Exercise 0.1.4(Centralizer of $M$)

Question

Find conditions on $p,q,r,s \in \mathbb R$ which determine precisely when $\begin{pmatrix} p & q \ r & s\end{pmatrix} \in \mathcal B$.

Abdullah Zubair · Accepted Answer

Let $Q = \begin{pmatrix} p & q \ r & s \end{pmatrix}$, where $p,q,r,s \in \mathbb R$. Then $QM = \begin{pmatrix}p & p + q \ r & r + s \end{pmatrix}$ and
	$MQ = \begin{pmatrix} p + r & q + s \ r & s\end{pmatrix}$. We have that $QM = MQ$ precisely when $r = 0$ and $p = s$. Therefore, any matrix of the form
	$\begin{pmatrix} p & q \ 0 & p \end{pmatrix}$ lies in $\mathcal B$.

Richard Ganaye · Answer

\begin{proof} 
Put $X = \begin{pmatrix} p & q \ r & s\end{pmatrix}$. Then
\begin{align*}
 X \in {\mathcal B} & \iff \begin{pmatrix} p & q \ r & s\end{pmatrix} \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}\begin{pmatrix} p & q \ r & s\end{pmatrix} \
 &\iff \begin{pmatrix} p & p+ q \ r & r+ s\end{pmatrix} = \begin{pmatrix} p+ r & q+s \ r & s\end{pmatrix}\
 &\iff
 \begin{cases}
 p = p+r\
 p+q = q+s\
 r+s = s
 \end{cases}\
 &\iff
 \begin{cases}
 r = 0\
 p = s.
 \end{cases}
\end{align*}
Therefore any matrix $X \in {\mathcal A}$ is in $\mathcal {B}$ if and only if 
$$X = \begin{pmatrix} p & q \ 0 & p\end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1\end{pmatrix} + q \begin{pmatrix} 0 & 1 \ 0 & 0\end{pmatrix} = pI +  q (M-I) = (p-q) I + qM,$$ for some $p,q \in \mathbb{R}$. Then $M \in \mathrm{Span}(I,M)$. Conversely, we have verified in the solution of Exercise 1 that every linear combination of $I$ and $M$ is in ${\mathcal B}$.

For any $X \in {\mathcal A}$,
$$X \in {\mathcal B} \iff \exists (a,b)\in \mathbb{R}^2,\ X = aI + b M,$$
or equivalently
$$ {\mathcal B}  =  \mathrm{Span}(I,M).$$
\end{proof}

Exercise 0.1.4(Centralizer of $M$)

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Exercise 0.1.4(Centralizer of $M$)

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