Let $V$ be a finite-dimensional vector space over $F$ with $char(F) = 0$  and $ T: V \rightarrow V $ a linear map. Suppose that $Tr(T^n)=0$ for all $n≥1.$  Show that $T$ nilpotent. I have seen a proof that uses the Fitting's lemma. I believe there should be some straight-forward proof from scratch. Could you help me with some suggestions? Thanks so much.
16172
Let $A$ be an $n\times n$ complex nilpotent matrix. Then we know that because all eigenvalues of $A$ must be $0$, it follows that $\text{tr}(A^n)=0$ for all positive integers $n$.  What I would like to show is the converse, that is,      if $\text{tr}(A^n)=0$ for all positive integers $n$, then $A$ is nilpotent.   I tried to show that $0$ must be an eigenvalue of $A$, then try to show that all other eigenvalues must be equal to 0. However, I am stuck at the point where I need to show that $\det(A)=0$.  May I know of the approach to show that $A$ is nilpotent?
33615
Let $G$ be a finite group and $N$ be a normal subgroup of $G$ such that $G/N$ is nilpotent.  a) Prove that there exists a nilpotent subgroup $K$ of $G$  such that $G=NK.$  b) Suppose that $N$ is abelian and $Z\left(G\right) =\lbrace e \rbrace.$ Show that if $K$ is nilpotent and $G=NK$ then $K=N_G\left(K\right)$ and $K \cap N =\lbrace e \rbrace.$  Could anyone help me to give a solution for the problem! thank you!
109008
Let $k$ be an algebraically closed field and $V$ be a finite-dimensional $k$-vector space of dimension $n$.  Let $T:V \rightarrow V$ be a $k$-linear endomorphism of $V$. A vector $v \in V$ is called a cyclic vector for $T$ if the set of vectors $\{T^nv: n \in \mathbb{Z}, n \geqslant 0\}$ span $V$.  1 Show that if $v \in V$ is a cyclic vector, then $\{v, Tv,\cdots, T^{n-1}v\}$ form a basis for $V$.  2 If $T$ admits a cyclic vector, and $A:V\rightarrow V$ is a linear map commuting with $T$, show that there exists a polynomial $P(x) \in k[x]$ such that $A=P(T)$.  3 Show that a cyclic vector for $T$ exists if and only if the minimal polynomial of $T$ is equal to the characteristic polynomial of $T$.
107106
Let $f:V\times V\rightarrow F$ be a bilinear form in a finite inner product space V.  If $F=R$, how can I prove that there exists a single linear transformation $T:V \rightarrow V$ so that for each $v,u\in V$, $f(u,v) = (Tu, v)$ ?  Would that sill be the case if $F=C$?
99852
I'm reading about linear algebra and I came across with the following theorem where I have a problem convincing myself:      Theorem 2.1 $\,$ Every linear operator on a finite-dimensional complex vector space has an eigenvalue.      Proof:       To show that $T$ (our linear operator on $V$) has an   eigenvalue, fix any nonzero vector $v \in V$. The vectors $v, Tv,  T^2v,..., T^nv$ cannot be linearly independent, because $V$ has   dimension $n$ and we have $n + 1$ vectors. Thus there exist complex   numbers $a_0,...,a_n$, not all $0$, such that      $$a_0v + a_1Tv + ··· + a_nT^nv = 0.$$ Make the $a$’s the coefficients   of a polynomial, which can be written in factored form as $$a_0 + a_1z  + ··· + a_nz^n = c(z − r_1)\cdots(z − r_m),$$ where $c$ is a non-zero complex number, each $r_j$ is complex, and the equation holds for all   complex $z$. We then have      $${\color{red}{ 0=(a_0I + a_1T + ··· + a_nT^n)v= c(T − r_1I)\cdots(T −  r_mI)v}},$$ which means that $T − r_j$ I is not injective for at least   one $j$. In other words, $T$ has an eigenvalue.$\;\blacksquare$   I'm having trouble with the factorized form of the matrix polynomial (in red). I understood that the factorization holds for a polynomial by the fundamental theorem of algebra but why does this also hold for matrices?  In other words, why is the the part I highlighted true? Does such an factorization always exist?    Could I have some help to see this? Thank you =)   P.S.  is my reference (page 3).  UPDATE:   Someone else also has asked the  before it seems.
