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Last semester, I was following a Linear Algebra course and in the exam of that course, the following question was asked:

Of the following 5 matrices A, how many of them satisfy $A^4 = I$ ? $$\begin{bmatrix}1&0\\0&-1\end{bmatrix}, \begin{bmatrix}\frac{\sqrt2}{2}&\frac{\sqrt2}{2}\\-\frac{\sqrt2}{2}&\frac{\sqrt2}{2}\end{bmatrix}, \begin{bmatrix}\frac{\sqrt2}{2}&\frac{\sqrt2}{2}\\-\frac{\sqrt2}{2}&-\frac{\sqrt2}{2}\end{bmatrix}, \begin{bmatrix}\frac{1}{2}&-\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&\frac{1}{2}\end{bmatrix}, \begin{bmatrix}0&-1\\1&0\end{bmatrix} $$

I answered this question by doing all the calculations to the power of 4, and as you would expect, this took way too much time but I eventually figured out the correct answer which is 2 of them, which are the following matrices:

$$\begin{bmatrix}1&0\\0&-1\end{bmatrix} and \begin{bmatrix}0&-1\\1&0\end{bmatrix} $$

I'm taking the exam again and was wondering if there's a different and easier way to solve such a question?

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    $\begingroup$ A possible lead would to see if $det(A)^4 = 1$. But this may not give you the correct answer but can point in you in the right direction. $\endgroup$ – Good Morning Captain Aug 13 '18 at 8:04
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    $\begingroup$ The key is that all of them are rotational matrices. $\endgroup$ – Kenny Lau Aug 13 '18 at 8:05
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    $\begingroup$ Except the one in the middle. That one has rank 1 (duplicated column) so its fourth power cannot be the identity matrix. $\endgroup$ – Kenny Lau Aug 13 '18 at 8:06
  • $\begingroup$ @KennyLau Can you maybe elaborate on how that helps? $\endgroup$ – user481197 Aug 13 '18 at 8:11
  • $\begingroup$ @KennyLau I was about to point out that too, but according to wikipedia, the first one doesn't follow the rotation matrix pattern either. $\endgroup$ – F.Carette Aug 13 '18 at 8:13
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Here's how I would have approached the problem.

(1) Since this matrix is diagonal, we can calculate $$ \begin{bmatrix}1&0\\0&-1\end{bmatrix}^4=\begin{bmatrix}1^4&0\\0&(-1)^4\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix} $$ so this matrix satisfies $A^4=I$.

(2) For this matrix, I notice that $\sqrt{2}/2$ appears a lot, so my brain goes to the fact that $\sin(\pi/4)=\cos(\pi/4)=\sqrt{2}/2$. Thinking of this matrix as corresponding to a linear transformation in the plane, I took a look at the form for a rotation matrix: $$ \begin{bmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{bmatrix}. $$ This tells me that our matrix corresponds to a rotation by $-\pi/4$, so $A^4$ corresponds to composing this rotation by itself four times, resulting in rotating by $-\pi$. Since, in general, rotating a vector by $-\pi$ doesn't result in the same vector, it follows that $A^4\neq I$.

(3) Here I noticed that the bottom row is a scalar multiple of the top row, so the matrix cannot be invertible. You can justify this by realizing this means the rows are not linearly independent, or simply by computing the determinant and seeing that it's zero. Since the matrix is not invertible, it is impossible that $A^4=I$. We can see this in a few ways: first, the properties of determinants tell me that $A^4=I$ implies that $\det(A)^4=\det(I)=1$ and so $\det(A)\neq{0}$, which we know isn't true for our $A$ since it's not invertible; second, $A^4=I$ can be written as $A(A^3)=I$, and this would imply that $A$ is invertible with $A^{-1}=A^3$, but $A$ isn't invertible.

(4) Like with (2), I see $1/2$ and $\sqrt{3}/2$, so I think of rotation matrices and see that this matrix corresponds to a rotation by $\pi/3$. Again, since for a vector $v\in\mathbb{R}^2$, $A^4v$ is an application of $A$ to the vector $v$ four times, it follows that $A^4$ corresponds to a rotation by $4\pi/3$. Thus, $A^4v\neq v$ for a general $v$.

(5) Correcting my answer thanks to the comments below, if one did not wish to do the matrix multiplication, one could again realize that the matrix is a rotation matrix corresponding to $\theta=\pi/2.$ As applying this rotation four times to any vector results in a rotation by $2\pi$, or, equivalently, no rotation at all, we have that $A^4=I$.

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  • $\begingroup$ Great answer. Although it has been accepted, I have trouble understanding how you solved (5). You seem to use $({A^T})^4=I \implies A^4=I^T=I$. Does it always stand, or did I miss something? $\endgroup$ – F.Carette Aug 13 '18 at 8:51
  • $\begingroup$ Good question. I'll add some details. $\endgroup$ – Ryan Gibara Aug 13 '18 at 8:59
  • $\begingroup$ I just realised that the $(A^T)^n=(A^n)^T$ property can be trivially deduced from the product property $(AB)^T=B^{T}A^T$. My bad, and thanks for the added details $\endgroup$ – F.Carette Aug 13 '18 at 9:17
  • $\begingroup$ There, you got it. No problem! $\endgroup$ – Ryan Gibara Aug 13 '18 at 9:44
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    $\begingroup$ Careful! The transpose of matrix 5 is not diagonal: $\begin{bmatrix}0&-1\\1&0\end{bmatrix}^T = \begin{bmatrix}0&1\\-1&0\end{bmatrix}$. But you could use the fact that the square is $-I$. $\endgroup$ – Rob Aug 13 '18 at 10:56
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The first one is trivial: it is a diagonal matrix, so $\begin{bmatrix}1&0\\0&-1\end{bmatrix}^4 = \begin{bmatrix}1^4&0\\0&(-1)^4\end{bmatrix} = I_2$.

