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Let $T=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ be a non-scalar matrix.

If $S=\begin{pmatrix}e&f\\g&h\end{pmatrix}$ be such that $TS=ST$. Why there exists $\alpha,\beta\in \mathbb{C}$ such that $$S=\alpha T+\beta I\;?$$

Note that $TS-ST=0$ is equivalent to

$$\begin{bmatrix}bg-fc & af+bh-eb-fd\\ ce+dg-ga-hc & fc-bg\end{bmatrix} = \begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$$ This implies that $$\begin{cases} bg-fc = 0,\\ af+bh-eb-fd = 0,\\ ce+dg-ga-hc = 0,\\ fc-bg = 0. \end{cases}$$ Since $T$ is non scalar, then $b\neq 0$ or $c\neq 0$ or $a\neq d$. However, I cannot find $\alpha$ and $\beta$.

Schüler
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  • Dear professor. According to Lemaa 2.3 this result works only when $\text{dim}(H)=2$ (please see https://books.google.tn/books?id=lJbXcjpY4QsC&pg=PA75&hl=en&redir_esc=y#v=onepage&q&f=false ) I find contradiction between this result and the provided answer. Thank you for the explanation. – Schüler Feb 09 '19 at 07:40
  • I'm Ok. The problem is his comment he says:''The proof works in every vector space on any field K.'' – Schüler Feb 09 '19 at 08:56
  • Thank you very much for your explanation. – Schüler Feb 09 '19 at 09:07

1 Answers1

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EDIT. We assume that $T\in M_2(K)$, where $K$ is a field. $C(T)=\{S\in M_2(K);TS=ST\}$ is a vector space containing $\{I,T\}$, that are linearly independent; then it suffices to show that $dim(C(A))\leq 2$, that is, the entries of such a matrix $S=\begin{pmatrix}p&q\\r&s\end{pmatrix}$ depend at most on $2$ parameters.

$\textbf{Proof}$. There is a vector $u$ s.t. $\{u,Tu\}$ is a basis of $K^2$; otherwise, let $v,w$ be a basis of $K^2$; one has $Tv=av,Tw=bw,T(v+w)=c(v+w)$, that implies that $T=aI$, a contradiction.

In the basis $\{u,Tu\}$, $T$ becomes $\begin{pmatrix}0&a\\1&b\end{pmatrix}$ and

$(TS)_{1,1}=(ST)_{1,1},(TS)_{2,1}=(ST)_{2,1}$ iff

$q=ar,p=-br+s$; finally, the entries of $S$ depend at most on the $2$ parameters $r,s$ and we are done.

  • Thank you for your answer. I think you don't proceed as my attempt. Please why $T(Tu))=au+bTu$? – Schüler Feb 04 '19 at 16:06
  • Please have you seen my question ? Thanks – Schüler Feb 05 '19 at 08:20
  • I just read it. $a,b$ exist because $u,Tu$ is a basis. Note also that the proof works on every field $K$ and not only over $\mathbb{C}$. –  Feb 05 '19 at 08:31
  • I think that you used $a$ and $b$ in my question. I think also that the proof work in an arbitrary Hilbert space such that its dimension is 2. Do you agree with me? – Schüler Feb 05 '19 at 14:17
  • The proof works in every vector space on any field $K$. In particular, $K$ does not need to contain the eigenvalues of $A$. Do you understand why $u$ exists ? –  Feb 06 '19 at 15:13
  • I think you mix everything; if you want to talk about another topic, then ask another question in a new file. –  Feb 09 '19 at 15:17
  • Now, my only problem is that $T$ should be in $M_2(K)$ and not in $M_n(K)$. – Schüler Feb 09 '19 at 15:23
  • Of course, $n=2$; I wrote too fast. –  Feb 09 '19 at 15:25