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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1Hint: if $A$ is the diagonal matrix $\alpha I$, what is $A^{-1}$? – TonyK Nov 27 '12 at 16:32
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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
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I meant that $A = [a_{ij}]_{n \times n}$ – Nov 27 '12 at 19:16
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The easiest way is to let $A$ be a multiple of the identity. These are easy to invert.
Ross Millikan
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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.
EuYu
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$AA$ 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
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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 $. – Nov 27 '12 at 21:03
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Ok, please do not re-define existing notation without at least telling us what it means. There is no one who could've possibly guessed what you wanted $A^2$ to be. If you want your $"A^2"$ to equal $A$, then each entry needs to satisfy $a^2 = a$. Each entry therefore needs to be $0$ or $1$. – EuYu Nov 27 '12 at 21:55
