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$$A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$$

When $A^3 = O$ (zero matrix) is satisfied, what condition do the four real numbers $a, b, c, d$ meet?

I cubed $A$:

$A^3 = \begin{bmatrix} a(a^2+bc)+bc(a+d) & b(a^2+bc) + bd(a+d) \\ ac(a+d)+c(bc+d^2) & bc(a+d)+d(bc+d^2)\end{bmatrix}$

I don’t know how it is going to be.

Post script

Thank you for the replies, I might have found the answers:

$$ \begin{eqnarray} A^3_{11} * d - A^3_{12} * c = 0 \\ (ad-bc)(a^2+bc) = 0 \\ ad-bc = 0 \end{eqnarray} $$

But why $ a^2 + bc \neq 0$?

Rodrigo de Azevedo
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yudukikun5120
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  • What are your thoughts? What, for instance, is $\det(A)$? – lulu Mar 29 '21 at 13:15
  • Yes, I got this equation: $A^3 = \begin{pmatrix} a(a^2+bc)+bc(a+d) & b(a^2+bc) + bd(a+d) \ ac(a+d)+c(bc+d^2) & bc(a+d)+d(bc+d^2)\end{pmatrix}$. But I don’t know how I can get the answer. – yudukikun5120 Mar 29 '21 at 13:17
  • Please edit your post to indicate your efforts (and cubing the matrix definitely counts). As a Hint: suppose that $a\neq 0$. Then we can divide by $a$ to get a non-zero constant times a matrix with a $1$ in one entry. What cvann you say about the other entries? – lulu Mar 29 '21 at 13:20
  • by $O$ I take it you mean the zero matrix? – Rhys Hughes Mar 29 '21 at 13:23
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    Side note: it helps a lot if you note that $A^3=0\iff A^2=0$. See, e.g., this – lulu Mar 29 '21 at 13:26
  • Yes, $O$ means zero matrix. – yudukikun5120 Mar 29 '21 at 13:27

2 Answers2

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Let $f(x)$ be the characteristic polynomial of $A.$ Since $A^3$ is zero, $f(x)$ divides $x^3.$ Therefore, $f(x)=x^2.$ Thus, $A$ has both trace and determinant zero. In other words, $d=-a$ and $bc=a^2.$ (If $a=0,$ observe that either $b$ or $c$ must zero.)

David
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  • This is the way to go. I think you can improve it by noting that if trace and determinant are both zero then the characteristic polynomial is $X^2$ and so the condition is necessary and sufficient. – ancient mathematician Apr 02 '21 at 06:40
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It seems the nontrivial solution is

$$\left( \begin{array}{cc} a & b \\ -\frac{a^2}{b} & -a \\ \end{array} \right)$$

yarchik
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