# Is every self-inverse matrix diagonalizable?

If $A=A^{-1}$, is there always a matrix C such that $C^{-1}AC$ is a diagonal matrix (containing only -1 and 1 in the main diagonal) ?

How can I check with PARI/GP, if a given matrix is diagonalizable ?

I only found out that $A=A^{-1}$ implies $C^{-1}AC=(C^{-1}AC)^{-1}$ for any invertible C, but this does not answer my question.

Additional question : If A has integer entries, is there always a matrix C with INTEGER values such that $C^{-1}AC$ is a diagoal matrix and $C^{-1}$ has also integer
values ? For example, I did not find such a matrix for A = [ [11,-16,-4] [9,-13,-3] [-6,8,1] ] yet.

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OK, the theoretical part is done. What about PARI/GP, can I calculate diagonalizations with it ? – Peter Jun 17 '14 at 13:37
You didn't ask for diagonalizations; you asked if it was diagonalizable. – Eric Towers Jun 17 '14 at 13:39
No, I only want to check the diagonalizability, but if you know how I can do the diagonalization in PARI/GP, this would be wonderful. – Peter Jun 17 '14 at 13:40
(The answer to the diagonalizable question is below.) For this second question, assuming a is square "matdiagonal(mateigen(a,1)[1])". – Eric Towers Jun 17 '14 at 13:42
Duplicate of math.stackexchange.com/q/496338/18880 which was however closed. I think there are other duplicates, but cannot find them right now. Management summary: if the characteristic is not$~2$, this is OK. I characteristic$~2$ it fails miserably. – Marc van Leeuwen Jun 17 '14 at 14:15

You can rewrite that condition as $A^2 = I_2$. Therefore $A$ is a root of $X^2 - 1$.

• Case 1: the characteristic of the base field is not $2$ (for example $\mathbb{R}$, $\mathbb{C}$...), then this polynomial is split with simple roots, so $A$ is diagonalizable. The eigenvalues will be roots of $X^2 - 1$ so they will indeed only be either $1$ or $-1$.
• Case 2: the characteristic is $2$. Then $\bigl(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\bigr)^2 = I_2$ is a counterexample.
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$A$ is a root of the polynomial $f = x^{2} -1$, which has distinct roots. Thus its minimal polynomial, which divides $f$, has distinct roots. It follows that $A$ is diagonalizable.

This assumes that $1 \ne -1$. that is, the characteristic of the underlying field is not $2$. In characteristic $2$, the matrix $$A = \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}$$ satisfies $A = A^{-1}$, but it is not diagonalizable.

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Hint: in this case, $$A\times A=I\implies P(A) = 0, \\P = (X-1)(X+1)$$

I assume that $1+1\neq 0$. In that case, $X\pm 1$ are different factors.

Some theorem (Lemme des noyaux in french, does anyone know the english term?) states the second inequality in: $$E = \ker P(A) =\ker (A-I) \oplus \ker (A+I)$$

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+1 for giving the explicit diagonalization in terms of kernels. – Andreas Caranti Jun 17 '14 at 13:35
OK if characteristic is not $2,$ but does not work for characteristic $2$. – Geoff Robinson Jun 17 '14 at 13:38
yes, I have read the others answers. – mookid Jun 17 '14 at 13:40

Answer to second question: "matrank(mateigen(a)) = matsize(a)[1]", if you know the matrix a is square. If not, append "= matsize(a)[2]"

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According to PARI, [[0,-1],[0,1]] is a counterexample. The rank is 1 not equal matsize(a)[1] (=2), but [[1,-1],[0,1]] leads to the diagonal matrix [[0,0],[0,1]]. – Peter Jun 17 '14 at 13:54
@Peter: Which of the two matrices you wrote are you actually talking about? – Eric Towers Jun 17 '14 at 14:26
@Peter: Never mind, both things you wrote are incorrect. The rank of [[0,-1],[0,1]] is clearly 1 because the second row is a multiple of the first. Also the diagonalization of [[1,-1],[0,1]] is [[1,0],[0,1]]. – Eric Towers Jun 17 '14 at 14:28
I do not mean the diagonalization of [[1,-1],[0,1]], but using this matrix [[0,-1],[0,1]] is transformed to [[0,0],[0,1]]. – Peter Jun 17 '14 at 14:46
... which has rank one. No idea what you're getting at. – Eric Towers Jun 17 '14 at 14:47