Fall 2026: Math 590 Quiz 6 Solutions

Problem 1

Let \(A\) be an \(n\times n\) matrix over \(\mathbb{R}\) and suppose \(\lambda\) is an eigenvalue of \(A\). Define the algebraic and geometric multiplicities of \(\lambda\). (5 points)

Solution. The characteristic polynomial \(p_A(x) = \det(xI - A)\) has \(\lambda\) as a root, so we can write

p_A(x) = (x - \lambda)^{e}\, q(x), \qquad q(\lambda) \neq 0.

The algebraic multiplicity of \(\lambda\) is this exponent \(e\), i.e., the number of times \((x - \lambda)\) occurs as a factor of \(p_A(x)\).

The eigenspace of \(\lambda\) is \(E_\lambda = \mathrm{nullspace}(A - \lambda I) = \{ v \in \mathbb{R}^n : Av = \lambda v\}\). The geometric multiplicity of \(\lambda\) is \(\mathrm{dim}\, E_\lambda\), the maximum number of linearly independent eigenvectors for \(\lambda\). Equivalently, it is \(n - \mathrm{rank}(A - \lambda I)\).

Problem 2

Determine whether or not the matrix \(A = \begin{pmatrix} 2 & 2\\0 & 2\end{pmatrix}\) is diagonalizable. You must fully justify your answer. (5 points)

Solution. \(A\) is upper triangular, so

p_A(x) = \det\begin{pmatrix} x-2 & -2\\0 & x-2\end{pmatrix} = (x-2)^2.

Thus \(\lambda = 2\) is the only eigenvalue, with algebraic multiplicity \(2\).

To find the eigenspace, solve \((A - 2I)v = 0\):

A - 2I = \begin{pmatrix} 0 & 2\\0 & 0\end{pmatrix}.

This says \(2y = 0\), so \(y = 0\) and \(x\) is free. Hence

E_2 = \ker(A - 2I) = \mathrm{Span}\left\{ \begin{pmatrix} 1\\0\end{pmatrix} \right\}, \qquad \mathrm{dim}\, E_2 = 1.

So the geometric multiplicity of \(\lambda = 2\) is \(1\), which is less than its algebraic multiplicity \(2\). Every eigenvector of \(A\) is a multiple of \((1,0)^T\), so there is no basis of \(\mathbb{R}^2\) consisting of eigenvectors of \(A\). Therefore \(A\) is not diagonalizable.

(Alternatively: if \(A = PDP^{-1}\) with \(D\) diagonal, the diagonal entries of \(D\) are the eigenvalues of \(A\), so \(D = 2I\) and \(A = P(2I)P^{-1} = 2I\). But \(A \neq 2I\), a contradiction.) \(\square\)