The matrix A below has charachteristic polynomial (λ+1)2(λ-2)2. Compute bases for the(+2)eigenspace and the (-1)eigenspace and explain why A is not diagonalizable
Hello,
As in your previous thread, solve for X $\displaystyle \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}$ in :
$\displaystyle AX= \lambda X$ (definition of eigenvalue, that will give you eigenspaces for each eigenvalues)
So solve for X in $\displaystyle AX=2X$ -> (+2) eigenspace
$\displaystyle AX=-X$ -> (-1) eigenspace
If the dimension of the (+2) eigenspace plus the dimension of the (-1) eigenspace is inferior to 4, then the matrix is not diagonalizable.