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How can I construct a basis and find the dimension of the following subspace?

$$U = \{ M \in \textsf{M}_{2\times 2} :\, (\forall J \in \textsf{M}_{2\times 2} )( MJ=JM^t ) \}$$

My original intuition was to let $ M = \left[ \matrix{a & b \\ c & d} \right]$ and $J = \left[ \matrix{e & f \\ g & h} \right]$. Then I would find $MJ$, $M^{T}$, and $JM^{T}$ and then equate the two together: $MJ = JM^{T}$. From there I'm not quite sure how to proceed and find a basis and dimension.

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  • $\begingroup$ In fact (assuming the $T$ is a transpose), $U = \{0\}$ so the basis will be the empty set. $\endgroup$
    – Ben Grossmann
    Commented Sep 22, 2019 at 20:37
  • $\begingroup$ Isn’t the identity matrix in $U$, @Omnomnomnom? $\endgroup$
    – J. W. Tanner
    Commented Sep 22, 2019 at 20:42
  • $\begingroup$ @J.W.Tanner whoops, I missed the $J$ and the $M$ in the definition $\endgroup$
    – Ben Grossmann
    Commented Sep 22, 2019 at 20:54

1 Answer 1

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I assume that $\textsf{M}_{2\times 2}$ denotes the ring of $2\times 2$ matrices over a field $F$ (for instance, we can pick $F=\Bbb R$). Let $E_{12}=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}\in \textsf{M}_{2\times 2}$ and $M=\begin{pmatrix} m_{11} & m_{12}\\ m_{21} & m_{22}\end{pmatrix}\in \textsf{M}_{2\times 2}$. Then $ME_{12}=\begin{pmatrix} 0 & m_{11}\\ 0 & m_{21}\end{pmatrix}$ and $ E_{21}M^t=\begin{pmatrix} m_{12} & m_{22}\\ 0 & 0\end{pmatrix}$. Thus $M E_{21}=E_{21}M^t$ iff $m_{12}=m_{21}=0$ and $m_{11}=m_{22}$, thus iff $M=m_{11}I$. Conversely, if $M=mI$ for some $m\in F$ then $MJ=mIJ=mJ=Jm=JM^t$ for each $J\in \textsf{M}_{2\times 2}$. Thus $U=\{mI: m\in F\}$, which is one-dimensional linear space over $F$ with a basis $\{I\}$.

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