Brock Brown

2021-12-31

For any square matrix A, we can define sinA using the formal power series as follows:

$\mathrm{sin}A=\sum _{n=0}^{\mathrm{\infty}}\frac{{(-1)}^{n}}{(2n+1)!}{A}^{2n+1}$

Prove or disprove: there exists a$2\times 2$ real matrix A such that

$\mathrm{sin}A=\left[\begin{array}{cc}1& 2020\\ 0& 1\end{array}\right]$

Prove or disprove: there exists a

deginasiba

Beginner2022-01-01Added 31 answers

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.

So we would have to find a matrix for cosine such that

Juan Spiller

Beginner2022-01-02Added 38 answers

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

$\stackrel{\u2015}{A}=\left[\begin{array}{cc}\lambda & 1\\ 0& \lambda \end{array}\right]$

And

$\mathrm{sin}\stackrel{\u2015}{A}=\left[\begin{array}{cc}\mathrm{sin}\lambda & \mathrm{cos}\lambda \\ 0& \mathrm{sin}\lambda \end{array}\right]$

Hence$\lambda$ belongs to $\frac{\pi}{2}+2\pi \mathbb{Z}$ and

$\mathrm{sin}\stackrel{\u2015}{A}=\left[\begin{array}{cc}\mathrm{sin}\lambda & 0\\ 0& \mathrm{sin}\lambda \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

is diagonalizable. A contradiction.

is not diagonalizable. If it was 1 would be the only eigenvalue and B would be equal to the identity matrix.

Therefore if

And

Hence

is diagonalizable. A contradiction.

Vasquez

Expert2022-01-09Added 669 answers

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