Practice Final #2
davidbendebury
Part A. \\ The determinant expression is equal to:\\ (1-λ)\left| \begin{array}{ccc} 1-λ & 1 \\ 0 & 1-λ \end{array} \right| = (1-λ)(1-λ)(1-λ) \\ Therefore, the only eigenvalue is λ=1. Since the term it comes from is raised to a power of 3, it has an algebraic multiplicity of 3. Plugging this eigenvalue into ker(A-λI)=0, we find that only one vector can solve this. So the geometric multiplicity is 1.\\ Part B.\\ The determinant of this matrix is simply the diagonal values multiplied together. This makes it (1-λ)(2-λ)(3-λ)(4-λ)(5-λ). So the eigenvalues are 1,2,3,4,5.\\ Since there are five eigenvectors and this matrix has five columns, it is diagonalizable.
omar77w: June 5, 2015, 8:39 p.m.
Yup that's what I got.