Let $A$ be a $3×3$ matrix such that $A^2 = 2A$. Prove that $det(A)$ must be either $0$ or $8$.

Question

I have no clue about how to start this problem. Could someone help me with this?

Answer 1

Taking the determinant of both sides of the equation $A^2=2A$ gives $\operatorname{det}(A)^2=2^3 \cdot \operatorname{det}(A)=8 \cdot \operatorname{det}(A)$ (here, it is crucial that $A$ be a $3 \times 3$-matrix). Subtracting $8 \cdot \operatorname{det}(A)$ from both sides, we get $\operatorname{det}(A)^2-8 \cdot \operatorname{det}(A)=0$. Factoring the left hand side, we get $\operatorname{det}(A)(\operatorname{det}(A)-8)=0$. Finally, setting each factor equal to zero gives that either $\operatorname{det}(A)=0$ or $\operatorname{det}(A)=8$.

More generally, if $A^2=mA$ where $A$ is an $n \times n$-matrix, then either $\operatorname{det}(A)=0$ or $\operatorname{det}(A)=m^n$. Even more generally, if $A^k=mA$, then either $\operatorname{det}(A)=0$ or $\operatorname{det}(A)=\sqrt[k-1]{m^n}$ or, if $k$ is odd, $\operatorname{det}(A)=-\sqrt[k-1]{m^n}$.

Answer 2

Here are the two properties of the determinant you will need:

• $\det(A^m)=[\det(A)]^m$
• $\det (cA)=c^n \det A$, where $n$ is the order of your matrix.

Answer 3

Since $A^2 - 2A = 0$, the minimal polynomial of $A$ divides $X^2 - 2X = X(X-2)$. Thus, any eigenvalue of $A$ is either 0 or 2. The determinant is the product of the eigenvalues.