Eigenvectors are directions preserved by a linear operator. Invariant subspaces are the broader objects: regions of a vector space on which the operator can be studied independently.
Invariant subspaces
Definition. Let \(T:V\to V\) be linear. A subspace \(U\subset V\) is \(T\)-invariant if \(T(U)\subset U\). If \(U\) is invariant, the restriction \(T|_U:U\to U\) is a linear operator, so one can study part of \(T\) inside \(U\).
Example. For any polynomial \(p\), the kernel \(\ker p(T)\) and image \(\operatorname{im}p(T)\) are \(T\)-invariant because \(T\) commutes with \(p(T)\). Eigenspaces, generalized eigenspaces, and kernels of powers \((T-\lambda I)^k\) are all invariant.
Eigenvalues and eigenspaces
Definition. A scalar \(\lambda\in F\) is an eigenvalue of \(T\) if there exists \(v\ne0\) such that \(Tv=\lambda v\). Such a vector is an eigenvector. The eigenspace is \[ E_\lambda=\ker(T-\lambda I). \]
Proposition. Eigenvectors belonging to distinct eigenvalues are linearly independent.
Proof. Suppose \(v_1, \ldots,v_r\) have distinct eigenvalues \(\lambda_1, \ldots, \lambda_r\), and choose a shortest nontrivial relation \(\sum_i a_iv_i=0\). Applying \(T-\lambda_r I\) removes the \(v_r\)-term and gives \(\sum_{i<r}a_i(\lambda_i- \lambda_r)v_i=0\), a shorter nontrivial relation unless all \(a_i=0\) for \(i<r\). Then also \(a_r=0\), contradiction.
Characteristic polynomial and multiplicity
Definition. For a matrix \(A\in M_n(F)\), the characteristic polynomial is \[ \chi_A(t)=\det(tI-A). \] The algebraic multiplicity of \(\lambda\) is its multiplicity as a root of \(\chi_A\). The geometric multiplicity is \(\dim E_\lambda\).
Theorem. A scalar \(\lambda\) is an eigenvalue of \(A\) if and only if \(\chi_A(\lambda)=0\). For each eigenvalue, \[ 1\le \dim E_\lambda\le \text{algebraic multiplicity of }\lambda. \]
Proof. The equation \(Av=\lambda v\) with \(v\ne0\) is equivalent to \((\lambda I-A)v=0\) having a nonzero solution, which is equivalent to \(\lambda I-A\) being singular, hence determinant zero. The geometric multiplicity bound follows by choosing a basis adapted to \(E_\lambda\); the matrix of \(A\) begins with a \(\lambda I\) block, so \((t- \lambda)^{\dim E_\lambda}\) divides \(\chi_A(t)\).
Real and complex fields
Counterexample. The real rotation matrix \[ R=\begin{pmatrix}0&-1\\1&0\end{pmatrix} \] has characteristic polynomial \(t^2+1\), so it has no real eigenvalue. Over \(\mathbb C\), it has eigenvalues \(i\) and \(-i\). Eigenvalue questions depend on the field.
Trace and determinant through eigenvalues
Proposition. If the characteristic polynomial of \(A\) splits with eigenvalues \(\lambda_1, \ldots,\lambda_n\) counted with algebraic multiplicity, then \[ \operatorname{tr}A=\sum_{i=1}^n\lambda_i, \qquad \det A=\prod_{i=1}^n\lambda_i. \] These identities hold even when \(A\) is not diagonalizable.
Proof. Over a splitting field, triangularize \(A\). The diagonal entries of the triangular matrix are the eigenvalues counted with algebraic multiplicity. Trace and determinant are unchanged by similarity, and for a triangular matrix they are the sum and product of diagonal entries.
Invariant subspaces from polynomials
Proposition. If \(p(T)q(T)=0\) and \(p,q\) are coprime, then \(V=\ker p(T)\oplus\ker q(T)\). This is the algebraic mechanism behind decomposing a space into spectral pieces when the characteristic polynomial factors.
Remark. Eigenvalues are not merely roots of a polynomial. They signal a decomposition problem: can the whole space be assembled from invariant one-dimensional or generalized invariant pieces?