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How to find an example of matrix $A$ that satisfies $A^{-1} = \frac{1}{n} A$, where $A \in n \times n$? For example if $A= \left( \begin{array}{ccc} 1 & 1 & 1\\ 1 & i & i^2\\ 1 & i^2 & i^4 \\ \end{array} \right)$, then $A^{-1}=\frac{1}{3} \overline{A} $

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Hint: if $A$ is the diagonal matrix $\alpha I$, what is $A^{-1}$? –  TonyK Nov 27 '12 at 16:32
    
You said "$A\in n\times n$". I have never seen notations like this. Did you mean $A$ is an $n\times n$ matrix, or its entries are positive integers in $\{1,2,\ldots,n\}$, or something else? –  user1551 Nov 27 '12 at 17:49
    
I meant that $A = [a_{ij}]_{n \times n}$ –  laovultai Nov 27 '12 at 19:16
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Your matrix satisfies the polynomial $A^2 - nI = 0$. Therefore the minimal polynomial of your matrix is either $A \pm \sqrt{n}I = 0$ where the only matrices which satisfy the criteria are $A=\pm\sqrt{n}I$ or the minimal polynomial splits as $(A+\sqrt{n}I)(A-\sqrt{n}I) = 0$. In this case, you can take any matrix similar to a diagonal matrix with $\pm\sqrt{n}$ on the diagonals.

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followup question: give counterexample that $AA = A^2$( $A \in n \times n$) –  laovultai Nov 27 '12 at 16:54
    
@alvoutila: What is AA? –  Phira Nov 27 '12 at 16:56
    
$AA=A \cdot A$ where $\cdot$ is matrix product. –  laovultai Nov 27 '12 at 19:12
    
$AA$ is always $A^2$. $A^2$ is shorthand for $AA$ so I'm not sure what you mean. –  EuYu Nov 27 '12 at 19:41
    
to me $A^2= \left( \begin{array}{ccc} 1^2 & 1^2 & 1^2\\ 1^2 & i^2 & i^4\\ 1^2 & i^4 & i^8 \\ \end{array} \right) \neq A \cdot A $. –  laovultai Nov 27 '12 at 21:03
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The easiest way is to let $A$ be a multiple of the identity. These are easy to invert.

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Hadamard matrices all have this property.

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