# All invariant subspaces of a linear operator $T$

Let $T \in \mathscr L(\mathbb F^n)$ such that $T(x_1,x_2,...,x_n)=(x_1,2x_2,...,nx_n)$. Then find all the invariant subspaces of $T$. Clearly, $Null$ $T$ and $Range$ $T$ are two invariant subspaces.

Also, all the subspaces spanned by the eigen vectors form $1$-dimensional invariant subspaces. In this case the eigen values of $T$ are $i$, where $i$$\in\{1,2,...,n\} and the corresponding eigen vector is of form (0,...,0,a,0,...,0) where a(\neq 0)\in \mathbb F is the i^{th}-component of the vector. But what about the invariant subspaces of other dimensions? For instance, if W is an invariant subspace of T of dimension k, then T|_W is a linear operator on W. So, for any w\in W, Tw\in W. If w\in span(e_1,e_2,...,e_k) then Tw\in span(e_1,2e_2,...,ke_k). Since, dim W= dim Tw=k, we can say that Tw=span(e_1,2e_2,...,ke_k)\in span(e_1,e_2,...,e_k)=W i.e Tw\in W. So, can we conclude from here that there will be invariant subspaces of all dimensions under T, given by span(e_1,e_2,...,e_k), where 1\leqslant k\leqslant n (which are precisely 2^n in number)? • What do you think the invariant subspaces are? – Demophilus Oct 22 '17 at 13:32 • @Demophilus If W is an invariant subspace of T of dimension k, then T|_W is a linear operator on W. So, for any w\in W, Tw\in W. If w\in span(e_1,e_2,...,e_k) then Tw\in span(e_1,2e_2,...,ke_k). Since, dim W= dim Tw=k, we can say that Tw=span(e_1,2e_2,...,ke_k)\in span(e_1,e_2,...,e_k)=W i.e Tw\in W. So, can we conclude from here that there will be invariant subspaces of all dimensions under T, given by span(e_1,e_2,...,e_k), where 1\leqslant k\leqslant n? – JackT Oct 22 '17 at 13:52 ## 2 Answers I'll work in characteristic zero to avoid quirks due to the field. Suppose that V is an invariant subspace, and let v\in V. Assume for a moment that all entries of V are nonzero. We have that v,Tv,T^2v,T^3v,\ldots,T^{n-1}v\in V. If \alpha_0T^0v+\cdots+\alpha_{n-1}T^{n-1}v=0, we get a linear system with matrix$$ A=v_0v_1v_2\cdots v_{n-1}\begin{bmatrix} 1&1&1&\cdots&1\\1&2&3&\cdots&n-1\\ 1&2^2&3^2&\cdots&(n-1)^2\\ \vdots&\vdots&\vdots&\cdots&\vdots\\ 1&2^{n-1}&3^{n-1}&\cdots&(n-1)^{n-1} \end{bmatrix} $$This is (one of) the well-known Vandermonde Matrix, which is invertible. So the only possible solution of the system is given by \alpha_0=\alpha_1=\cdots=\alpha_{n-1}=0, and thus v,Tv,\ldots,T^{n-1}v are linearly independent, which means that \dim V=n and so V=\mathbb F^n. So any invariant subspace will be made of vectors with at least one entry equal to zero. On the "nonzero part" of the subspace we can repeat the above reasoning, to conclude (as you suggested in your comment) that any invariant subspace of T is of the form$$ W=\text{span}\,\{e_j:\ j\in K\} $$for some K\subset\{1,\ldots,n\}. • :Thanks a lot... – JackT Oct 22 '17 at 14:07 Well, you can get 2^n (instead of just n) subspaces off the bat, by considering \operatorname{span}(B), where B ranges over all subsets of \lbrace e_1, \ldots, e_n \rbrace. Included in this are the whole space and the trivial space (spanned by the empty set), which are the range and nullspace respectively (unless the field has non-zero characteristic). The question is, are there any others? First, suppose \mathbb{F} = \mathbb{C}, and U is an invariant subspace, so that T|_U is an operator. It also will be diagonalisable. To see this, suppose \lambda is an eigenvalue for T|_U, and consider$$\operatorname{null}(T|_U - \lambda I|_U)^2 \subseteq \operatorname{null}(T - \lambda I)^2 = \operatorname{null}(T - \lambda I) = \operatorname{null}(T_U - \lambda I|_U),$$so the generalised eigenspace is no larger than the eigenspace. Hence a basis for$U$, consisting of eigenvectors for$T|_U$(and hence$T$) exists, so$U$must be of the above form. If$\mathbb{F} = \mathbb{R}$, the same argument mostly works. You just need to tiptoe around the complex case. If you extend the map to the complexification of the space, the argument works exactly as is. If$\mathbb{F}\$ is a different field, I'm not sure if I can help.

• You are absolutely correct. – JackT Oct 22 '17 at 14:11