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8 votes
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Show that the operator has a chain of invariant subspaces.

For an elementary argument not requiring the Jordan canonical form: we argue by induction on $n$. For the base case $n=0$, the statement is trivial: the required chain is $V_0 = \{ 0 \} = V$. (If ...
Daniel Schepler's user avatar
7 votes
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Is there a connection between my density formula and an invariant mean defined by a folner sequence of rational numbers?

This answer might be long, but I think it will be helpful since there is a lot going on here (I also will try to make it easy to read). First of all, it's very impressive and beneficial that you are ...
mathworker21's user avatar
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7 votes
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Invariant subspaces in Lie groups and algebras representations

The representation $\Phi$ of $G$ induces a representation of $\mathfrak g$ defined as follows: if $X\in\mathfrak g$ and $v\in V$, then$$\varphi(X)(v)=\left.\frac{\mathrm d}{\mathrm dt}\right|_{t=0}\...
José Carlos Santos's user avatar
6 votes
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Is every invariant subspace equal to some $\text{null}(T-\lambda I)^n$?

No. For one thing, $V$ is always $T$-invariant, and unless $T$ has only one eigenvalue, will not be equal to a generalized eigenspace. For another example, consider $T\colon\mathbb{C}^3\to\mathbb{C}^...
Arturo Magidin's user avatar
5 votes
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If a linear map $T$ has a $k$-dimensional invariant subspace, does it admit an $n-k$ invariant subspace?

The result is true over any field. Let $V$ be an $n$-dimensional vector space over an arbitrary field $\mathbb{F}$, let $T \colon V \rightarrow V$ be an operator and let $U$ be a $k$-dimensional $T$-...
levap's user avatar
  • 65.9k
5 votes

If a linear map $T$ has a $k$-dimensional invariant subspace, does it admit an $n-k$ invariant subspace?

I hope I'm not mistaken but I believe this result is true for real numbers. In summary, we prove the following (which yelds a positive answer to our question as I'll explain) : If $T$ has a real ...
Joel Cohen's user avatar
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5 votes
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Do there exist two-dimensional subspaces $W$ and $Y$ of $\mathbb{R^4},$ both invariant under $\alpha,$ such that $\mathbb{R^4} = W \oplus Y$?

Elegant Solution or Cryptic Hint, depending on who's reading: Previous. If $T:\Bbb R^2\to\Bbb R^2$ is linear and $T^3=0$ then $T^2=0$. Proof: The minimal polynomial $m(x)$ is a factor of $x^3$, ...
David C. Ullrich's user avatar
4 votes

What's the point of invariant subspaces

Let $T \colon V \rightarrow V$ be a linear operator. A $T$-invariant subspace $W \subseteq V$ is a subspace such that $T(W) \subseteq W$. This is a very restrictive condition on $W$. For example, if $...
levap's user avatar
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4 votes
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Hermitian Operators and the Spectral Theorem

I think that this is a great question that is usually unasked and unanswered. This is quite unfortunate because it has a very simple answer: We can use the adjoint operator $T^{*}$ to detect $T$-...
levap's user avatar
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4 votes

Finding all the invariant subspaces of an operator $\ T(x_1, x_2, \ldots, x_n) = (x_1, 2x_2, 3x_2,\ldots, n x_n)$

Since all eigenvalues are different then any set of eigenvectors define an invariant subspace. So, there are $2^n$ such invariant subspaces.
Leox's user avatar
  • 8,186
4 votes
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Find "a'' basis for $W = [ p(x) ∈ P_5(R) | p(−x) = p(x)]$

NOTE: The condition $p(x)=p(-x)$ is for even functions. Thus only polynomials that will be even functions (linear combination of even functions) will survive. But here is a rather detailed way of ...
Anurag A's user avatar
  • 41.2k
4 votes
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Nilpotent Operator and Invariant Subspace

Let $$a_0v+a_1T(v)+\cdots+a_{k-1} T^{k-1}(v) =0\hspace{1cm}(1)$$ Then $$a_0T(v)+a_1T^2(v)+\cdots+a_{k-2} T^{k-1}(v) =0$$ which implies $$a_0T^2(v)+a_1T^3(v)+\cdots+a_{k-3} T^{k-1}(v) =0$$ Proceeding ...
Sahiba Arora's user avatar
  • 10.9k
4 votes
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Hoffman Kunze, Lemma concerning triangulable operators

Let's tackle the second part first. By definition, the $T$-conductor of $\alpha$ into $W$ is the smallest degree monic polynomial $q$ where $q(T)\alpha\in W$, i.e. $q$ is the monic generator of $S(\...
Christian Sykes's user avatar
4 votes
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Are $V=<(1,0,1),(1,1,1)>$ and $W=<(1,0,1),(0,1,0)>$ equivalent?

For example, are subspaces $\color{blue}{V}=\{(1,0,1),(1,1,1)\}$ and $W=\{(1,0,1),(0,1,0)\}$ analogous? It's easier to use different names so I renamed the first one to $V$. Note that in the usual ...
StackTD's user avatar
  • 27.9k
4 votes
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Problem understanding invariant subspaces and foliations

I think you're asking this in the context of the controlled dynamical system $(\mathcal{U},\Sigma): \dot{x}(t)=Ax(t)+Bu(t)$, but i'll try to keep the context general. First of all, your definition of ...
Siddhartha's user avatar
  • 2,701
4 votes
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How do we know that the eigenspaces of $T$ are the only $T$-invariant subspaces?

