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}4 - \lambda & 1\\2 & 3 - \lambda\end{matrix}\right] = 0

   Subtract λ from the diagonal.

3. \lambda^{2} - 7 \lambda + 10 = 0

   Expand the determinant.

4. \left(\lambda - 5\right) \left(\lambda - 2\right) = 0

   Factor.

5. \lambda = 5, \lambda = 2

   Eigenvalues (with multiplicity).

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

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

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

   Solve (A − 5I)v = 0 for a basis eigenvector.