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Thread: A matrix squared is the identity

  1. #1
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    A matrix squared is the identity

    Hello,

    Let $A$ be a $(n \times n)$ matrix over the complex numbers be such that $A^2 = I$. Show that $A$ is diagonalizable.

    So I know that $\det (A^2 - \lambda I) = \det (I - \lambda I) = (1-\lambda)^n$ and the minimal polynomial, call it $n(x)$, of $A^2$ must divide $(1-\lambda)^n$.

    Because 1 is the only complex number which solves the characteristic polynomial I also have $A^2(v) = \lambda v = 1 v = v$. $A^2 = I$ so $A^2$ should have rank $n$. So $n(x) = (1- \lambda)^n$?

    The standard way to show that $A$ is diagonalizable is to show that it's minimal polynomial is equal to its characteristic polynomial. I was trying to prove that the characteristic polynomial of $AB$ must be divisible by either the minimal or the characteristic polynomials of A and/or B but I came up with a counterexample immediately.

    My only other idea involves $A^2(v) = \lambda v$ implies $A(v) = \sqrt{\lambda} v$ but I also think that's false.

    If you think of $A$ as ranging over values in $M_n(\mathbb{C}$, then the set of solutions for $A^2 - I$ is obviously $\{-I, I\}$ but now the trouble is in showing that no other matrices would satisfy that equation.

    Does anyone have any advice?
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  2. #2
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    Re: A matrix squared is the identity

    the minimal polynomial must divide $x^2-1$

    The standard way to show that A is diagonalizable is to show that it's minimal polynomial is equal to its characteristic polynomial.
    Not True
    Last edited by Idea; Sep 21st 2019 at 10:40 PM.
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  3. #3
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    Re: A matrix squared is the identity

    I got a better reference (I think) and it should be,

    Let $\displaystyle A:V \to V$ be a linear operator. Then $A$ is diagonalizable if and only if its minimal polynomial in $F[T]$ splits in $F[T]$ and has distinct roots.

    The reference is 14 pages so I'll read that and then try again.
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  4. #4
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    Re: A matrix squared is the identity

    I found,

    Let $A$ be a linear transformation of finite order on a Complex vector space: if $A^m = \text{Id}_V$ for $m$ a positive integer, then $A$ is diagonalizable.

    The proof is,

    Since $A^m = \text{Id}_V$, $A$ is killed by $T^m - 1$, therefore the minimal polynomial of $A$ is a factor of $T^m -1$. The polynomial $T^m-1$ has distinct roots over $\mathbb{C}$.

    I can prove that the roots of unity in $\mathbb{C}$ are all distinct part so that's fine.

    How do I calculate $T^m -1$? This is the characteristic polynomial, yes? It should be $\det(A^m - \lambda I) = \det(I - \lambda I) = (1- \lambda)^n?$.

    Also, is there a way to compute the characteristic polynomial of a linear operator without finding a matrix representation?
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  5. #5
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    Re: A matrix squared is the identity

    A is a (nn) matrix

    The characteristic polynomial is of degree $n$ which is not necessarily the same as $m$
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