Exercise · ZW-EIGENSPACE-009 · ★★★ Advanced
The characteristic polynomial of a \(3\times 3\) matrix is \((\lambda-2)^2(\lambda-5)\). What is the dimension of the eigenspace of \(\lambda=5\)?
Choose one option.
Show answer
- C. \(1\)
Explanation
The dimension of \(E_\lambda\) is at least 1 and at most the multiplicity of the root; for \(\lambda=5\) (a simple root) it can only be 1.
Review the concept: Eigenspace
The eigenspace \(E_\lambda\) of a square matrix \(A\) for the eigenvalue \(\lambda\) is zero together with the eigenvectors of \(\lambda\), where \(I\) is the identity of the size of \(A\).
\[E_\lambda=\ker(A-\lambda I)\]
More practice problems: eigenspace
- ZW-EIGENSPACE-001 Let [4 1; 2 3]. Which vector spans the eigenspace of λ=5?
- ZW-EIGENSPACE-002 Let A=diag(2,2,3). What is the dimension of the eigenspace of λ=2?
- ZW-EIGENSPACE-003 Let A=3I of size 2× 2. What is the eigenspace of λ=3?
- ZW-EIGENSPACE-004 Let [2 1; 0 2]. What is the dimension of the eigenspace of λ=2?
- ZW-EIGENSPACE-005 Which statement about the eigenspace E_λ is true?
- ZW-EIGENSPACE-006 Let [1 2; 2 4]. A nonzero vector lies in the eigenspace of λ=0 if and only if it is a multiple of:
- ZW-EIGENSPACE-007 Let [1 1 0; 0 1 0; 0 0 2]. What is the dimension of the eigenspace of λ=1?
- ZW-EIGENSPACE-008 Let A=diag(1,2) and v=(1,1), the sum of an eigenvector for λ=1 and one for λ=2. Is v an eigenvector…