Megoldani a matematikai problémákat

Lépésről lépésre

1. \det(A - \lambda I) = 0

   Eigenvalues are the roots of the characteristic polynomial.

2. \det\left[\begin{matrix}- \lambda & 1\\-1 & - \lambda\end{matrix}\right] = 0

   Subtract λ from the diagonal.

3. \lambda^{2} + 1 = 0

   Expand the determinant.

4. \lambda = - i, \lambda = i

   Eigenvalues (with multiplicity).

5. \lambda = - i:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}i\\1\end{matrix}\right]

   Solve (A − - iI)v = 0 for a basis eigenvector.

6. \lambda = i:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}- i\\1\end{matrix}\right]

   Solve (A − iI)v = 0 for a basis eigenvector.