup$

So is the following correct:

Check under scalar multiplication:

Matrix operations give $(cA)^2=c^2A^2=c^2A$ since $A^2=A$.

Check by letting $A=\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$

$kA=2\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$ where k=2 and is an element of the set of real numbers

$kA=\begin{pmatrix} 2 & 0 \\ 0 & 2\end{pmatrix}$ which is not a vector in U as the rule that must be satisfied is $k^2A=kA$ but

$k^2A=\begin{pmatrix} 4 & 0 \\ 0 & 4\end{pmatrix} \neq kA$

If this is correct, is it a requirement to use numbers to disprove the law does not hold?

$\endgroup$
2
  • $\begingroup$ @coldnumber is this correct? $\endgroup$ – Angie01 Aug 3 '15 at 2:07
  • $\begingroup$ @KittyL is this correct? $\endgroup$ – Angie01 Aug 3 '15 at 2:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.