93730
I need help with this question:  Let $V$ be a finite vector space where $ \dim V = n $, over the complex numbers  and let $ T: V\to V $ be a linear transformation. Prove that $ V = \ker(T^n) \oplus Im(T^n) $
27432
I have to prove the following: If $R$ is a finite-dimensional algebra over a field $F$, then $J(R)$ is nilpotent.  I thought about this, but there are some gaps:    Because $R$ is in particular a finite dimensional vector space, there is a composition series $0=R_0\subset R_1\subset\dots\subset R_k=R$ of VECTOR SPACES. Do I know that this is a composition series of $R$-modules? If yes, why? Or how can I find some?  If that is the case, I would proceed as follows: $J(R)R_n/R_{n-1}\subseteq R_n/R_{n-1}$. Now, I would like to use, that the quotient is simple, so if the above is a strict inclusion (how to show that?), then $J(R)R_n/R_{n-1}=0$. Now, by induction it is easy to see that $J(R)^k=\{0\}$.
19585
Let $T$ be a linear operator on the finite dimensional vector space $V$. Suppose $T$ has a cyclic vector. Prove that if $U$ is any linear operator which commutes with $T$, then $U$ is a polynomial in $T$.
23947
I have come across a question about determining possible minimal polynomials for a rank one linear operator and I am wondering if I am using the correct proof method.  I think that the facts needed to solve this problem come from the section on Nilpotent operators from Hoffman and Kunze's "Linear Algebra".     Question: Let $V$ be a vector space of dimension $n$ over the field $F$ and consider a linear operator $T$ on $V$ such that $\mathrm{rank}(T) = 1$.  List all possible minimal polynomials for $T$.   Sketch of Proof:   If $T$ is nilpotent then the minimal polynomial of $T$ is $x^k$ for some $k\leq n$.  So suppose $T$ is not nilpotent, then we can argue that $T$ is diagonalizable based on the fact that $T$ must have one nonzero eigenvalue otherwise it would be nilpotent (I am leaving details of the proof of diagonalization but it is the observation that the characteristic space of the nonzero eigenvalue is the range of T and has dimension $1$).  Thus the minimal polynomial of $T$ is just a linear term $x-c$.    Did I make a mistake assuming that $T$ can have only one nonzero eigenvalue?  Thanks for your help
25809
I'm looking for the shortest and the clearest proof for this following theorem:  For $V$ vector space of dimension $n$ under $\mathbb C$ and $T: V \to V $ linear transformation , I need to show $V= \ker T^n \oplus  $ Im $T^n$.  Any hints? I don't know where to start from.  Thank you.
76732
I know that if $V$ is a finite-dimensional vector space and $T$ is a linear operator on $V$, then $T$ is a direct sum of a nilpotent operator and an invertible operator.   My question is: Is there an infinite-dimensional vector space $V$ over a field $\mathbb{F}$ such that every linear operator $T$ on $V$ can be written as a direct sum of a nilpotent linear operator and an invertible linear operator?  I mean "There are subspaces such that $V=M\oplus N$ where $T$ is invertible on $M$ and $T$ is nilpotent on $N$"?
54366
Suppose $T$ is in $L(V)$, $m$ is a positive integer, and $v$ in vector space $V$ is such that $(T^{m-1})v \neq 0$, and $(T^m)v = 0$. Prove that  $[v, Tv, T^2v, ... , T^{m-1}v]$  is linearly independent  I get that $(T^j)v \neq 0\ \forall\ j &lt; m$. Additionally, since $T$ is nilpotent, $V$ has a basis with respect to which the matrix of $T$ has $0$s on and below the diagonal. However, I'm not sure if these can be used to show linear independence or if they're even relevant to the problem at all. Any help is appreciated!
