What matrix A with dimension n x n is always true given A - A^2 = O where O is the zero matrix with dimension n x n and I with dimension n x n is the identity matrix.

A) A is a diagonal matrix
B) A = A^2
C) A = I
D) A = O
E) I = A^2

Just got out of an exam with this question and I chose B. I thought about this question for a little bit longer after I handed in my exam and I came to the conclusion that A = I.

A - A^2 = O
A = A^2
AA^-1 = (A^-1)(A^2)
I = A

Can someone confirm if this is right?

  • $\begingroup$ What is the definition of a "direction matrix"? $\endgroup$ – Omnomnomnom Nov 8 '14 at 1:54
  • 1
    $\begingroup$ You can only say that $A = I$ if $A$ is non-singular. Note that $A = O$ also satisfies this equation. $\endgroup$ – Omnomnomnom Nov 8 '14 at 1:54
  • $\begingroup$ Oops, it's suppose to be "diagonal matrix" $\endgroup$ – Wade Nov 8 '14 at 2:00
  • $\begingroup$ Should this say "What statement about a matrix $A$ with dimension $n\times n$ is always true..."? $\endgroup$ – alex.jordan Nov 8 '14 at 4:56
  • $\begingroup$ Or perhaps "What statement about a matrix $A$ with dimension $n\times n$ would always make it true that..."? $\endgroup$ – alex.jordan Nov 8 '14 at 5:00

It seems to me that the other answers have misinterpreted the question. My understanding is that the question asks:

Given that $A$ is a matrix satisfying $A - A^2 = 0$, which of the following is always true?

The only correct answer to this question is B. Certainly, if $A = A^2$, then we can say that $$ A = (A - A^2) + A^2 = 0 + A^2 = A^2 $$ As a counterexample to all of the other choices, consider the matrix $$ A = \pmatrix{1&1\\0&0} $$

  • $\begingroup$ The question may be so trivial? Unbeliveable! $\endgroup$ – Przemysław Scherwentke Nov 8 '14 at 6:05

It is right (answer C, the proof needs a little modification), but D) is also true. A counterexample to E): $$ A=\begin{bmatrix} 0&1\\ 1&0 \end{bmatrix} $$ And for example $2I$ is a counterexample to A).

Edit: And corected version of your proof: $A - A^2 = O$, $A = A^2$, and now, instead of

$AA^{-1} = (A^{-1})(A^2)$

either $AA^{-1} = A^2(A^{-1})$ or $A^{-1}A = (A^{-1})(A^2)$

and finally $I = A$.

  • $\begingroup$ But is D always true? $\endgroup$ – Wade Nov 8 '14 at 2:03
  • $\begingroup$ @Wade If $A$ is zero matrix, so is $A^2$. $\endgroup$ – Przemysław Scherwentke Nov 8 '14 at 2:05
  • $\begingroup$ I could say the same if A is an identity matrix. So would both C and D be correct? $\endgroup$ – Wade Nov 8 '14 at 2:08
  • $\begingroup$ @Wade Yes, they are. And the argument is simplier than yours. (Well, in fact, you should multiply by $A^{-1}$ either on left or on right). $\endgroup$ – Przemysław Scherwentke Nov 8 '14 at 2:12

I think $B$, $C$ and $D$ are all correct.

Infact $B$ is the question rewritten in another way. And as already established $C$ and $D$ are correct.

  • $\begingroup$ This was a multiple choice question. Is there one answer that is more correct than the other? $\endgroup$ – Wade Nov 8 '14 at 2:09
  • $\begingroup$ @Wade It depends on the rules of your exam. The correct answer is B), C), D), but subset of it may be partially scored. $\endgroup$ – Przemysław Scherwentke Nov 8 '14 at 2:21

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