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This is from Lang's introduction to Linear Algebra page no 61.

Determine all $2\times 2$ matrices $A$ such that $A^2 = 0$.

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

$A^2=\begin{pmatrix} a^2+bc & ab+bd \\ ac+cd & d^2+cb \end{pmatrix}$

Equating all the four terms of the matrix to zero and solving,

$a^2+bc=0$ and $d^2+bc=0$ gives $a=\pm d$

Solving $ab+bd=0$ and $ac+cd=0$ gives $a=-d$ OR $b=c$, which gives $a=d=0$

My question: I know that if $A$ is an $n\times n$ matrix with all the diagonal elements and all the elements below it equal to zero then $A^n=0$, so in this question \begin{equation} \begin{pmatrix} 0 & x \\ 0 & 0 \end{pmatrix} \end{equation} qualifies as $A$, but it does not agree with $b=c$ as in above solution, so where did I go wrong? And if I go with $b=c$, then $A^2$ does not equal $0$.

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Do you know what determinants, eigenvalues, characteristic equations are? –  Calvin Lin Apr 19 '13 at 14:48
    
@CalvinLin, this question comes before all the topics mentioned by you in the text –  Vikram Apr 19 '13 at 14:50

2 Answers 2

up vote 7 down vote accepted

When you take the difference of 2 equations, it does not mean that solutions to the new equation will satisfy both of the old equations.

For example, if $a=d=0$, then this satisfies your conclusion of $a= \pm d$, but doesn't always satisfy your initial conditions of $a^2+bc = 0, d^2 + bc = 0$ (in particular if $b=c\neq0$).

You can read Proof that 0=1 to see another example of how this method of solving equations can go wrong.

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thanx for the insight, pls suggest some light reading that throws some light on the point you mentioned, taking your suggestion I tried back-substitution and got the answer just in 3-4 steps :) –  Vikram Apr 19 '13 at 16:02

Well you have your 4 equations:

$$\cases{a^2+bc=0 \\ ab+bd=0 \\ ac+cd=0 \\ d^2+bc=0}$$

And you know that if $A^2=0$ then $det(A^2)=det(A)^2=0$ and thus $det(A)=0$. So you have a fifth equation:

$$ad-bc=0 \Rightarrow ad=bc$$

Now replace this in your system and factorize it:

$$\cases{a^2+ad=a(a+d)=0 \\ ab+bd=b(a+d)=0 \\ ac+cd=c(a+d)=0 \\ d^2+ad=d(a+d)=0 \\ ad=bc}$$

  • If $a+d\neq0$ then $a=b=c=d=0$
  • If $a+d=0$ then all these equations are equivalent and you're left with two equations (you already know you'll probably have 2 degrees of liberty):

$$\cases{a+d=0 \\ ad=bc}\Rightarrow\cases{d=-a \\ -a^2=bc}$$

  • If $b=0$ then $a=d=0$ and you're just left with $c\in\Bbb{R}$
  • If $b\neq0$ then $\cases{d=-a \\ c=-\frac{a^2}{b}}$

Now this means that you have two possible matrices s.t. $A^2=0$:

$A=\left(\begin{array}{cc} a && b \\ -\frac{a^2}{b} && -a \end{array}\right),(a,b)\in\Bbb{R}\times\Bbb{R}^*$ or $A=\left(\begin{array}{cc} 0 && 0 \\ c && 0 \end{array}\right),c\in\Bbb{R}$

You also have their transposes $\left((A^T)^2=A^TA^T=(AA)^T=(A^2)^T=0^T=0\right)$

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As a side note, $ad-bc=0$ doesn't 'magically appear', but can be found from the initial 4 equations as $0 = (a^2+bc)(d^2+bc)-(ab+bd)(ac+cd) = (ad-bc)^2$. Of course, this is calculating the discriminant in disguise. –  Calvin Lin Apr 19 '13 at 18:22
    
Haha, yes you're right I could've written it this way ! And as you said, it's just saying that $(a^2+bc)(d^2+bc)-(ab+bd)(ac+cd)=det(A^2)=det(A)^2=(ad-bc)^2=0$ :) –  Dolma Apr 21 '13 at 19:09

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