Calculation Results
1. Definition & formulas
The matrix sine of a square matrix $A$ is a matrix function defined by the usual Taylor series:
Alternative (complex-exponential) expression:
If $A$ is diagonalizable $A = VDV^{-1}$, then $\sin(A)=V\sin(D)V^{-1}$ where $\sin(D)$ applies $\sin(\cdot)$ to diagonal entries. The series always converges for every square matrix $A$.
2. Useful computation remarks
- Diagonal / diagonalizable matrices: compute $\sin$ on eigenvalues and transform back: $\sin(A)=V\sin(D)V^{-1}$.
- Special nilpotent matrices: if $N^m=0$ then series truncates and computation is finite.
- Powers that square to $-I$: if $A^2=-\alpha^2 I$ (rotation-like), then $\sin(A)=\frac{\sin(\alpha)}{\alpha}A$ (special case with $\alpha\neq0$).
- Numerical methods: use Schur decomposition or scaling-and-squaring with Padé approximants for stability in code.
3. Worked examples
Example 1 — diagonal matrix
Compute $\sin(A)$ for
Diagonal case: apply sine to diagonal entries
Input hint for calculator: enter matrix rows as π/2 0
and 0 π.
Example 2 — rotation-generator matrix
Let
Observe $A^2=-I$, so powers cycle: $A^{2k+1}=(-1)^k A$. Thus the series becomes
So
Input hint: enter 0 -1 and 1 0 for matrix
A.
4. Common mistakes
- Applying sine elementwise: $\sin(A)\neq(\sin(a_{ij}))$ in general — use powers or diagonalization.
- Assuming $\sin(A+B)=\sin(A)+\sin(B)$: matrix sine is not linear in that way (unless $A$ and $B$ commute and extra conditions hold).
- Truncating series carelessly: for numerical work, naive truncation can cause large errors; prefer stable algorithms.
- Using diagonalization without checking: not all matrices are diagonalizable; use Schur/Jordan-based methods if needed.
- Mishandling scaling factors: for $A^2=-\alpha^2 I$ use $\sin(A)=\frac{\sin(\alpha)}{\alpha}A$ (avoid division by zero when $\alpha=0$ — treat separately).
5. Practice problems (answers hidden)
Try these exercises. Click Show Answer to reveal the result.
Exercise 1
Show Answer
Because $A^2=-I$ and $\sin(A)=A\sin(1)$ (see Example 2).
Exercise 2
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Diagonal case: apply sine to each eigenvalue.
Exercise 3
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Here $A^2=0$, so the sine series truncates:
$$ \sin(A)=A=\begin{pmatrix}0 & 1\\[6pt]0 & 0\end{pmatrix}. $$Nilpotent matrices often yield finite polynomial results.
Exercise 4
Show Answer
Diagonal example with common angle values.
If you are exploring matrix sine, you may also want to check matrix rank, SVD, and RREF.
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