If there was, it would have to be $1$-dimensional. In other words, it would be equal to $\Bbb Cv$, for some vector $v\ne0$. But asserting that $\Bbb Cv$ is invariant is the same thing as asserting ...
José Carlos Santos's user avatar
4 votes
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$T$ is a normal operator, prove any eigenspace of $T+T^*$ is invariant under $T$

Suppose $\lambda$ is an eigenvalue of $T+T^*$ and $V_{\lambda}$ is its eigenspace. We have to show that if $v\in V_{\lambda}$ then $T(v)\in V_{\lambda}$ as well. And indeed: $(T+T^*)(Tv)=(T^2+T^*T)(v)=...
Mark's user avatar
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4 votes
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Invariant subspace of $T$ (normal) is also an invariant subspace of $T^\ast$.

$\newcommand{\Tr}{\mathrm{Tr}}$ $\newcommand{\diag}{\mathrm{diag}}$ Suppose $\dim(W) = k, 1 \leq k < n$ (where $n = \dim(V)$). Then there exists an orthonormal basis $\{\alpha_1, \ldots, \alpha_k, \...
Zhanxiong's user avatar
  • 14.3k
4 votes
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Determine all subspaces $U$ of $\mathbb{R}^3$ such that $\phi (U)=U$.

What you did is fine. And now, if $v_1$, $v_2$ and $v_{-1}$ are eigenvalues corresponding to the eigenvalues $1$, $2$, and $-1$ respectively, then the subspaces $U$ of $\Bbb R^3$ such that $\phi(U)=U$ ...
José Carlos Santos's user avatar
4 votes
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Invariant subspace problem for $\ell^2(\mathbb{N})$

The answer is, we don't know. The invariant subspace problem is known for non separable Hilbert spaces and for finite dimensional ones, e.g. see this answer here. As noted by Disintegrating By Parts, ...
Zahlenteufel's user avatar
4 votes
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Is Sequence Convergence a topological Invariant?

Your idea to set up a homeomorphism is good. But you are confusing a few concepts. Let's start at this point in your proof (slightly paraphrased): Suppose $B \subseteq Y$ is sequentially closed. We ...
K. Jiang's user avatar
  • 8,304
3 votes
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T-invariant linear independence

Since you're given that $\;\dim U=n+1\;$ and we have $\;n+1\;$ vectors in $\;B:=\{\,u=T^0u, Tu,...,T^nu\,\}\;$ , all we have to do is to prove $\;B\;$ is linearly independent . So assume it isn't, and ...
DonAntonio's user avatar
  • 212k
3 votes

Do all linear operators with an invariant space, other than the kernel and range, have an upper triangular matrix from?

The relation between invariant subspaces and having an upper triangular form is best expressed using the notion of a flag. Given a finite dimensional vector space $V$ of dimension $n$, a (full) flag ...
levap's user avatar
  • 65.9k
3 votes
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Every linear operator on $\mathbb{R}^5$ has an invariant 3-dimensional subspace

This is an easy consequence of the existence of the real Jordan normal form of the matrix of the endomorfism. That matrix is similar to a block diagonal matrix, with each block being a real Jordan ...
José Carlos Santos's user avatar
3 votes
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Understanding T-conductor in Linear Algebra

The vector $\alpha$ belongs to $V$. Your remark would be correct if we were assuming that $\alpha\in W$. In particular, if $\alpha\in W$, $S_T(\alpha;W)$ is indeed the set of all polynomials.
José Carlos Santos's user avatar
3 votes

Why invariant subspaces are important?

For instance: $T$ is invertible if and only if each $T|_{U_i}$ is invertible; $T$ is diagonalizable if and only if each $T|_{U_i}$ is diagonalizable. But the $U_i$'s are smaller spaces. Therefore, ...
José Carlos Santos's user avatar
3 votes
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Every $k$ dimensional subspace is $S$-invariant implies $S$ is a multiple of the identity

Let $v\in V$ and let $X(v)$ be the collection of $k$-dimensional subspaces of $V$ that contain $v$. Then $$\langle v\rangle=\bigcap_{W\in X(v)}W,$$ and since each $W$ is $S$-invariant, also $\langle ...
Servaes's user avatar
  • 63.6k
3 votes
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Why are there two non-trivial T-invariant subspaces and why the restriction to T is cyclic?

The minimal polynomial is reducible: $\mu_T(x) = (x-2)(x^2 + 2x + 4)$. Each divisor of the minimal polynomial gives you a different invariant subspace; for instance, the divisor $x-2$ gives you the ...
arkeet's user avatar
  • 6,880
3 votes

Prove that $W$ is $g(T)$-invariant for any polynomial $g(t)$.

Since $W$ is invariant, $T(W)\subset W$. But then$$T^2(W)=T\bigl(T(W)\bigr)\subset T(W)\subset W.$$So,$$T^3(W)=T\bigl(T^2(W)\bigr)\subset T(W)\subset W$$and so on. And since$$(\forall n\in\mathbb{Z}^+)...
José Carlos Santos's user avatar

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