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 $A^2=A$
 A: 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 $A_1,A_2\in M_{22}$.
A: Try the identity matrix: $I^2=I$, so $I\in U$. What can you say about $2I$?
A: 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.
A: 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?
