Assume that $A \in M_{n}(F)$ is a matrix, such that for each $B \in M_n(F)$ we have $AB=BA$.
Prove that there exists $\lambda \in F$ such that $A=\lambda.I_n$
Note 1 : $F$ is a field.
Note 2 : This question should be solved without the use of vectors and determinants. Unfortunately, I don't know where to start ! I mean, in general form, When we want to prove that something exists, We should find it. But i don't know how !
Consider the set of matrix units $\{E_{kj}\}$, where $E_{kj}$ is the matrix with a $1$ in the $k,j$ entry and zeroes elsewhere. The hypothesis is that $A$ commutes with all these. Note also that $$ A=\sum_{k,j}a_{kj}E_{kj} $$ and that $$ E_{kj}E_{st}=\delta_{j,s}\,E_{kt},\ \ \ \sum_kE_{kk}=I. $$ So, for instance, for fixed $k$ and $j$ with $k\ne j$, $$ a_{kj}E_{kj}=E_{kk}AE_{jj}=E_{kk}E_{jj}A=0. $$ So $a_{kj}=0$. As we can do this for all $k\ne j$, we get that $A$ is diagonal: $$ A=\sum_ka_{kk}E_{kk}.$$ Now, for any $k$, $$ a_{11}E_{k1}=E_{k1}A=AE_{k1}=a_{kk}E_{k1} $$ and then $a_{kk}=a_{11}$. Thus, $$ A=\sum_ka_{11}E_{kk}=a_{11}\,I. $$