[Question]:
Suppose A is a $2\times 2$ matrix, $A^2=I$.
If A $\ne$ $I,-I$ , find $tr(A)$ and $det(A)$.
[My solution]:
There are $3$ cases for the eigenvalues of A.
case(i) $1,1$.
case(ii) $-1,-1$.
case(iii) $1,-1$.
I guess only the case(iii) holds.
But I have no idea whether case(i) and (ii) hold.
I know that $tr(A)$ is the sum of the sum of eigenvalues and $det(A)$ is the product of eigenvalues.
So the answer is $tr(A)=0$ , $det(A)=-1$ ?
Thanks.
The minimal polynomial of $A$ is a factor of $X^2-1$ and so there are three possibilities:
$X-1$, and then $A=I$.
$X+1$, and then $A=-I$.
$X^2-1$, and then it coincides with the characteristic polynomial.
By hypothesis, the only possibility is the last one.
Since the characteristic polynomial of a $2\times2$ matrix is $X^2-\operatorname{tr}(A)X+\det(A)$, we conclude that $\operatorname {tr}(A)=0$ and $\det(A)=-1$.
We can avoid talking about minimal polynomials as follows.
From $A^2=I$ and $A^2-\operatorname{tr}(A)A+\det(A)I=0$ we get $\operatorname{tr}(A)A=(1+\det(A))I$. If $\operatorname{tr}(A)\ne0$, then $A=\lambda I$ and so $\lambda=\pm1$, which is not allowed. Therefore, $\operatorname {tr}(A)=0$ and so $\det(A)=-1$.
Let $J$ be the Jordan normal form of your matrix, so there exists a matrix $P$ such that $PAP^{-1} = J$. Then $J^2 = PA^2P^{-1} = PP^{-1} = I$. Since the Jordan normal form retains the trace and determinant, it suffices to consider these.
We indeed know that the eigenvalues must be $\pm 1$. Since $A \notin \{I,-I\}$, there are only $3$ possible Jordan normal forms $J$: $$\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}, \begin{pmatrix}-1 & 1\\ 0 & -1\end{pmatrix}, \begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix}$$ However, the square of the first matrix is $$ \begin{pmatrix}1 & 2\\ 0 & 1\end{pmatrix}$$ and the square of the second matrix is $$ \begin{pmatrix}1 & -2\\ 0 & 1\end{pmatrix}$$ So the only Jordan normal form that works is the third one, which has $\det(A) = -1$ and $\operatorname{tr}(A) = 0$.
Since $A$ is a "small" matrix, I find easier a brute-force approach than using big theorems:
Let $$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ Then $$A^2=\begin{pmatrix}a^2+bc&ab+bd\\ac+cd&bc+d^2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$$
Suppose that $a+d=\text{Tr}(A)\neq 0$. Then, from cells $(1,2)$ and $(2,1)$ we get $b=c=0$. Now, from the main diagonal we get $a^2=d^2=1$, and since $A$ is not $I$ nor $-I$, we get that $ad=-1$, a contradiction. So $\text{Tr}(A)=0$, that is, $a=-d$.
Now $$\det(A)=ad-bc=-a^2-bc=-1$$
For diagonalisable $A$, you always have $$ A=UDU^{-1} $$ for a diagonal matrix $D$.
If $\lambda_1=\lambda_2=1$, then $$ D=1 $$ which would imply $A=1$.
By a similar line of reasoning, if $\lambda_1=\lambda_2=-1$, then $$ D=-1 $$ which would imply $A=-1$.
As you know that $1\neq A\neq -1$, the rest of the solution in the OP follows.
Because $A \ne I$ and $A \ne -I$, then there are vectors $x_1$, $x_2$ such that $$ y_1= (A-I)x_1 \ne 0 \\ y_2 = (A+I)x_2 \ne 0. $$ Because $A^2=I$, it follows that $(A+I)y_1=0$ or $Ay_1=-y_1$. Likewise $Ay_2=y_2$. So the vectors $y_1,y_2$ are a basis of eigenvectors of $A$ with eigenvalues $\pm 1$. So $A$ is similar to $\left[\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right]$, which gives you $\mbox{tr}(A)=0$, $\mbox{det}(A)=-1$.
From Cayley-Hamilton theorem we have that $ A^{2}-tr(A)\cdot A+det(A)\cdot I_{2}=0\Leftrightarrow I_{2}-tr(A)\cdot A+det(A)\cdot I_{2}=0 (*) $.
$ A^{2}=I_{2}\Leftrightarrow det(A^{2})=det(I_{2})\Leftrightarrow (det(A))^{2}=1 $, so $ det(A)=\pm 1 $.
Suppose $ det(A)=1 $. From $ (*) $ we get $ tr(A)\cdot A=2I_{2}\Leftrightarrow tr(tr(A)\cdot A)=tr(2I_{2})\Leftrightarrow tr(A)\cdot tr(A)=2tr(I_{2})\Leftrightarrow (tr(A))^{2}=4, $ hence $ tr(A)=\pm2 $.
If $ tr(A)=2 $ we obtain that $ 2A=2I_{2}\Leftrightarrow A=I_{2} $ , contradiction.
If $ tr(A)=-2 $ we obtain that $ 2A=-2I_{2}\Leftrightarrow A=-I_{2} $ , contradiction.
So $ det(A)=-1 $ which implies $ -tr(A)\cdot A=0 \Leftrightarrow tr(A)=0 $.