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Let $\mathrm{A}$ a $2 \times 2$ matrix such that $\mathrm{I}\neq\mathrm{A}\neq\mathrm{-I}$, where $\mathrm{I}$ is the $2 \times 2$ identity matrix. If $\mathrm{A}=\mathrm{A}^{-1}$, find the trace of $\mathrm{A}.$

N. F. Taussig
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Harold
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  • What have you tried? Please include your work and explain why you are stuck so that we can write a response appropriate to your skill level. – N. F. Taussig Jan 23 '15 at 23:27

2 Answers2

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$$A=A^{-1}\implies A^2=I\implies (A+I)(A-I)=0\implies m_A(x)=p_A(x)=(x+1)(x-1)$$

with $\;m_A\,,\,\,p_A\;$ the minimal and characteristic polynomials, resp., and thus our matrix is similar to the diagonal one

$$\begin{pmatrix}\!\!-1&0\\0&1\end{pmatrix}\implies \text{Tra}\,A=0$$

The result also follows directly from the first line if you know that Tr.$\,A$ is minus the linear coefficient of the characteristic polynomial.

Timbuc
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if $A = A^{-1}$ then the eigenvalues of $A$ also satisfy $\lambda = \frac{1}{\lambda}.$ that means $\lambda = \pm 1$ so the trace of $A$ is the sum of the eigenvalues which is zero.

we will deal with the case where $A \neq I$ and has eigenvalues $1,1.$ in this instance $A$ is similar to the nilpotent matrix $N = \pmatrix{1&1\\0&1}.$ but then $A \neq A^{1-}$ because $N \neq N^{-1}.$ the other case is similar.

abel
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