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The original question was as follows. Define $T\in\mathcal{L}(F^n)$ by $$T(x_1,x_2,x_3,\cdots,x_n)=(x_1,2x_2,3x_3,\cdots,nx_n).$$(a) Find all eigenvalues and eigenvectors of $T.$

(b) Find all invariant subspaces of $T$.

Part (a) is easy, for which the answer is for each $\lambda=n$, we have an eigenvector of $(0,\cdots,a_i,\cdots,0)$ where $a_i$ is at $i^{th}$ position in the list.

But I need help in verification of my proof in Part (b). Here is my proof:

Proof: Since we have $n$ eigenvectors and each of the eigenvectors can form an invariant subspaces under $T$, where the invariant subspaces corresponding to each eigenvector is denoted by $U_1,\cdots,U_n$. By theorem, we know the sum of the invariant subspaces is still invariant under T. Hence $U_1+\cdots+U_n$ is invariant under $T$. But $U_1,\cdots,U_n$ actually forms a basis of $F^n$ (since the eigenvectors are actually standard basis of $F^n$) Hence we can conclude that the set of any linear combinations of $u_1+\cdots+u_n$ for each $u_j\in U_j$ is the invariant subspaces of $T$. In other words, $span(u_1+\cdots+u_n)$ is the invariant subspaces of $T$.

Am I correct with this proof?

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No. The invariant subspace are those subspaces $U$ of $F^n$ such that $T(U)\subset U$. In your case, the invariant subspaces are those subspaces of $F^n$ of the type $\langle S\rangle$, where $S$ is a subset of the standard basis. In particular, there are $2^n$ invariant subspaces (one of which is $\{0\}$ and another of which is $F^n$; these two are always invariant). But, if $\{e_1,\ldots,e_n\}$ is the standard basis, every space $Fe_j$ is invariant; this provides $n$ invariant subspaces. And every space $Fe_j\oplus Fe_k$, with $j\neq k$ is invariant too; this provides $\frac{n(n-1)}2$ invariant subspaces. And so on…

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  • $\begingroup$ I understand $\{0\}$ and $F^n$ is always an invariant subspaces. But I don't understand what you mean by "there are $2^n$ invariant subspaces", can you elaborate more about this? $\endgroup$ – Ling Min Hao Jan 28 '18 at 13:20
  • $\begingroup$ @LingMinHao I have expandaded my answer. Is it clear now? $\endgroup$ – José Carlos Santos Jan 28 '18 at 13:24
  • $\begingroup$ Yes I get what you mean. But according to my proof, I also found $2^n$ possible invariant subspaces as $span(u_1+\cdots+u_n)$ also have $2^n$ possible combinations. What I mean is each of these combinations represents an invariant subspace. For example, if we have $span(v_1,v_2,v_3)$ , then we can write it as either $\{av_1\}$, $\{bv_2\}$, $\{cv_3\}$, $\{av_1+bv_2\}$, $\{av_1+cv_3\}$, $\{bv_2+cv_3\}$, $\{av_1+bv_2+cv_3\}$ and $\{0\}$, where each of them represents a subspace. $\endgroup$ – Ling Min Hao Jan 28 '18 at 14:03
  • $\begingroup$ Maybe i should write "Let $W$ be invariant subspaces of $T$. Then $\cdots\cdots\cdots$ we have $W\in span(u_1,\cdots,u_n)$" instead? $\endgroup$ – Ling Min Hao Jan 28 '18 at 14:06
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    $\begingroup$ @LingMinHao Instead of writing “linear combinations of $u_1+u_2+\cdots+u_n$”, you should have written “linear combinations of some elements of $\{u_1,u_2,\ldots,u_n\}$”. $\endgroup$ – José Carlos Santos Jan 28 '18 at 14:59

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