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 ?

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    $\begingroup$ If $v$ is an eigenvector of $T$, then $T(v)=\lambda v$, if $\lambda\neq 0$ then the space spanned by $v$ gets mapped to the space mapped by $v$. That could be very useful. $\endgroup$ – Mathematician 42 May 9 '16 at 18:18
  • $\begingroup$ @Mathematician 42 To start with abasis which has eigen vectors as its elements such that T(v)=λv ?? $\endgroup$ – Shona May 9 '16 at 18:27
  • $\begingroup$ Not every linear operator allows a basis of eigenvectors. But start easy, suppose there is such a basis, can you do it? If yes, is there something more generalized that could be useful? $\endgroup$ – Mathematician 42 May 9 '16 at 18:31
  • $\begingroup$ If there is such abasis it implies T is diagonalizable $\endgroup$ – Shona May 9 '16 at 18:34
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    $\begingroup$ See Schur decomposition. $\endgroup$ – lhf May 9 '16 at 18:58

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.

  • $\begingroup$ Its confising can you elaborate ? $\endgroup$ – Shona May 10 '16 at 11:29
  • $\begingroup$ Not really. Can you say what part is confusing? Have you come across quotient spaces before? $\endgroup$ – Aaron May 10 '16 at 18:23
  • $\begingroup$ The last part 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. $\endgroup$ – Shona May 10 '16 at 18:44
  • $\begingroup$ We started with $W$ as an invariant subspace, and we found a $v$ such that $\mathbb Cv+W$ is also invariant. This has dimension one larger than $W$ (if $W$ is spanned by $w_1, \ldots, w_k$, then our new space is spanned by $w_1,\ldots, w_k, v$. Note that this doesn't quite work if $W=V$, because then $V/W$ is $0$ dimensional, and only linear maps of positive dimensional spaces to themselves have eigenvectors. So induction gives us invariant subspaces of dimensions $0, 1, 2, \ldots, \dim V$, but then the process stops. $\endgroup$ – Aaron May 11 '16 at 4:30

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.

  • $\begingroup$ Every eigen space will have eigen vectors so there will always be some W which would be invariant .. right ? $\endgroup$ – Shona May 9 '16 at 18:56
  • $\begingroup$ Right, but it could happen that some eigenspace has a lower dimension than the algebraic multiplicity of the corresponding eigenspace, in other words, it could happen that the space spanned by all eigenvectors is not all of $V$. So you cannot use the above proof to show that for any $1\leq k\leq \dim(V)$ there is an invariant subspace of dimension $k$. But using a basis of generalized eigenvectors (which exists) you can do it. $\endgroup$ – Mathematician 42 May 9 '16 at 19:00
  • $\begingroup$ If we say in this proof we take W as a basis of generalised eigenvectors where k= 1 , ... , dim(V) . Then it would do ? $\endgroup$ – Shona May 9 '16 at 19:05
  • $\begingroup$ I'll leave that up to you to figure out, a generalized eigenspace is certainly an invariant subspace, however generalized eigenvectors are not eigenvectors, hence the argument will be less trivial. You will have to modify the basis of generalized eigenvectors to make it work. Ihf's comment is actually a more direct approach, you can also try to understand that decomposition. $\endgroup$ – Mathematician 42 May 9 '16 at 19:33
  • $\begingroup$ How to modify ? By taking distinct eigen vectors ? $\endgroup$ – Shona May 9 '16 at 19:42

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