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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 $X = \begin{pmatrix} \lambda & 0\\ 0 & \lambda \end{pmatrix}$ 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 $Y = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}$ I think.

But that's not enough to prove it right? : /

kathelk
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  • Show that if $X$ commutes with a basis for $\mathbb{R}^{2\times 2}$ then it commutes with all of the space. Afterward, find your favorite basis and check what $X$ need to satisfy in order to commute with each of its elements. – Ofir Jan 10 '17 at 07:52
  • $XY=YX$ is supposed to hold for all possible $Y$ simultaneously. – Arthur Jan 10 '17 at 07:53

4 Answers4

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Expanding the comment of Ofir: $$ \begin{pmatrix} a & 0\\ c & 0 \end{pmatrix} = \begin{pmatrix} a & b\\ c & d \end{pmatrix} \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} \begin{pmatrix} a & b\\ c & d \end{pmatrix} = \begin{pmatrix} a & b\\ 0 & 0 \end{pmatrix} $$ implies $b = c = 0$. $$ \begin{pmatrix} 0 & a\\ 0 & c \end{pmatrix} = \begin{pmatrix} a & b\\ c & d \end{pmatrix} \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} \begin{pmatrix} a & b\\ c & d \end{pmatrix} = \begin{pmatrix} c & d\\ 0 & 0 \end{pmatrix} $$ implies $a = d$.

Martín-Blas Pérez Pinilla
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Be $X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$. Now choose $Y=\begin{pmatrix}0&1\\1&0\end{pmatrix}$. Then you have $$XY = \begin{pmatrix}b&a\\d&c\end{pmatrix} \stackrel!= YX=\begin{pmatrix}c&d\\a&b\end{pmatrix}$$ Therefore we get $a=d$ and $b=c$, that is, $$X=\begin{pmatrix}a&b\\b&a\end{pmatrix}$$ Now choose $Y=\begin{pmatrix}1&0\\0&0\end{pmatrix}$. Then you have $$XY = \begin{pmatrix}a&0\\b&0\end{pmatrix} \stackrel!= YX = \begin{pmatrix}a&b\\0&0\end{pmatrix}$$ Clearly this is only possible if $b=0$. Therefore we now have $$X=\begin{pmatrix}a&0\\0&a\end{pmatrix}$$ It is easy to check that this matrix indeed commutes with all other matrices, therefore we've found the general form.

celtschk
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$A,B,C,D$ can be any number you like. $$\begin{bmatrix} \lambda &0 \\ 0& \lambda \end{bmatrix}\begin{bmatrix} A&B \\ C&D \end{bmatrix}=\begin{bmatrix} A\lambda &B\lambda \\ C\lambda & D\lambda \end{bmatrix}=\begin{bmatrix} A&B \\ C&D \end{bmatrix}\begin{bmatrix} \lambda &0 \\ 0& \lambda \end{bmatrix}$$ hence for all $Y \in \mathbb{R}^{2x2}$, it satisfies the relation $X \cdot Y = Y \cdot X$.

Renascence_5.
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Since $X = \begin{pmatrix} \lambda & 0\\ 0 & \lambda \end{pmatrix}$ and $XY = YX$ then define $Y=\begin{pmatrix} \ y_1 & y_2\\ y_3 & y_4 \end{pmatrix}$

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

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.

kub0x
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