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# The eigenspace of a linear transformation is a nontrivial subspace.

Put content here**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., \(E_\lambda 
eq \{\mathbf{0}\}\)).
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**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.
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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.
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**Consequence:** The dimension of \(E_\lambda\) (called the *geometric multiplicity* of \(\lambda\)) is at least 1.

# Parents

* Eigenvalues and eigenvectors