Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $0$.

share|cite|improve this question
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
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.

share|cite|improve this answer
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 $\left((A^T)^2=A^TA^T=(AA)^T=(A^2)^T=0^T=0\right)$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.