# 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?

• Do you know about eigenvalues and eigenvectors? If so, then notice you have an annihilator polynomial which is factorized with simple roots. Jul 26, 2022 at 6:43
• Also, $A-3$ does not make sense, you mean $A-3I$. Jul 26, 2022 at 6:44
• I do know about eigenvalues and eigenvectors but I don't know what an annihilator polynomial is Jul 26, 2022 at 6:48
• A polynomial $P$ with $P(A)=0$. Jul 26, 2022 at 6:50
• As for your final question: though to some extent polynomials in a single square matrix behave somewhat like polynomials in $X$, solving polynomial equations over matrices is rather more difficult than solving the same equations in (say) $\Bbb C$. Jul 26, 2022 at 9:03

1. $$A-3$$ 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 $$A-3I$$ .

1. $$A(A-3I) =0$$ 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 $$A=3I$$

• How would you go about finding counterexamples? Jul 26, 2022 at 6:51
• Eigenvalues, basically! Jul 26, 2022 at 6:52
• That's the part which I don't really understand Jul 26, 2022 at 6:54
• Yes. Minimal polynomial divides annihilating polynomial.If the given polynomial is the minimal polynomial. Then eigenvalues are $0, 3$ and $A$ is diagonalizable. Choose $A$ diagonal matrix with diagonal entries $0$ and $3$ Jul 26, 2022 at 6:55
• $A-3$ does make sense under the convention that in a given (unitary) ring, any integer $n$ stands for $n$ times the ring unit. In this context that makes $A-3$ stand for $A-3I_2$, and it is actually a very convenient convention, as long you one knows that it is being used. Jul 26, 2022 at 9:06

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?

• $A$ will be similar to $\begin{pmatrix}\lambda_1&a\\0&\lambda_2\end{pmatrix}$ with $\lambda_i\in\{0,3\}$ and $a\in\{0,1\}$ anyway, so I don't think solving the coefficients for all transition matrices is of interest.
– zwim
Jul 26, 2022 at 12:14
• @zwim Well, I think it's an easy an instructive way to find all solution, in a simple case as two-by-two Jul 26, 2022 at 13:19

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 $$\alpha\in\mathbb R$$.