Diagonalizability and Repeated Eigenvalues

00:00 5 marks

Consider the matrix \(A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}\).

(a) Find the eigenvalues of \(A\) by solving its characteristic equation.

(b) Determine whether \(A\) is diagonalizable, and explain how the multiplicity of each eigenvalue influences your conclusion.

Part a

4 minutes 3 marks

Question

Find the eigenvalues of \(A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}\) by solving the characteristic equation \(\det(A – \lambda I) = 0\).

Hint Hint

Hint

Compute \(\det\begin{pmatrix} 2-\lambda & 1 & 0 \\ 0 & 2-\lambda & 0 \\ 0 & 0 & 3-\lambda \end{pmatrix}\) exploiting the upper triangular block structure.

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Complete Answer

The eigenvalues are \(\lambda = 2\) (with algebraic multiplicity 2) and \(\lambda = 3\).

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Part b

4 minutes 2 marks

Question

Determine whether \(A\) is diagonalizable by comparing the algebraic and geometric multiplicities of its eigenvalues. Justify your answer.

Hint Hint

Hint

For \(\lambda = 2\), find the dimension of the null space of \(A-2I\). Compare this geometric multiplicity with the algebraic multiplicity.

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Complete Answer

The matrix \(A\) is not diagonalizable because the geometric multiplicity of the eigenvalue \(\lambda = 2\) is less than its algebraic multiplicity.

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