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}2 - \lambda & 0 & 0\\0 & 3 - \lambda & 4\\0 & 4 & 9 - \lambda\end{matrix}\right] = 0

   Subtract λ from the diagonal.

3. - \lambda^{3} + 14 \lambda^{2} - 35 \lambda + 22 = 0

   Expand the determinant.

4. - \left(\lambda - 11\right) \left(\lambda - 2\right) \left(\lambda - 1\right) = 0

   Factor.

5. \lambda = 2, \lambda = 11, \lambda = 1

   Eigenvalues (with multiplicity).

6. \lambda = 1:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}0\\-2\\1\end{matrix}\right]

   Solve (A − 1I)v = 0 for a basis eigenvector.

7. \lambda = 2:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}1\\0\\0\end{matrix}\right]

   Solve (A − 2I)v = 0 for a basis eigenvector.

8. \lambda = 11:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}0\\\frac{1}{2}\\1\end{matrix}\right]

   Solve (A − 11I)v = 0 for a basis eigenvector.