# Trace of $A$ if $A =A^{-1}$

Suppose $I\neq A\neq -I$, where $I$ is the identity matrix and $A$ is a real $2\times2$ matrix. If $A=A^{-1}$ then what is the trace of $A$? I was thinking of writing $A$ as $\{a,b;c; d\}$ then using $A=A^{-1}$ to equate the positions but the equations I get suggest there is an easier way.

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What does the first sentence mean? –  Julian Kuelshammer Nov 1 '12 at 19:08
@JulianKuelshammer - Probabliy that $char(F)\neq 2$ –  Belgi Nov 2 '12 at 11:18
@Belgi now the appearence changed, the $\neq$ signs did not show up before –  Julian Kuelshammer Nov 2 '12 at 12:04

If ${\bf A} = {\bf A}^{-1}$ we may assume that $\det({\bf A}) \neq 0$. (Otherwise, what is ${\bf A}^{-1}$?) We can multiply the right-hand side of both sides of the equation by ${\bf A}$, giving:

$${\bf A} = {\bf A}^{-1} \iff {\bf A}{\bf A} = {\bf A}^{-1}{\bf A} \iff {\bf A}^2 = {\bf E} \, .$$ where ${\bf E}$ notes the $n \times n$ identity matrix. The characteristic polynomial of ${\bf A}$ is given by:

$$\det({\bf A}-\lambda {\bf E}) = \det \left[ \begin{array}{cc} a_{11}-\lambda & a_{12} \\ a_{21} & a_{22}-\lambda \end{array} \right] =(a_{11}-\lambda)(a_{22}-\lambda)-a_{12}a_{21}$$ $$=a_{11}a_{22} - \lambda a_{11} - \lambda a_{22} + \lambda^2 - a_{12}a_{21} = \lambda^2 - (a_{11} + a_{22})\lambda + (a_{11}a_{22} - a_{12}a_{21}) \, .$$ If $\text{tr}({\bf A})$ denotes the trace of ${\bf A}$, then $\det({\bf A}-\lambda {\bf E}) = \lambda^2 - \text{tr}({\bf A})\lambda + \det({\bf A}).$

We usually consider the eigenvalues, $\lambda$, given by $\det(A-\lambda E) = 0$. In other words, we usually consider the equation $\lambda^2 - \text{tr}({\bf A})\lambda + \det({\bf A}) = 0$. There is a great result due to Hamilton which says that if we replace the number $\lambda$ in the numerical equation $\det({\bf A}-\lambda {\bf E}) = 0$ by the matrix ${\bf A}$ then we get a matrix equation:

$${\bf A}^2 - \text{tr}(A){\bf A} + \det(A){\bf E} = {\bf 0} \, .$$

Next, we ask ourselves: what os $\det({\bf A})$? Well, since ${\bf A}^2 = {\bf E}$ it follows that $\det({\bf A}^2) = \det({\bf E})$ and, since $\det({\bf XY}) = \det({\bf X})\det({\bf Y})$, we have $\det({\bf A})^2 = \det({\bf E})$ which tells us $\det({\bf A})^2 = 1$. Clearly, $\det({\bf A}) = \pm 1$. Thence:

$${\bf A}^2 - \text{tr}(A){\bf A} \pm \det(A){\bf E} = {\bf 0} \implies {\bf E} - \text{tr}(A){\bf A} \pm {\bf E} = {\bf 0}\, .$$

At this point, we should separate into the $\pm$ cases:

$${\bf E} - \text{tr}(A){\bf A} - {\bf E} = {\bf 0} \ \ \text{or} \ \ {\bf E} - \text{tr}(A){\bf A} + {\bf E} = {\bf 0} \, .$$

In the first case, we get $\text{tr}(A){\bf A} = 0$. Since $\det({\bf A}) \neq 0$ it follows that $\text{tr}({\bf A}) = 0.$ In the second case, $-\text{tr}(A){\bf A} = -2{\bf E}$. This is only possible if ${\bf A} \propto {\bf E}$. For more detail, notice that:

$$-\text{tr}(A){\bf A} = -2{\bf E} \implies \text{tr}\left[-\text{tr}(A){\bf A}\right] = \text{tr}\left[-2{\bf E}\right] \implies \text{tr}(A)^2 = 4 \implies \text{tr}(A) = \pm 2.$$

It follows that the necessary conditions are that $\text{tr}(A) = -2, 0 , 2.$

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Let $t=\text{tr} A$. Since $A^2=I_2$, $\det A=\pm1$ and Cayley-Hamilton theorem yields $A^2-t\cdot A+\det A \cdot I_2=O_2$, that is, $I_2-t\cdot A\pm I_2=O_2$. Thus $t\cdot A=O_2$ or $t\cdot A=2I_2$. Taking the trace again, this implies that $t^2=0$ or $t^2=4$.

Hence $t=\text{tr} A$ is in $\{-2,0,+2\}$ and these three cases are realized by $A=\begin{pmatrix}\pm1 & 0\\ 0& \pm1\end{pmatrix}$.

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why are u putting zeros in the off-diagonal positions? –  Vaolter Nov 2 '12 at 13:37
Because these matrices work and because squares of diagonal matrices are easy to compute. –  Did Nov 2 '12 at 15:31

From $A=A^{-1}$ you will know that all the possible eigenvalues are $\pm 1$, so the trace of $A$ would only be $0$ or $\pm 2$. You may show that all these three cases are realizable.

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The condition $A\ne\pm I$ didn't show up when I was answering the question. With the additional assumption $A\ne\pm I$, the trace of $A$ must be $0$. –  23rd Nov 14 '13 at 15:23

Since $A=A^{-1}$, then $\mathrm{det}A=\pm 1$. This simplifies your approach.

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Let $J = V^{-1}A V$ be a Jordan normal form of $A$. Then $J = \mathbb{diag}(J_{\lambda_1,m_1},...,J_{\lambda_k,m_k})$, where $J_{\lambda,i}$ is a Jordan block with eigenvalue $\lambda$ and size $i \times i$.

Then from $A^2=I$ we have $J^2 = I$, from which we get $J_{\lambda_l,m_l}^2 = I$ for each block. It follows from this that $\lambda_l^2 = 1$, or equivalently, $\lambda_l = \pm 1$.

A brief computation shows that if $m_l>1$, then $[J_{\lambda_l,m_l}^2]_{12} = 2 \lambda_l \neq 0$, from which it follows that $m_l = 1$ for all $l$, and consequently, $J$ is diagonal with entries $\pm 1$.

Consequently, $A^2 = I$ iff $A$ has the form $V \mathbb{diag}(\pm 1,..., \pm 1)V^{-1}$.

Since the trace is invariant to similarity transformations, and the dimension is $2$, there are three possible values corresponding to the values $\{(\pm 1) + (\pm 1) \}$. However the values $\{-2, 2\}$ correspond to the matrices $-I$ and $I$ respectively, eliminating them leaves $0$ as the only possible value.

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Generalizing if $A^2=\operatorname{Id}$ we have that $\operatorname{tr}(A)= \dim(V_1)-\dim(V_{-1})$.

That's because $A$ is diagonalizable and trace is invariant for conjugacy.

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