Assertion

Suppose \(V\) is a finite-dimensional vector space over \(\mathbf{C}\) with \(dimV = n\). Suppose \(T \in \mathcal{L}(V)\). Therefore, \(\forall i \leq n, \exists U \subset V\) such that \(dimU = i\) and \(U\) is invariant under \(T\).

Proof

Suppose \(V\) is a finite-dimensional vector space over \(\mathbf{C}\) with \(dimV = n\). Suppose \(T \in \mathcal{L}(V)\).

By Induction,

Base Case:

Suppose \(dimV = 1\). \(V \subset V\) and \(V\) is invariant under \(T\) by \(T \in \mathcal{L}(V)\). This demonstrates the conclusion for the case \(dimV = 1\).

Inductive Case:

Assume \(\forall V\) such that \(dimV \leq n\) the conclusion holds. Suppose \(V\) is a finite dimensional vector space over \(\mathbf{C}\) such that \(dimV = n + 1\), and \(T \in \mathcal{L}(V)\). Because \(T \in \mathcal{L}(V)\) and \(V\) is a complex vector space, \(\exists \lambda \in \mathcal{C}\) such that \(\lambda\) is an eigenvalue of \(V\).

Let \(U = range(T - \lambda I)\)

Suppose \(u \in U\). \(\implies Tu = T(u) = Tu - \lambda u + \lambda u = (T - \lambda I)u + \lambda u).\)

\((T- \lambda I) u \in range(T- \lambda I) \implies (T - \lambda I ) u \in U\).

\(\lambda u \in U\) because \(range(T - \lambda I)\) is vector space and, thus, closed under scalar multiplication.

Therefore, \(T(u) \in U\).

\(\implies T_{\|U} \in \mathcal{L}(U)\).

\(U\) is a subspace of \(V\) which is finite dimensional. \(\implies \exists u_{1},...,u_{j} \in U \subset V\) such that \((u_{1},...,u_{j})\) is a basis of \(U\). \(T - \lambda I\) is not surjective because all eigenvectors are in \(null( T - \lambda I)\), so \(dimU < dimV\). Thus, by assumption,

\(\forall i \leq j, \exists (u_{k_{1}}, ..., u_{k_{i}})\) that forms an invariant subspace of \(U\). Thus, for \(i \leq j\), V has an invariant subspace of dimension \(i\).

\(dimU < dimV \implies \exists (v_{1}, ... v_{m}) \in V - U\) such that \((u_{1},...,u_{j},v_{1},...,v_{m} )\) is a basis of V.

Let \(x \in (u_{1}, ... , u_{j}, v_{1}, ... v_{t}), k \leq m\).

\(\implies \exists a_{1},...,a_{j},b_{1},...,b_{t} \in \mathcal{C}\) such that \(x = a_{1}u_{1} + ... + a_{j}u_{j} + b_{1}v_{1} + ... + b_{t}v_{t}\).

Let \(t \leq m\).

\(Tv_{t} = Tv_{t} - \lambda v_{t} + \lambda v_{t} = (T - \lambda I ) v_{t} + \lambda v_{t}\).

Thus, \(Tv_{t} \in span(u_{1}, ..., u_{j}, v{t} )\).

Thus, \(Tx = T(a_{1}u_{1} + ... + a{j}u_{j}) + b_{1}Tv_{1} + ... + b_{t}Tv_{t}\) .

\(T(a_{1}u_{1} + ... + a{j}u_{j}) \in span( T(u_{1} + ... ,u_{j})\) by \(U\) invariant under \(T\).

\(b_{1}Tv_{1} \in span(u_{1}, ..., u_{j}, v_{1})\) by \(Tv_{1} \in span(u_{1}, ..., u_{j}, v_{1})\) …. …. \(b_{1}Tv_{t} \in span(u_{1}, ..., u_{j}, v_{1}, ..., v_{t})\) by \(Tv_{t} \in span(u_{1}, ..., u_{j}, v_{1}, ... , v_{t})\)

Thus, \(Tx \in span(u_{1},...,u_{j},v_{1},...,v_{t})\), so \((u_{1},...,u_{j},v_{1},...,v_{t})\) is invariant under \(T\) for \(t \leq m\).

Therefore, \(V\) is has an invariant subspace for all \(i \leq n + 1\).

Q.E.D.