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Prove that $AB=BA$ for all $B \implies A=c \cdot I $ for some $c$ where $I$ is an identity matrix.

Both $A,B\in M_{n \times n}.$ I don't know how attempt proof.

anon
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Mario
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  • Quantifiers, please. – Git Gud Mar 03 '15 at 22:32
  • Could you elaborate? – Mario Mar 03 '15 at 22:34
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    As such, your assertion is false. – Bernard Mar 03 '15 at 22:35
  • What you want to say is: If $AB = BA$ for EVERY matrix $B$. The "for every" is a quantifier (often written $\forall$ or, "for all"). – David Wheeler Mar 03 '15 at 22:36
  • The exact problem is : Let $A \in M_{n \times n}$ Show if the only matrix at $M_{n \times n}$ similar to $A$ is $A$ then $A=c \cdot I$ – Mario Mar 03 '15 at 22:37
  • You want to ask how to prove that $$\forall A\in M_{n\times n}\left(\forall B\in M_{n\times n}(AB=BA)\implies \exists c\in \mathbb C(A=cI_n)\right),$$ but you didn't. Instead you omitted quantifiers which leaves people guessing at what's is being asked. For instance taking $$A=\begin{bmatrix} 1 & 0\ 0 & 1\end{bmatrix}, B=\begin{bmatrix} 0 & 0\ 0 & 1\end{bmatrix},$$ you get a counter-example to a statement that looks like what you wrote. – Git Gud Mar 03 '15 at 22:38
  • Look up the proof of Schur's lemma. – anon Mar 03 '15 at 22:55
  • I meant $A=\begin{bmatrix} 1 & 0\ 0 & 0\end{bmatrix}$ above. – Git Gud Mar 03 '15 at 23:35

1 Answers1

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Use special matrices. If $E_{h}$ is the matrix having $1$ in place $(h,h)$ and $0$ elsewhere, then $AE_{h}$ is the matrix $$ \begin{matrix} [\,0&\dots&0&a_h&0&\dots&0\,]\\ &&&\uparrow\\ &&& \scriptstyle h \end{matrix} $$ having all zero columns except for the $h$-th column, which is the $h$-th column of $A$. Similarly, the matrix $E_{h}A$ is the matrix having all zero rows except for the $h$-th row, which is the $h$-th row of $A$.

In particular, $a_{hj}=0$ for $j\ne h$. Since $h$ is arbitrary, we conclude that $A$ is diagonal.

Now consider the matrix $E_{hk}$ that's obtained from the identity by switching the $h$-th row with the $k$-th row (for $h\ne k$). It's easy to show that $E_{hk}$ is the matrix obtained from $A$ by switching the $h$-th row with the $k$-th row. Similarly, $AE_{hk}$ is obtained from $A$ by switching the $h$-th column with the $k$-th column.

Thus, by comparing $E_{1k}A$ with $AE_{1k}$, we conclude that the coefficient in place $(h,h)$ of $A$ is the same as the coefficient in place $(1,1)$.

Therefore all coefficients on the diagonal are equal. Call this common coefficient $c$ and we have proved that $A=cI$.

Note that this uses only matrix multiplication where the special matrices used have only coefficients $0$ and $1$, so it holds for matrices over any commutative ring.

egreg
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