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

Question

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

deginasiba · Accepted Answer

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=(0−404000) which is not possible.

Juan Spiller · Accepted Answer

Notice thatB=[1202001]is not diagonalizable. If it was 1 would be the only eigenvalue and B would be equal to the identity matrix.Therefore if sin⁡A=B, A isn&#039;t diagonalizable either. The (complex) Jordan normal form of A would beA―=[λ10λ]Andsin⁡A―=[sin⁡λcos⁡λ0sin⁡λ]Hence λ belongs to π2+2πZ andsin⁡A―=[sin⁡λ00sin⁡λ]=[1001]is diagonalizable. A contradiction.

Vasquez · Accepted Answer

A=[1101]⇒B=An=[1n01]⇒sin⁡A=[sin⁡1cos⁡10sin⁡1] Use sin⁡A=∑n=0∞(−1)nA2n+1(2n+1)!

For any square matrix A, we can define sinA using

Answered question

Answer & Explanation

New Questions in Trigonometry