The eigenspace of a linear transformation is a nontrivial subspace.
Created almost 9 years ago, updated 5 months ago
ID24051
Theorem: For any eigenvalue (\lambda) of a linear transformation (T: V o V), the eigenspace (E_\lambda = \ker(T - \lambda I)) is a nontrivial subspace of (V) (i.e., (E_\lambda
eq {\mathbf{0}})).
Proof: Since (\lambda) is an eigenvalue, by definition there exists a nonzero vector (\mathbf{v}) such that (T(\mathbf{v}) = \lambda \mathbf{v}), which means ((T - \lambda I)(\mathbf{v}) = \mathbf{0}), so (\mathbf{v} \in \ker(T - \lambda I) = E_\lambda). Thus (E_\lambda) contains at least one nonzero vector.
Since the kernel of any linear transformation is a subspace, (E_\lambda) is a subspace. Combined with the existence of a nonzero vector, it is a nontrivial subspace.
Consequence: The dimension of (E_\lambda) (called the geometric multiplicity of (\lambda)) is at least 1.
**Theorem:** For any eigenvalue \(\lambda\) of a linear transformation \(T: V o V\), the eigenspace \(E_\lambda = \ker(T - \lambda I)\) is a nontrivial subspace of \(V\) (i.e., \(E_\lambda
eq \{\mathbf{0}\}\)).
**Proof:** Since \(\lambda\) is an eigenvalue, by definition there exists a nonzero vector \(\mathbf{v}\) such that \(T(\mathbf{v}) = \lambda \mathbf{v}\), which means \((T - \lambda I)(\mathbf{v}) = \mathbf{0}\), so \(\mathbf{v} \in \ker(T - \lambda I) = E_\lambda\). Thus \(E_\lambda\) contains at least one nonzero vector.
Since the kernel of any linear transformation is a subspace, \(E_\lambda\) is a subspace. Combined with the existence of a nonzero vector, it is a nontrivial subspace.
**Consequence:** The dimension of \(E_\lambda\) (called the *geometric multiplicity* of \(\lambda\)) is at least 1.