For any square matrix A, we can define sinA using

Brock Brown

Brock Brown

Answered question

2021-12-31

For any square matrix A, we can define sinA using the formal power series as follows:
sinA=n=0(1)n(2n+1)!A2n+1
Prove or disprove: there exists a 2×2 real matrix A such that
sinA=[1202001]

Answer & Explanation

deginasiba

deginasiba

Beginner2022-01-01Added 31 answers

The relation sin(A)2+cos(A)2=I also hold for matrices
So we would have to find a matrix for cosine such that C2=(0404000) which is not possible.
Juan Spiller

Juan Spiller

Beginner2022-01-02Added 38 answers

Notice that
B=[1202001]
is not diagonalizable. If it was 1 would be the only eigenvalue and B would be equal to the identity matrix.
Therefore if sinA=B, A isn't diagonalizable either. The (complex) Jordan normal form of A would be
A=[λ10λ]
And
sinA=[sinλcosλ0sinλ]
Hence λ belongs to π2+2πZ and
sinA=[sinλ00sinλ]=[1001]
is diagonalizable. A contradiction.
Vasquez

Vasquez

Expert2022-01-09Added 669 answers

A=[1101]B=An=[1n01]sinA=[sin1cos10sin1]
Use
sinA=n=0(1)nA2n+1(2n+1)!

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