Question

Let \[ J = \begin{pmatrix} 2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 3 \end{pmatrix}. \] Then the geometric multiplicity of the eigenvalue \(2\) of \(J\) is equal to ________. (Answer in integer)

Hint:

Write J - 2I and solve the homogeneous system directly, or note that J splits into 2x2 diagonal blocks and count how many of those blocks belong to eigenvalue 2.

Answer 1

Correct Answer: 2

Step 1: Understand what "geometric multiplicity" means.
The geometric multiplicity of an eigenvalue \(\lambda\) of a matrix \(A\) is the dimension of its eigenspace, that is the dimension of the null space of \(A - \lambda I\), or the number of linearly independent eigenvectors for that eigenvalue.

Step 2: Notice the block structure of J.
\(J\) is a \(6 \times 6\) upper triangular matrix that splits into three \(2 \times 2\) blocks along the diagonal, with all other entries zero:
\[ J_1 = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}, \quad J_2 = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}, \quad J_3 = \begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix} \] Here \(J_1\) and \(J_2\) are both \(2\times2\) Jordan blocks with eigenvalue \(2\), and \(J_3\) is a \(2\times2\) Jordan block with eigenvalue \(3\).

Step 3: Find the nullity of \(J - 2I\) block by block.
For a single \(2\times2\) Jordan block with eigenvalue \(2\), namely \(\begin{pmatrix}2 & 1\\0&2\end{pmatrix}\), subtracting \(2I\) gives \(\begin{pmatrix}0&1\\0&0\end{pmatrix}\), which has rank \(1\) and nullity \(2-1=1\).
So each of \(J_1\) and \(J_2\) contributes a nullity of \(1\) to the eigenspace of \(2\) (each block is one Jordan chain, giving exactly one independent eigenvector).
The block \(J_3\) has eigenvalue \(3\), not \(2\), so \(J_3 - 2I = \begin{pmatrix}1&1\\0&1\end{pmatrix}\) is invertible, contributing nullity \(0\) to the eigenspace of \(2\).

Step 4: Add up the contributions and give the final answer.
Since \(J\) is block diagonal, the nullity of \(J - 2I\) is the sum of the nullities of each block: \(1 + 1 + 0 = 2\).
So the geometric multiplicity of the eigenvalue \(2\) is \(2\), coming from two separate \(2\times2\) Jordan blocks, each giving one independent eigenvector.
\[ \boxed{2} \]