Show that if matrix $X$ satisfies the relation $X \cdot Y = Y \cdot X$ it has a specific form Show that: If $X \in \mathbb{R}^{2x2}$ is a matrix which satisfies the relation $X \cdot Y = Y \cdot X$ for all $Y \in \mathbb{R}^{2x2}$, then $X$ has the form $X = \begin{pmatrix}
\lambda &amp; 0\\ 
0 &amp; \lambda
\end{pmatrix}$ with $\lambda \in \mathbb{R}$.



I don't know how to solve this task? Maybe we are allowed to set specific matrix for $Y$? But I'm not sure because it's saying "for all" so we must stay general?

Else I would say we can satisfy this relation when we set $Y = \begin{pmatrix}
1 &amp; 0\\ 
0 &amp; 0
\end{pmatrix}$ I think. 

But that's not enough to prove it right? : /
Expanding the comment of Ofir:
$$
\begin{pmatrix}
a &amp; 0\\ 
c &amp; 0
\end{pmatrix} =
\begin{pmatrix}
a &amp; b\\ 
c &amp; d
\end{pmatrix}
\begin{pmatrix}
1 &amp; 0\\ 
0 &amp; 0
\end{pmatrix} =
\begin{pmatrix}
1 &amp; 0\\ 
0 &amp; 0
\end{pmatrix}
\begin{pmatrix}
a &amp; b\\ 
c &amp; d
\end{pmatrix} =
\begin{pmatrix}
a &amp; b\\ 
0 &amp; 0
\end{pmatrix}
$$
implies $b = c = 0$.
$$
\begin{pmatrix}
0 &amp; a\\ 
0 &amp; c
\end{pmatrix} =
\begin{pmatrix}
a &amp; b\\ 
c &amp; d
\end{pmatrix}
\begin{pmatrix}
0 &amp; 1\\ 
0 &amp; 0
\end{pmatrix} =
\begin{pmatrix}
0 &amp; 1\\ 
0 &amp; 0
\end{pmatrix}
\begin{pmatrix}
a &amp; b\\ 
c &amp; d
\end{pmatrix} =
\begin{pmatrix}
c &amp; d\\ 
0 &amp; 0
\end{pmatrix}
$$
implies $a = d$.
Since $X = \begin{pmatrix}
\lambda &amp; 0\\ 
0 &amp; \lambda
\end{pmatrix}$ and $XY = YX$ then define $Y=\begin{pmatrix}
\ y_1 &amp; y_2\\ 
y_3 &amp; y_4
\end{pmatrix}$

Then we have $$\begin{pmatrix}
\lambda &amp; 0\\ 
0 &amp; \lambda
\end{pmatrix}\cdot \begin{pmatrix}
\ y_1 &amp; y_2\\ 
y_3 &amp; y_4
\end{pmatrix} = \begin{pmatrix}
\ y_1 &amp; y_2\\ 
y_3 &amp; y_4
\end{pmatrix}\cdot \begin{pmatrix}
\lambda &amp; 0\\ 
0 &amp; \lambda
\end{pmatrix}$$

This gives us the following system of linear equations:

$$\begin{cases}\lambda y_1 = \lambda y_1 \\ \lambda y_2 = \lambda y_2 \\ \lambda y_3 = \lambda y_3 \\ \lambda y_4 = \lambda y_4\end{cases}$$ 

This system of linear equations tells you that $X$ and $Y$ commute.