If A is a $2×2 $ matrix, does $A^2=3A$ imply $ A=0$ or $A=3I$? 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: *

*$A-3$ doesn't makes sense. We can add two matrix and we can multiply a matrix by a scalar( one at a time, known as linear combination).

$A-3$ should be $A-3I$ .


*$A(A-3I) =0$ doesn't imply $A=0$ or $A=3I$ . Cancellation doesn't holds.$M_2(F) $ is not an integral domain (contains zero divisors)

Counter example:
$A=\begin{pmatrix}0&0\\0&3\end{pmatrix}$
Then $A^2-3A=0$ but neither $A=0$ nor $A=3I$
A: 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.
A: 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?
A: 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. $AB=BA$.
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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 $\det(A)=\det(B)=0$.
A: 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 $\alpha\in\mathbb R$.
