Brock Brown

2021-12-31

For any square matrix A, we can define sinA using the formal power series as follows:
$\mathrm{sin}A=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}}{\left(2n+1\right)!}{A}^{2n+1}$
Prove or disprove: there exists a $2×2$ real matrix A such that
$\mathrm{sin}A=\left[\begin{array}{cc}1& 2020\\ 0& 1\end{array}\right]$

deginasiba

The relation ${\mathrm{sin}\left(A\right)}^{2}+{\mathrm{cos}\left(A\right)}^{2}=I$ also hold for matrices
So we would have to find a matrix for cosine such that ${C}^{2}=\left(\begin{array}{cc}0& -4040\\ 0& 0\end{array}\right)$ which is not possible.

Juan Spiller

Notice that
$B=\left[\begin{array}{cc}1& 2020\\ 0& 1\end{array}\right]$
is not diagonalizable. If it was 1 would be the only eigenvalue and B would be equal to the identity matrix.
Therefore if $\mathrm{sin}A=B$, A isn't diagonalizable either. The (complex) Jordan normal form of A would be
$\stackrel{―}{A}=\left[\begin{array}{cc}\lambda & 1\\ 0& \lambda \end{array}\right]$
And
$\mathrm{sin}\stackrel{―}{A}=\left[\begin{array}{cc}\mathrm{sin}\lambda & \mathrm{cos}\lambda \\ 0& \mathrm{sin}\lambda \end{array}\right]$
Hence $\lambda$ belongs to $\frac{\pi }{2}+2\pi \mathbb{Z}$ and
$\mathrm{sin}\stackrel{―}{A}=\left[\begin{array}{cc}\mathrm{sin}\lambda & 0\\ 0& \mathrm{sin}\lambda \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

Vasquez

$A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]⇒B={A}^{n}=\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]⇒\mathrm{sin}A=\left[\begin{array}{cc}\mathrm{sin}1& \mathrm{cos}1\\ 0& \mathrm{sin}1\end{array}\right]$
Use
$\mathrm{sin}A=\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{A}^{2n+1}}{\left(2n+1\right)!}$

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