Question

Prove that if AB = BA for every N by N matrix A, then B = cI, for some constant c.

Answer 1

Question 1.

Prove that if $AB=BA$ for every $N$ by $N$ matrix $A$, then $B=cI$ for some constant $c$.

Solution. It follows from $AB=BA$ that $A$ should also be square of size $N$. For each $i=1,\ldots,N$ consider the diagonal matrix $C_{i}$, whose $(i,i)$-th entry is $1$ and all the other entries are zero. Since $AC_{i}=C_{i}A$, then for all $j\neq i$

$(AC_{i})(i,j)=(C_{i}A)(i,j)\Leftrightarrow 0=1\cdot A(i,j)\Leftrightarrow A(i,j)=0.$

Thus, we proved that $A$ is diagonal. Now show that $A(i,i)=A(j,j)$ for all $1\leq i,j\leq N$. Consider the matrix $D_{ij}$, whose unique nonzero entry is $D_{ij}(i,j)=1$. Then $AD_{ij}=D_{ij}A$ implies

$(AD_{ij})(i,j)=(D_{ij}A)(i,j)\Leftrightarrow A(i,i)\cdot D_{ij}(i,j)=D_{ij}(i,j)\cdot A(j,j)\Leftrightarrow A(i,i)=A(j,j).$

So, $A$ is diagonal and all the diagonal entries of $A$ are equal, i. e. $A=cI$ for some constant $c$. $\Box$