# Vector Subspace

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 $A^2=A$

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

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 $A_1,A_2\in M_{22}$.

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

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.

• 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. – Angie01 Aug 3 '15 at 0:33
• 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 $U$. – coldnumber Aug 3 '15 at 0:35

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?

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