Answer to Question #274154 in Linear Algebra for Dhruv rawat

Question #274154

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

1
Expert's answer
2021-12-07T09:57:40-0500

Counterexample

Let A=[100020002].A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{bmatrix}. The matrix AA is diagonal, hence diagonisable.

Find the eigenvalues of AA


AλI=[1λ0002λ0002λ]A-\lambda I=\begin{bmatrix} 1-\lambda & 0 & 0 \\ 0 & 2-\lambda & 0 \\ 0 & 0 & 2-\lambda \\ \end{bmatrix}

det(AλI)=1λ0002λ0002λ\det(A-\lambda I)=\begin{vmatrix} 1-\lambda & 0 & 0 \\ 0 & 2-\lambda & 0 \\ 0 & 0 & 2-\lambda \\ \end{vmatrix}

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

λ1=1,λ2=λ3=2.\lambda_1=1, \lambda_2=\lambda_3=2.

The matrix AA has eigenvalues 1,2,21,2,2 (not all distinct), but is diagonisable.


False.


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