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$A$ is $n\times n$ matrix. How to prove whether it is true or false $$A^2=I \implies A=\pm I$$ I was trying on $2\times 2$ case...multiplying general entries and then equating them to the identity requirements...but it is not a proof...

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  • $\begingroup$ It means eigenvalues are $\large \pm 1$. $\endgroup$ – Felix Marin Nov 30 '13 at 19:31
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Hint: It is false. Complete the following matrix suitably: $\begin{bmatrix}\clubsuit & 0\\0 &\spadesuit\end{bmatrix}$.


More generally, given natural numbers $m,n$ and $A\in \Bbb C^{n\times }$ such that $A^m=I_n$, it is true that the eigenvalues of $A$ are $m$ roots of $1$, (not necessarily all of them).

Furthermore $A$ is necessarily diagonalizable.

This facts allows one to easily build counter examples for the statement $A^m=I_n\implies A=\pm I_n$, if $m\ge 2$. For instance, if $m=n$, let $A=\operatorname{diag}(e^{\large{2\pi i/m}}, e^{\large{4\pi i/m}}, \ldots,e^{\large{2m\pi i/m}})$.

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  • $\begingroup$ It's better that change one of the star's to another symbol in the matix in your hint. $\endgroup$ – R Salimi Nov 30 '13 at 19:21
  • $\begingroup$ @RSalimi I decided to follow your advice. $\endgroup$ – Git Gud Nov 30 '13 at 19:28
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For a slightly more geometric approach, consider the definition of a matrix as a linear transformation; a mapping $A : \mathbb{R}^n \rightarrow \mathbb{R}^n$ which fixes the origin (and has other properties: in general, $A(au+bv) = aA(u) + bA(v)$, where $u, v$ are vectors and $a, b$ are scalars). Since we are working with a $2 \times 2$ matrix, $n = 2$, and we are mapping two dimensional space (the plane) onto itself.

A reflection across the line $y = ax$ is a linear transformation in two dimensional space, and can be written as $$A = \left[ \begin{array} {cc} \frac{a^2 - 1}{a^2 + 1} & \frac{2a}{a^2 + 1} \\ \frac{2a}{a^2+1} & \frac{1-a^2}{a^2+1} \end{array}\right].$$ Clearly, applying a reflection twice will map a vector to itself, so $A^2 = I$. This can be verified algebraically, as well.

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Try $A=\begin{bmatrix}0&1\\1&0\end{bmatrix}$.

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  • 1
    $\begingroup$ A transposition! :) $\endgroup$ – Belgi Nov 30 '13 at 18:41
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Any time that you get back to where you started by doing an operation twice, you have the situation $A^2=I$. Such operations are called involutions. So, in the plane, reflection about any axis that passes through the origin will have this property. Turning your shirt inside-out is probably not a linear operation, so it won’t have a matrix, but it is an involution. These operations are everywhere, even though you can’t unring a bell.

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  • $\begingroup$ I'm not really sure what this answer means, but I like it; +1! $\endgroup$ – Robert Lewis Dec 1 '13 at 5:49
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Hint:

$$A=\begin{pmatrix}0&-1\\-1&\;0\end{pmatrix}$$

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  • $\begingroup$ Is there sort of a "way" to see the the counter case? $\endgroup$ – user7777777 Nov 30 '13 at 18:40
  • $\begingroup$ @user7777777 Search for the counter case starting from simple matrices, like this one or the one proposed by Git Gud. $\endgroup$ – hhsaffar Nov 30 '13 at 18:42
  • $\begingroup$ What "required transpose"?? $\endgroup$ – DonAntonio Nov 30 '13 at 18:48
  • $\begingroup$ But how does it lead me to the needed counter cases? $\endgroup$ – user7777777 Nov 30 '13 at 18:52
  • $\begingroup$ You have two direct answers where $A^2=I\rlap{\;\;\;\;/}\implies A=\pm I\;$ . These two are counter examples! $\endgroup$ – DonAntonio Nov 30 '13 at 18:55
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It is in fact quite easy and straightforward to work out and present a general solution to this problem, using some ordinary algebra and a tad of information about matrices, in the event that $n = 2$. In the following we shall derive general formulas for all solutions of $A^2 = I$ in the $2 \times 2$ case, and see in so doing that the answer to the question, "does $A^2 = I \Rightarrow A = \pm I$?" is most definitely negative. To wit:

