Question

If some eigenvalues of a matrix are repeated, the matrix is not diagonisable.true or false with full explanation

Answer 1

Counterexample

Let $A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{bmatrix}.$ The matrix $A$ is diagonal, hence diagonisable.

Find the eigenvalues of $A$

$$A-\lambda I=\begin{bmatrix} 1-\lambda & 0 & 0 \\ 0 & 2-\lambda & 0 \\ 0 & 0 & 2-\lambda \\ \end{bmatrix}$$

$$\det(A-\lambda I)=\begin{vmatrix} 1-\lambda & 0 & 0 \\ 0 & 2-\lambda & 0 \\ 0 & 0 & 2-\lambda \\ \end{vmatrix}$$

$$=(1-\lambda)(2-\lambda)^2=0$$

$\lambda_1=1, \lambda_2=\lambda_3=2.$

The matrix $A$ has eigenvalues $1,2,2$ (not all distinct), but is diagonisable.

False.