90641
(a) Prove that a square matrix $T$ of rank one has $\text{tr}(T)=0$ if and only if $T^2=0$.       (b) Consider a matrix $A$ of the form $A=aI+T$, where $a\ne0$, $I$ is the identity matrix,   and $T$ has rank one and zero trace. Find the inverse and the determinant of $A$.       (c) Find the inverse of $A$ as above when $T$ has rank one but nonzero trace $\text{tr}(T)=b$.      For which value of $b$ is $A$ not invertible?    I'm still stuck on part (a), but campus buildings are closing soon, so I'll be working from home but would love to get some hints / comments on this question.  I'll have limited access to this site - on my phone.  For part (a), I've been trying to look at the SVD of matrix $A$, since one can read off the rank very easily - by looking at the number of non-zero singular values of $A$.  Then I am trying some block matrix multiplication to see whether $T^2 = 0$, from assuming that $\text{tr}(T) =0$.  So far, no luck.  Do you think I should stick with this SVD approach, or is it better to play around with the definition and properties of nilpotent operators?  Any other hints for the other parts of the question would be greatly appreciated.  Thanks!
83064
So I've got a bijection. It clearly has an inverse, but how exactly do I prove that the inverse is a linear map as well?     Suppose that the linear map $T:U\to V$ is a bijection. So $T$ has an inverse map $T^{-1}:V\to U$. Prove that $T^{-1}$ is a linear map.
33276
I am self-studying Hoffman and Kunze's book Linear Algebra. This is exercise $4$ from page $115$. It is in the section of The transpose of a Linear Transformation.     Let $V$ be a finite-dimensional vector space over the field   $\mathbb{F}$ and let $T$ be a linear operator on $V$. Let $c$  be a   scalar and suppose there is a non-zero vector $\alpha$ in $V$ such   that $T\alpha=c\alpha.$ Prove that there is a non-zero linear   functional $f$ on $V$ such that $T^{t}f=cf.$ ($T^{t}$ is the transpose of $T$.)   I tried to solve this question by induction on $\dim V$. I was able to show the base case, that is, when $\dim V=1$, but I got stuck in the  inductive step. If $\alpha $ is a non-zero vector, then we can find a base $\mathcal{B}=\{\alpha,\alpha_{1},\ldots\alpha_{m}\}$ of $V$. We can write $V=W_{1}\oplus W_{2}$, where $W_{1}=\langle \alpha \rangle$ and $W_{2}=\langle \alpha_{1},\ldots,\alpha_{m}\rangle.$ I can not show that $T(W_{2})\subset W_{2}.$ Anyway, $\alpha \notin W_{2}.$  EDIT: If $T$ is a linear transformation from $V$ to $W$, then $T^{t}$ is the linear transformation from $W^{\star}$ into $V^{\star}$ such that $$(T^{t}f)(\alpha)=g(T\alpha)$$ for every $f\in W^{\star}$ and $\alpha \in V$.
5203
I have a question related to the Primary Decomposition Theorem for a vector space under a linear transformation.  The following question is a direct corollary of this theorem.  However, I wish to prove it without calling upon theorem.  I am currently studying for a comprehensive exam and ran across this question.  I never really felt comfortable with the proof I saw of the Primary Decomposition Theorem so I hoping with your help with this problem I will understand this particular problem and better understand that proof as I believe their proofs will be essentially the same techniques.   Question:  Suppose we have a linear transformation $T: \mathbb{R^n}\to \mathbb{R^n}$ with minimal polynomial $m(x)=f(x)g(x)$ and $f(x),g(x)$ are relatively prime.  Show that we can write $\mathbb{R^n}=M\oplus N$ where $M$ and $N$ are $T$-invariant subspaces s.t $f(x)$ is the minimal polynomial for $T|_M$ and $g(x)$ is the minimal polynomial for $T|_N$.   So I know that I want to use $M=ker(f(T))$ and $N=ker(g(T))$ but after that I really don't have much.  Since this isn't a homework problem for a class I am essentially looking for someone to present me a proof of this simpler question and walk me through their logic so that I can hopefully understand the general theorem and its proof.  Thanks in advance for the help!
8151
Prove: If a linear map, $φ: U→V$  is bijective, then $φ(x)=0$ has only the solution $x=0$.  Either a proof or explanation will be appreciated.