Question

a. Let \(A\) be a diagonalizable \(n \times n\) matrix. Show that if the multiplicity of an eigenvalue \(\lambda \) is \(n\), then \(A = \lambda I\).

b. Use part (a) to show that the matrix \(A =\left({\begin{aligned}{*{20}{l}}3&1\\0&3\end{aligned}}\right)\) is not diagonalizable.

Short answer

a. It is proved that if the multiplicity of an eigenvalue \(\lambda \)is \(n\), then\(A = \lambda I\).

b. By part (a), if A would be diagonalizable then A would be \(3I\). Since \(A \ne 3I\)It is not diagonalizable.

Step-by-step solution

Showing that if the multiplicity of an eigenvalue

If\(A\)is diagonalizable, then\(A = PD{P^{ - 1}}\), for some invertible matrix\(P\)and diagonal matrix\(D\), where entries in\(D\)are eigenvalues of\(A\).

Since multiplicity of\(\lambda \)is\(n\)and \(\lambda \)appears\(n\)times in every diagonal entry in\(D\).

In other words,\(D = \lambda I\), so:

\(\begin{aligned}{c}A &= PD{P^{ - 1}}\\ &= P\lambda I{P^{ - 1}}\\ &= \lambda PI{P^{ - 1}}\\ &= \lambda I\end{aligned}\)

Hence \(A = \lambda I\).

It is proved that if the multiplicity of an eigenvalue \(\lambda \)is \(n\), then\(A = \lambda I\).

Showing that the matrix \(A\) not diagonalizable

The matrix is triangular, so its eigenvalues are on its diagonal.

Only eigenvalue is 3 with multiplicity 2.

By part a), if\(A\)would be diagonalizable then\(A\)would be\(3I\).

Since \(A \ne 3I\), it is not diagonalizable.