Let $V$ is a finite dimensional vector space over $\mathbb{C}$ and $T$ be a linear operator on $V$ . How to prove $T$ has an invariant subspace of dimension $k$ for each $k = 1,2, \ldots ,\text{dim}V$ .
Can it be solved by induction ?
Let $V$ is a finite dimensional vector space over $\mathbb{C}$ and $T$ be a linear operator on $V$ . How to prove $T$ has an invariant subspace of dimension $k$ for each $k = 1,2, \ldots ,\text{dim}V$ .
Can it be solved by induction ?
Suppose that $W\subset V$ is invariant. Then $T$ descends to a well-defined linear operator on $V/W$. Since every linear operator over $\mathbb C$ has an eigenvector, we can find some $v+W\in V/W$ such that $T(v+W)=\lambda(v+W)$. Tracing through the definitions, we see that $\mathbb Cv+W=\operatorname{span}(v,W)$ is an invariant subspace of $V$. Thus, if we have an invariant subspaace of dimension $k$, then we also have one of dimension $k+1$. Taking $W=\{0\}$ for our base case, we have invariant subspaces for all dimensions.
Suppose that $T$ has admits a basis of eigenvectors $\left\{v_1, \dots, v_n\right\}$. Choose $k$ of these eigenvectors, say $\left\{v_1, \dots, v_k\right\}$. Let $W=\text{span}\left\{v_1, \dots , v_k\right\}$. Then $T(W)\subset W$, hence $W$ is an invariant $k$-dimensional subspace.
Now if $T$ is not diagonalizable, you can still decompose $V$ into generalized eigenspaces, you can do a very similar argument in that case. Notice that it's very important that we work over an algebraically closed field.