I have got the following task:
(1) For each $n \in \mathbb{N}$, specify matrices $A, B \in \mathbb{R}^{n \times n}$ for which $A B \neq B A$ is true.
(2) Determine $$ M:=\left\{A \in \mathbb{R}^{2 \times 2}: A B=B A \quad \forall B \in \mathbb{R}^{2 \times 2}\right\} $$ thus the matrices $A \in \mathbb{R}^{2 \times 2}$, which commute with all matrices $B \in \mathbb{R}^{2 \times 2}$.
Could someone please help me with this or give me an approach? That would be very helpful. Thanks.
(1) Take a look at matrices where one entry is $1$ and all other are $0$. For $n=3$ take e.g. $A=(a_{i,j})$ with $a_{1,2} = 1$ and $a_{i,j} = 0$ for $(i,j)\neq (1,2)$. Then take $B=(b_{i,j})$ with $b_{2,1}=1$ and $b_{i,j}=0$ for $(i,j)\neq (2,1)$.
Multiplying those matrices results in $$ (AB)_{i,j} = \begin{cases} 1, \text{for } (i,j) = (1,1) \\ 0 \text{ otherwise} \end{cases}, (BA)_{i,j} = \begin{cases} 1, \text{for } (i,j) = (2,2) \\ 0 \text{ otherwise} \end{cases} $$
Thus we have $AB \neq BA$. Can you generalize that for general $n$? Note that $AB=BA$ for all $A,B\in\mathbb{R}^{1\times 1}$, so one can specify those matrices only for $n>1$.
(2) You can write matrices $A,B\in\mathbb{R}^{2\times 2}$ as $A = \left( \begin{array} & a & b \\ c & d \end{array} \right), B = \left( \begin{array} & e & f \\ g & h \end{array} \right)$. Now you can multiply those matrices with general values and compare each entry: $$ AB = \left(\begin{array}&ae+bg & af+bh \\ ce+dg & cf+dh \end{array}\right) \overset{!}{=} \left(\begin{array}&ea+fc & eb+fd \\ ga+hc & gb+hd\end{array}\right) = BA $$
Now you're searching for all matrices that satisfy above equation for all matrices $B$, so for arbitrary $e,f,g,h$.
In detail: $(AB)_{1,1}=(BA)_{1,1}$ gives us $ae+bg = ea+fc$ which is equivalent to $bg=fc$. To make this equation hold for arbitrary $f,g$ we need $b=c=0$. The entry $(1,2)$ yields (using $b=0$) $af=df$ which means that we need $a=d$ to make this true for arbitrary $f$. Looking at the other two entries doesn't give us new conditions, this means we get
$$ M = Z(\mathbb{R}^{2\times 2}) = \mathbb{R}\cdot I_2 = \left\{\left(\begin{array} & \lambda & 0 \\ 0 & \lambda \end{array} \right):\lambda \in \mathbb{R}\right\} $$
so the center of $\mathbb{R}^{2 \times 2}$ are all multiples of the identity matrix. ($Z(.)$ denotes the center, $I_2$ the identity matrix of size $2\times2$)
For (1), just note that any diagonal matrix with distinct entries commutes only with diagonal matrices. So, it won't commute with a non-diagonal matrix.
For (2), I am pasting my solution for the generalized problem.
$\underline{Claim}$: $Z(GL_n(\mathbb{R}))=\{\lambda I_n \mid \lambda \in \mathbb{R}\backslash \{0\}\}$, for $n \geq 2$.
Here, $Z(G)$ denotes the center of the group $G$. Basically, the set which contains elements that commute with every other element.
Proof. It is obvious that $\lambda I_n \in Z(GL_n(\mathbb{R}))$, as for any $M \in GL_n(\mathbb{R})$, $\lambda I_nM=\lambda M=M\lambda I_n$. Suppose, $S \in Z(GL_n(\mathbb{R}))$. Then, $\forall M \in GL_n(\mathbb{R})$, $SM=MS$. Choose $M$ to be a diagonal matrix with all different non-zero real entries, say $\lambda_1,\dots,\lambda_n$. Then, using $SM=MS$, $(SM)_{ij}=(MS)_{ij}$. Since $(AB)_{ij}=\sum\limits_{k=1}^na_{ik}b_{kj}$, we have $\sum\limits_{k=1}^ns_{ik}m_{kj}=\sum\limits_{k=1}^ns_{kj}m_{ik}$. Since $M$ is diagonal, $m_{ab}=0$, if $a \neq b$. Thus, $\lambda_i s_{ij}=\lambda_j s_{ij}$. For $i \neq j$, $\lambda_i \neq \lambda_j \implies s_{ij}=0$. So, $S$ should be a diagonal matrix. Let $s_i=s_{ii}, 1 \leq i \leq n$. Can you proceed to show that $S$ should be a multiple of $I_n$, perhaps by choosing a suitable $M$? Note that even the zero matrix commutes with all matrices.
