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Let $Z_n$ be the set of all $n \times n$ matrices that commute with all $n \times n $ matrices. Show that $$Z_n = \{\lambda I_n \ | \ \lambda \in \mathbb R\}$$

($I_n$ is the $n \times n$ identity matrix)

I don't know how to use $E_{ij}$ (matrix with $1$ in $(i,j)$ and $0$ elsewhere) and the elementary matrix $P_{ij}$ to prove this question. Can anyone explain it please?

Ant
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hotterthanmath
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2 Answers2

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Notice that

$$E_{ij}E_{kl}=\delta_{jk}E_{il}$$ so if a matrix $$A=\sum_{1\le k,l\le n}a_{kl}E_{kl}$$ commutes with the all the matrices then it commutes with $E_{ij}$ hence we get

$$AE_{ij}=E_{ij}A\iff \sum_{k=1}^n a_{ki}E_{kj}=\sum_{l=1}^n a_{jl} E_{il}$$ so we see that

$$a_{ii}=a_{jj}=:\lambda\;\forall i,j\quad \text{and} \quad a_{ki}=0\;\forall k\ne i$$ hence $A=\lambda I_n$. Finally, it's trivial that $\lambda I_n$ does commute with all the matrices.

  • Thanks! There's another part that i don't know how to prove. How can I show that all diagonal entries are equal? – hotterthanmath Jan 26 '15 at 11:26
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    From the equality on the RHS of the equivalence and since $E_{ij}$ form a basis we see that only for $k=i$ and $l=j$ we may have a non zero term and that $a_{ii}=a_{jj}$ so all diagonal entries are equal. –  Jan 26 '15 at 11:30
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    I'm sorry, but this explanation is horrible. I'm reading over it for about the 10th time now and still can't get it, I feel like it skips way too many steps. – DavidsKanal Jan 14 '20 at 21:05
  • The answer is poorly written. Also, the whole question is closed when clearly the OP is asking for a very specific thing. – slhulk May 08 '20 at 11:59
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Hint: Compute $A E_{ij}$ and $E_{ij} A$. Force them to be equal.

lhf
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