If we set

$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}, \tag{1}$

then we find, upon writing out the entries of the matrix equation

$A^2 = I, \tag{2}$

that the following relationships hold amongst the $a_{ij}$:

$a_{11}^2 + a_{12}a_{21} = a_{22}^2 + a_{21}a_{12} = 1, \tag{3}$

$a_{11}a_{12} + a_{12}a_{22} = a_{21}a_{11} + a_{22}a_{21} = 0, \tag{4}$

and since the trace of the matrix $A$,

$\text{Tr}(A) = a_{11} + a_{22}, \tag{5}$

we see that (4) may be written

$a_{12} \text{Tr}(A) = a_{21}\text{Tr}(A) = 0. \tag{6}$

Since $\text{Tr}(A)$ is an important invariant of the matrix $A$, (6) suggests that we break down further analysis according to the value of $\text{Tr}(A)$; so suppose to start that

$\text{Tr}(A) \ne 0, \tag{7}$

then by (6) we have

$a_{12} = a_{21} = 0, \tag{8}$

and thus (3) becomes

$a_{11}^2 = a_{22}^2 = 1, \tag{9}$

whence

$a_{11} = \pm 1; \; \; a_{22} = \pm 1, \tag{10}$

and since (7) applies, we must in fact have

$a_{11} = a_{22} = \pm 1, \tag{11}$

so that in fact

$A = \pm I \tag{12}$

in the event that (7) holds. On the other hand, if we have

$\text{Tr}(A) = 0, \tag{13}$

then

$-a_{22} = a_{11} = \alpha, \tag{14}$

where we have introduced the parameter $\alpha$; next we observe that, by (3),

$a_{12}a_{21} = 1 - \alpha^2, \tag{15}$

which implies that for $\alpha^2 \ne 1$,

$a_{12} \ne 0 \ne a_{21}; \tag{16}$

we introduce a second parameter $\beta \ne 0$ in this case and set

$a_{12} = \beta; \tag{17}$

then from (15), (17):

$a_{21} = (1 - \alpha^2) / \beta; \tag{18}$

thus we see that in the event that (13) holds and $\alpha \ne 1$, $\beta \ne 0$, $A$ takes the general form

$A = \begin{bmatrix} \alpha & \beta \\ (1 - \alpha^2) / \beta & -\alpha \end{bmatrix}; \tag{19}$

in the event that $\text{Tr}(A) = 0$, $\alpha^2 = 1$, (15) shows that at least one of $a_{12}$, $a_{21}$ vanishes. In the event that both $a_{12}, a_{21} = 0$, then clearly $A$ takes one of the two forms $A_0$, with

$A_0 = \pm \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \tag{20}$

and the only remaining case is $\text{Tr}(A) = 0$, $\alpha^2 = 1$, with exactly one of $a_{12}$, $a_{21}$ zero. But then it is easy to see that $A$ may be expressed as

$A = A_0 + \beta N, \tag{21}$

with $\beta \ne 0$, where

$N = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \tag{22}$

or

$N = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}. \tag{23}$

the reader may easily verify that $A_0N + NA_0 = 0$ for $A_0, N$ as in (20), (22)-(23), and since $N^2 = 0$, we have

$A^2 = (A_0 + \beta N)^2 = A_0^2 + \beta (A_0N + NA_0) + \beta^2 N^2 = A_0^2 = I. \tag{24}$

Indeed, it is easy to see by direct computation that $A^2 = I$ for each of the cases presented above; there are in fact entire parametrized families of matrices $A \ne I$ such that $A^2 = I$.

Finally, I think it is worthy of remark that it is probably a fairly well-known fact that the matrix equation $A^2 = I$ has parametrized families of solutions $A$ with $A \ne I$; indeed, if we choose any nonsingular matrix $E$ with $EA \ne AE$, then $(EAE^{-1})^2 =EA^2E^{-1} = EIE^{-1} = I$, always; but $EA \ne AE$ implies $EAE^{-1} \ne A$, so families of such $E$ in a sense "generate" families of such $A$. In the event that $A$ is not a scalar multiple of $I$, such $E$ are known to exist.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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