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Problem: Let $A$ be a $2\times 2$ matrix, $O$ be the null matrix, and $I$ be the identity matrix. Is the following statement true? $$\text{if } A^2=3A \implies A=O\text{ or }3I$$
I tried proving this just by factorising $A^2-3A=O \implies A(A-3I)=O \implies A-3I=O, A=O$. However, I cannot tell if this is correct. Do polynomials work the same way with matricies as they do with real/complex numbers?
$A-3$ should be $A-3I$ .
Counter example:
$A=\begin{pmatrix}0&0\\0&3\end{pmatrix}$
Then $A^2-3A=0$ but neither $A=0$ nor $A=3I$
As soon as a polynomial $P$ has two distinct roots $\alpha,\beta$, the polynomial equation $P[A]=0$ for $A$ a $2\times2$ matrix has "mixed" solutions $$ A=\pmatrix{\alpha&0\\0&\beta} \quad\text{and}\quad A=\pmatrix{\beta&0\\0&\alpha} $$ (because doing arithmetic with diagonal matrices basically means doing that arithmetic in each diagonal position separately, while leaving the off-diagonal positions zero), as well as "pure" solutions $A=\alpha I$ and $A=\beta I$. moreover, the number of solutions is not even finite (if the field of scalars is infinite), because one can apply any change of basis $A\mapsto P^{-1}AP$ to the mixed solutions to get infinitely many different ones. (The pure solutions do not change under change of basis.)
So the answer is no, not just those two solutions.
The other answers have given general arguments. However, for a $2\times 2$ matrix, you can actually do an explicit computation: Write $$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\,.$$ Then your equation becomes $$A^2=\begin{pmatrix}a&b\\c&d\end{pmatrix}^2=\begin{pmatrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{pmatrix}=3A=\begin{pmatrix}3a&3b\\3c&3d\end{pmatrix}\,.$$ Now you can systematically explore the solutions (it probably helps to start with the off-diagonal entries).
Can you take it from here?
Tl;dr: NO
Two matrices $A, B$ which are individually non-null matrices, can still be multiplied to give a null matrix. For eg., for your specific problem, we’ll consider $2\times 2$ matrices. Let $$A=\left[\begin{matrix}1 & 1\\ 0 & 0\end{matrix}\right], B=\left[\begin{matrix}1 & 2\\ -1 & -2\end{matrix}\right]$$ so multiplying gives us $$AB=\left[\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right].$$ However, note that $$BA= \left[\begin{matrix}1 & 1\\ -1 & -1\end{matrix}\right]$$ and this just shows that matrix multiplication is not commutative, i.e. $AB=BA$. ______________________________________
However, if $AB=O, \ \text{but}\ A,B≠O,$ then we can say that $\det(A)=\det(B)=0$, i.e. both matrices are singular. To see this, note that if any one of them (say A) is non-singular, then its inverse must exist, so we have $$A^{-1}AB=A^{-1}O$$ so we get $B=O$ which is contradictory to our hypothesis that $B≠O$. Thus, extending a similar argument to $\det(B)$ we get that $\det(A)=\det(B)=0$.
From $A^2-3A=O$, you should only deduce that the polynomial $t^2-3t$ annihilates matrix $A$. Now, any polynomial which is a factor of this must be a good candidate for the minimal polynomial of $A$. The three possibilities are:
$t$ or $(t-3)$ or $t(t-3)$
From the first two, $A=O$ or $A=3I$ are feasible possibilities but the third opens many possibilities for you to write a matrix $A$ such that $tr(A)=3$ and $det(A)=0$. A general choice is given by $A=\begin{pmatrix}0&0\\\alpha&3\end{pmatrix}$ where $\alpha\in\mathbb R$.
