Prove that $A = aI+B$

Question

Let $A \in M_{n\times n}(F)$. Show that $A = aI+B$, where $a \in F, B \in M_{n\times n}(F)$ and $\operatorname{tr}(B) = 0$.

I just need a hint to start proving this.

Solved

Answer 1

If $\operatorname{char} F$ does not divide $n$, then you can let $a = \text{average of the diagonal elements of } A.$

In other words: $$a = \dfrac{1}{n}\operatorname{tr} A.$$

Then, let $B = A - aI$. Clearly, you have $A = aI + B$. Moreover, $$\operatorname{tr}B = \operatorname{tr}(A - aI) = \operatorname{tr}(A) - a\operatorname{tr}(I) = \operatorname{tr}A - na = \operatorname{tr}A - \operatorname{tr}A = 0.$$

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On the other hand, if $\operatorname{char} F$ does divide $n$, then you cannot find such $a, B$ in general. For example, let $F = \Bbb F_2$ and consider $$A = \begin{pmatrix}1&0\\0&0\end{pmatrix}.$$

Suppose that we did have $A = aI + B$ with $\operatorname{tr}B= 0$.
Note that $\operatorname{tr}I = \operatorname{tr}B = 0$ and thus, we'd get that $\operatorname{tr}A = 0$, a contradiction.

In such a case, we can find an $a$ and $B$ if and only if $\operatorname{tr}A = 0$.
The "only if" follows from an argument as the one given above.
The "if" part follows by letting $A = B$ and $a = 0.$

Answer 2

Hint: $Tr(A-aI)$ equals $ Tr(A)-na$.