Question

The maximum number of linearly independent eigenvectors of the matrix
\(\begin{pmatrix} 1 & 1 & 0 & 0 \\ 2 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 1 & 3 \\ \end{pmatrix}\)
is equal to _________.

Answer 1

Correct Answer: 3

To determine the maximum number of linearly independent eigenvectors of the given matrix, we need to find its eigenvalues and their respective multiplicities. Consider the matrix:
\(\begin{pmatrix} 1 & 1 & 0 & 0 \\ 2 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 1 & 3 \\ \end{pmatrix}\).

The characteristic polynomial is computed as \(\det(A-\lambda I)\).
For the matrix, \(\lambda I = \begin{pmatrix} \lambda & 0 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{pmatrix}\),
so \(A - \lambda I = \begin{pmatrix} 1-\lambda & 1 & 0 & 0 \\ 2 & 2-\lambda & 0 & 0 \\ 0 & 0 & 3-\lambda & 0 \\ 0 & 0 & 1 & 3-\lambda \end{pmatrix}\).

Calculating the determinant, we get:
\(\det(A-\lambda I) = ((1-\lambda)(2-\lambda) - 2) \cdot ((3-\lambda)^2) = 0\).
Thus, \((\lambda^2 - 3\lambda) \cdot ((3-\lambda)^2) = 0\).
The roots (eigenvalues) are \(\lambda = 0\) with algebraic multiplicity 1 and \(\lambda = 3\) with algebraic multiplicity 3.

The geometric multiplicity is determined by finding nullity of \((A-\lambda I)\) for each eigenvalue.
For \(\lambda = 0\), we solve \(A - 0 \cdot I\), yielding a nullity of 1.
For \(\lambda = 3\), solve \(A - 3I\) with resulting:
\(\begin{pmatrix} -2 & 1 & 0 & 0 \\ 2 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}\), giving nullity 2.

Thus, the maximum number of linearly independent eigenvectors is the sum of geometric multiplicities: \(1 + 2 = 3\).

This value lies within the given range [3,3]. Hence, the solution is consistent and correct: 3.