oxidricasbt7

2022-12-20

Eigenvalues of a $2\times 2$ matrix A such that ${A}^{2}=I$

I have no idea where to begin.

I know there are a few matrices that support this claim, will they all have the same eigenvalues?

I have no idea where to begin.

I know there are a few matrices that support this claim, will they all have the same eigenvalues?

Kendall Cortez

Beginner2022-12-21Added 4 answers

If v is an eigenvector of A with eigenvalue $\lambda $, then

$v=Iv={A}^{2}v=A(Av)=A(\lambda v)=\lambda (Av)={\lambda}^{2}v.$

Thus, if $\lambda $ is an eigenvalue of A and ${A}^{2}=I$ then ${\lambda}^{2}=1$. This gives only two possibilities for $\lambda $, $\pm 1$.

Notice, that we never assumed that A is $2\times 2$. Indeed, if A is any square matrix and ${A}^{2}=I$ then the only possible eigenvalues of A are $\pm 1$.

$v=Iv={A}^{2}v=A(Av)=A(\lambda v)=\lambda (Av)={\lambda}^{2}v.$

Thus, if $\lambda $ is an eigenvalue of A and ${A}^{2}=I$ then ${\lambda}^{2}=1$. This gives only two possibilities for $\lambda $, $\pm 1$.

Notice, that we never assumed that A is $2\times 2$. Indeed, if A is any square matrix and ${A}^{2}=I$ then the only possible eigenvalues of A are $\pm 1$.

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