Let $ S=\{A=[a_{ij}] : AB=BA \, \forall B \in M_{n×n} \},$ then $\text{dim}(S)$ equals to?
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2Welcome to MathStackExchange. To help others understand how to help you, please give a little more context. Is this a homework question? What are yours thoughts about approaching it? – Mefitico Aug 29 '19 at 12:46
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Take n=2 and then try to find a pattern. – Manoj Kumar Aug 29 '19 at 12:52
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Duplicate of https://math.stackexchange.com/q/896979 – Jean Marie Aug 29 '19 at 13:00
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I'm voting to close this question as off-topic because the OP must work before asking such question. – Aug 29 '19 at 14:09
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On one hand you can see that $\forall \lambda$ a scalar $\lambda I_n\in S$. We’ll demonstrate that $S=span(I_n)$ and so $dim(S)=1$.
Call $E_{ij}\in M_{n\times n}$ tha matrix such that all of his entries are equal $0$ but the entry $i,j$ equals $1$. Suppose $A\in S$ then you can see that $AE_{ij}$ is the matrix with all the entries $0$ but the $j$-esim column is equal to the $i$-esim column of $A$ and $E_{ij}A$ is the matrix everywhere $0$ but the $i$-esim row equal to $j$-esim row of $A$. Since $AE_{ij}=E_{ij}A$ $\forall 1\le i,j\le n$ you obtain $A$ must be diagonal.
Can you find another type of matrix with which to verify that all the diagonal entries must be equal?
Alessandro Cigna
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For $n=2$, on taking $ A= \begin{vmatrix} a & b \ c & d \end{vmatrix}$ and $ B= \begin{vmatrix} 1 & 0 \ 0 & 0 \end{vmatrix} ;$ I got $b=c=d=0,$ and $a$ arbitrary. So only $1$ free variable, i.e. $dim(S)=1$ – user388189 Aug 29 '19 at 13:26
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No: $\begin{bmatrix} 1 & 0 \ 0 & 0 \ \end{bmatrix}\begin{bmatrix} a & b \ c & d \ \end{bmatrix}=\begin{bmatrix} a & b\ 0 & 0 \ \end{bmatrix}$ and $\begin{bmatrix} a & b \ c & d \ \end{bmatrix}\begin{bmatrix} 1 & 0\ 0 & 0\ \end{bmatrix}=\begin{bmatrix} a & 0 \ c & 0 \ \end{bmatrix}$ so you have $c=b=0$ and $a$ and $d$ free parameters – Alessandro Cigna Aug 29 '19 at 14:13
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