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Let $A$ be a square matrix of order three or two, and $I$ be a unit matrix. How many solutions are possible for the equation $$A^2=I$$? In case the solutions are infinite, or very large, how do I check, without actual multiplication, if a given matrix is a solution to the given equation. In the general case, what about a factorisable matrix polynomial equation?

Shubham
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    You may find this interesting: https://ckrao.wordpress.com/2012/11/16/when-is-a-matrix-the-inverse-of-itself/ – MPW Jan 25 '15 at 14:07
  • Yes, probably the page talks about orthogonal matrix. But this solves only the given matrix equation. What about an equation like $A^3-3A^2+2I=0$? – Shubham Jan 25 '15 at 14:15
  • Are you trying to solve it for the real or complex numbers? – Fabian Jan 25 '15 at 14:50
  • Could somebody provide an easier solution, I mean one that uses less advanced concepts, say high school concepts? Or is it that these concepts are necessary. The problem is rather an elementary one. @fabian, for now I'm working on matrices with real elements. – Shubham Feb 01 '15 at 20:02
  • See also: http://math.stackexchange.com/questions/44341/finding-number-of-matrices-whose-square-is-the-identity-matrix – Martin Sleziak Feb 19 '15 at 17:28

2 Answers2

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A matrix $A$ will satisfy $p(A)=0$ for a polynomial $p$ if and only if the minimal polynomial of $A$ divides $p$.

In your case, we have $p(x) = x^2 - 1$. So, the minimal polynomial of $A$ must be $x^2-1,x-1,$ or $x+1$. In any case, we can say

$A^2 = I$ if and only if $A$ is diagonalizable and has eigenvalues $1$ or $-1$

In the size $2$ case, we have $A=I,A=-I$, or $$ A = S\pmatrix{1&0\\0&-1}S^{-1} $$ to check whether $A$ is one of these last matrices, we can verify that trace$(A) = 0$ and $\det(A) = -1$.

Ben Grossmann
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Notice in general (unless the field is finite) there could be uncountably infinite many solutions even for $2\times 2 $ matrices, e.g. $$\forall c\in\mathbb R:A= \left( \begin{array}{ccc} 1 & 0 \\ c & -1 \\ \end{array} \right) $$

R B
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