Prove that ::

If $A$ is an $n\times n$ matrix such that $AB = BA$ for any $n\times n$ matrix $B$, then $A=\lambda I_n$

  • $\begingroup$ Start by writing out what the matrices $A$ and $B$ must look like if $AB = BA$, then the proof should be easier to see. i.e. both $A$ and $B$ must be diagonalizable. $\endgroup$ – Anthony P Nov 29 '16 at 14:56
  • 2
    $\begingroup$ It might be easier to start with $B=E_{ij}$ where $E_{ij}$ is the matrix consistng only of zeroes and a $1$ in the $(i,j)$ entry. $\endgroup$ – ctst Nov 29 '16 at 14:58
  • $\begingroup$ See math.stackexchange.com/questions/478849/when-will-ab-ba for more guidance $\endgroup$ – Anthony P Nov 29 '16 at 14:58
  • $\begingroup$ I slightly reworded the statement you want to prove and reformatted. Note that the question is still not a complete sentence and is missing important details, like the work you've done on the problem. $\endgroup$ – Matthew Leingang Nov 29 '16 at 15:08
  • $\begingroup$ Schur's Lemma... $\endgroup$ – tired Nov 29 '16 at 15:35

Let $AB=BA$ for all matrices $B$.

To show that $A=\lambda I$.

Let $v(\neq 0)\in V$ be fixed.

Then $\{v,Av\}$ is linearly dependent.Otherwise if not then $\{v,Av\}$ can be extended to form a basis of $V$ say $\{v,Av,u_1,u_2\ldots u_n\}$.

Define $B$ as \begin{bmatrix}0 & 1& 0\ldots 0\\0 & 0&0\ldots 0\\0 &0 &0\ldots 0\\\ldots\\\ldots \\0&0&0\ldots 0\end{bmatrix}

Note that $BA(v)=v $ and $B(v)=0$.

Since $AB=BA$ so $v=0$ which is false.

So $\{v,Av\}$ is linearly dependent. Hence $Av=a_v v$

Now to show that $a_v$ is independent of $v$.

Let us consider two vectors $v,w\neq 0$. Then $Av=a_vv,Aw=a_ww$

CASE I:$\{v,w\}$ is linearly dependent.

Then $v=bw\implies Av=bAw=ba_w w=a_wv\implies a_vv=a_wv$ which is false as $v\neq 0$.


$\{v,w\}$ is linearly independent.

Then $a_vv+a_wv=Av+Aw=A(v+w)=a_{v+w}(v+w)=a_{v+w}v+a_{v+w}w$. (Try to claim that this is also not possible.)

Hence $a_v$ is independent of $v$. So $Av=\lambda v\forall v$

So $A=\lambda I_n$


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