The middle one is eliminated: it has duplicated column, so it is singular, so its fourth power is also singular, but the identity matrix is not singular.

Let $R_\theta$ denote the rotational matrix of angle $\theta$.

The thing with rotational matrices is that $R_\varphi R_\psi = R_{\varphi + \psi}$, so $(R_\theta)^4 = R_{4\theta}$.

Then, the rest of the matrices are $R_{-45^\circ}$, $R_{60^\circ}$, $R_{90^\circ}$.

Therefore, their fourth powers are respectively $R_{-180^\circ}$, $R_{240^\circ}$, $R_{360^\circ}$, and only the last one is the identity matrix.

Therefore the answer is the first and the fifth one.

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A more algebraic approach: $A^4=I$ means $A$ is root of the polynomial $$f(X):=X^4-1=(X+1)(X-1)(X+i)(X-i).$$

This is true iff $f$ is a multiple of the minimum polynomial $m_A$ of $A$. Since $m_A$ divides the characteristic polynomial $p_A$ of $A$, in this case the characteristic polynomials are of degree $2$, and there is no scalar matrix (the only ones with minimum polynomial of degree at most $1$), in fact $m_A=p_A$ (modulo some unit). Thus we can check if $f(A)=0$ by checking if $p_A|f$.

Matrix 1: $p_A(X)=(X-1)(X+1)$ divides $f$, so $f(A)=0$.

Matrix 2: $p_A(X)=-\frac{\sqrt2}2(X^2-2X+2)$ does not divide $f$, so $f(A)\neq0$.

Matrix 3: $p_A(X)= -\frac{\sqrt2}2(X^2-2)$ does not divide $f$.

Matrix 4: $p_A(X)=-\frac12(X^2-2X+\sqrt3+1)$ does not divide $f$.

Matrix 5: $p_A(X)=X^2+1$ divides $f$.

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I don't have much to add to the accepted answer, but I wanted to suggest that this problem is relatively easy if you start by thinking about what a matrix means as a linear transform. Draw both column vectors and look at them as transforms to $\hat{i}$ and $\hat{j}$, and it should be clear which transforms will result in no change when applied 4 times.

The first leaves $\hat{i}$ unchanged and negates $\hat{j}$. Applying it a second time flips it back to the original. Flipping it 2 more times again leaves us at the original.

The second is a clockwise rotation by 45°. Doing this 4x only takes us halfway around the circle.

The third maps both basis vectors to the same point. So we lose a dimension, determinant is 0, matrix is not invertible, yadda yadda yadda, the point is: applying it 4x certainly isn't going to get us back to the original.

The fourth is another rotation, again by an angle that doesn't get us back to the original.

The fifth is the classic counterclockwise rotation by 90°. Applying it 4x takes us all the way around the circle.

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The second, the fourth and the fifth matrices are of the form $\left(\begin{smallmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{smallmatrix}\right)$, for some $\theta\in\mathbb R$. So, $\left(\begin{smallmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{smallmatrix}\right)^4=\operatorname{Id}_2$ if and only if $4\theta\in2\pi\mathbb Z(\iff 2\theta\in\pi\mathbb Z)$.

The first matrix is a diagonal one, and so it is easy to work with.

The third matrix has determinant $0$, and therefore cannot be a solution.

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  • $\begingroup$ The first one isn't a rotational matrix. $\endgroup$ – Kenny Lau Aug 13 '18 at 8:17
  • $\begingroup$ @KennyLau I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Aug 13 '18 at 8:19
  • $\begingroup$ @KennyLau The first matrix is a rotation matrix; an improper rotation if you will, but a rotation nevertheless. It is not in the special orthogonal group $\mathrm{SO}(2)$, but it is in the rotation group $\mathrm O(2)$. $\endgroup$ – AccidentalFourierTransform Aug 13 '18 at 13:53
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In this case for $x=4$ it is more easy to just work out the matrix multiplication but for higher values of $x$ I suggest trying to diagonalize the matrix i.e.:

$A = CDC^{-1}$ with $D = diag(\lambda_1,...,\lambda_n)$ so the diagonal matrix with eigenvalues on the diagonal (assuming $n$ eigenvalues here) and $C = (a_1,...,a_n)$ where $A a_i = \lambda_i a_i $ so $a_i$ are eigenvectors in $C$ and $C^{-1}$ denotes the inverse (of course you should check invertibility, for example through $det(C) \neq 0$.

Now we show what $A^x$ gives for $x \in \{2,3,4,...\}$:

$A^x = AA....A = CDC^{-1}CDC^{-1} .... CDC^{-1}= CD^x C^{-1}$ using $C^{-1}C=I$ with $I$ the identity matrix.

So you can just try to calculate $A^x = CD^xC^{-1}$ since you can compare this matrix to standard rotation and reflection matrices you can immediately see the rotation axis and line of reflection.

E.g. the first one, we see that using a standard basis $\{e_1,e_2\}$ for $\mathcal{R}^2$ we have that $R(e_1)=e_1, R(e_2)=-e_2$. This means that we reflect all $x \in \mathcal{R}^2$ around $span\{e_1\}$. This means $\lambda_1 = 1, \lambda_2 = -1$ and corresponding eigenvectors are $e_1,e_2$ respectively. Now you can plug these in the equation I explained and you will see that it gives $A^4 = D^4$ therefore you can see that $D=diag(1^4,(-1)^4)=diag(1,1)=I_{2 \times 2}$.

The rest of the matrices can be calculated using the same approach. Try to ‘see’ eigenvalues and eigenvectors from your matrix (compare to standard rotations and so on) and use the formula I gave to you. Hope it helps, greets!

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