Let $V$ be a finite dimensional vector space over $F$ and $T$ be a linear operator defined on $V$. Then if $p(t)$ and $q(t)$ are polynomials defined over $F$, then the the operators $P(T)$ and $Q(T)$ commute.   Is the converse true? i.e     $AB=BA \implies \exists$ polynomials $p(t)$ and $q(t)$ over $F$ and a linear transformation $T$ such that:   $p(T)=A$ and $q(T)=B$   A hint in the right direction would suffice.
36949
Consider square matrices over a field $K$. I don't think additional assumptions about $K$ like algebraically closed or characteristic $0$ are pertinent, but feel free to make them for comfort. For any such matrix $A$, the set $K[A]$ of polynomials in $A$ is a commutative subalgebra of $M_n(K)$; the question is whether for any pair of commuting matrices $X,Y$ at least one such commutative subalgebra can be found that contains both $X$ and $Y$.  I was asking myself this in connection with frequently recurring requests to completely characterise commuting pairs of matrices, like . While providing a useful characterisation seems impossible, a positive anwer to the current question would at least provide some answer.  Note that in many rather likely situations one can in fact take $A$ to be one of the matrices $X,Y$, for instance when one of the matrices , or more generally if its  (so coincides with the characteristic polynomial). However this is not always possible, as can be easily seen for instance for diagonal matrices $X=\operatorname{diag}(0,0,1)$ and $Y=\operatorname{diag}(0,1,1)$. However in that case both will be polynomials in $A=\operatorname{diag}(x,y,z)$ for any distinct values $x,y,z$ (then $K[A]$ consists of all diagonal matrices); although in the example in  the matrices are not both diagonalisable, an appropriate $A$ can be found there as well.   I thought for some time that any maximal commutative subalgebra of $M_n(K)$ was of the form $K[A]$ (which would imply a positive answer) for some $A$ with minimal polynomial of degree$~n$, and that a positive answer to my question was in fact instrumental in proving this. However I was wrong on both counts: there exist (for $n\geq 4$) commutative subalgebras of dimension${}&gt;n$ (whereas $\dim_KK[A]\leq n$ for all $A\in M_n(K)$) as shown in , and I was forced to correct  in the light of this; however it seems (at least in the cases I looked at) that many (all?) pairs of matrices $X,Y$ in such a subalgebra still admit a matrix $A$ (which in general is not in the subalgebra) such that $X,Y\in K[A]$. This indicates that a positive answer to my question would not contradict the existence of such large commutative subalgebras: it would just mean that to obtain a maximal dimensional subalgebra containing $X,Y$ one should in general avoid throwing in an $A$ with $X,Y\in K[A]$. I do think these large subalgebras easily show that my question but for three commuting matrices has a negative answer.  Finally I note that  to the cited MO question mentions a result by Gerstenhaber that the dimension of the subalgebra generated two commuting matrices in $M_n(K)$ cannot exceed$~n$. This unfortunately does not settle my question (if $X,Y$ would generate a subalgebra of dimension${}&gt;n$, it would have proved a negative answer); it just might be that the mentioned result is true because of the existence of $A$ (I don't have access to a proof right now, but given the formulation it seems unlikely that it was done this way).  OK, I've tried to build up the suspense. Honesty demands that I say that I do know the answer to my question, since a colleague of mine provided a convincing one. I will however not give this answer right away, but post it once there has been some time for gathering answers here; who knows somebody will prove a different answer than the one I have (heaven forbid), or at least give the same answer with a different justification.
63325
May I please ask how to show the dual space of the vector space of all polynomials is isomorphic to the infinite-dimensional Euclidean vector space over the reals?  (i.e. Show that $(\Bbb{R}[X])^*$ and $\Bbb{R}^{∞}$ are isomorphic where $\Bbb{R}$ means the set of all reals)
47642
If $X$ is a Banach space, and $T:X \to X$ is a bounded linear operator with norm &lt; $1$, then $I-T$ has a bounded inverse defined by $(I-T)^{-1} = \sum_{n=0}^\infty T^n$.   Thinking in terms of a converse, if $T$ is any bounded linear operator defined on $X$, then does the existence of a bounded inverse $S=(I-T)^{-1}$ imply that $S$ can be represented as $S=\sum_{n=0}^\infty T^n$?
10934
Let $T : V → W $ be an injective linear transformation, where $V, W$ are vector spaces of dimension $n$ over the field $K$. If $\{e_1,\ldots , e_n\}$ is a basis for $V$ , show that $\{T(e_1),\ldots , T(e_n)\}$ is a basis for $W$.
71241
I'm trying to solve some problems in Hoffman and Kunze and I'm kind of stuck on this one. This is 6.5.3 on Hoffman and Kunze.  Here is the question:  Let $T$ be a a linear operator on an $n$-dimensional space, and suppose that $T$ has $n$ distinct characteristic values. Prove that any linear operator $U$ which commutes with $T$ is a polynomial in $T$.  My work so far: Since $T$ has $n$ distinct characteristic values and since the space it acts on is also $n$ dimensional, $T$ must be diagonalisable. Now using the fact $UT=TU$ I can show that $U$ must also be diagonalizable. So these two operators must be simultaneously diagonalizable. Now I'm stuck with where this is going. I'm thinking of simultaneous diagonalizability because this is section 6.5 of Hoffman and Kunze and it deals with simultaneous diagonalizability. Am I on the right track with this?. Can anybody help?  Thanks so much for your time and your answers.
6030
Prove that a linear transformation $P \colon V \to V$ of a finite dimensional vector space satisfies $P^2 = P$ if and only if there exists a basis with respect to which $P$ can be written as a block matrix   $$P = \begin{bmatrix} I &amp; 0 \\ 0 &amp; 0 \end{bmatrix}.$$  Hence determine the minimal and characteristic polynomials of $P$.  I don't know what to do with this problem at all, so some help would be great, Thanks.
115181
Question:  Suppose $V$ and $W$ are finite dimensional vector spaces over $\mathbb{C}$. $A$ is a linear transformation on $V$,  $B$ is a linear transformation on $W$. Then there exists a non-zero linear map $C:V\to W$ s.t. $CA=BC$ iff $A$ and $B$ have a same eigenvalue.  ===========This is incorrect==========  Clearly, if $CA=BC$, suppose $Ax=\lambda x$, then $B(Cx)=CAx=C(\lambda x)=\lambda (Cx)$, so $A$ and $B$ have same eigenvalue $\lambda$. On the other hand, if $A$ and $B$ have a same eigenvalue $\lambda$, suppose $Ax=\lambda x, By=\lambda y$. Define $C:V\to W$ s.t. $Cx=y$, then $BCx=By=\lambda y=C\lambda x=CAx$. But I don't kow how to make $BC=CA$ for all $x\in V$.   ======================================
104454
I'm trying to read Elias Stein's "Singular Integrals" book, and in the beginning of the second chapter, he states two results classifying bounded linear operators that commute (on $L^1$ and $L^2$ respectively).  The first one reads:  Let $T: L^1(\mathbb{R}^n) \to L^1(\mathbb{R}^n)$ be a bounded linear transformation. Then $T$ commutes with translations if and only if there exists a measure $\mu$ in the dual of $C_0(\mathbb{R}^n)$ (continuous functions vanishing at infinity), s.t. $T(f) = f \ast \mu$ for every $f \in L^1(\mathbb{R}^n)$. It is also true that $\|T\|=\|\mu\|$.   The second one says:  Let $T:L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ be bounded and linear. Then $T$ commutes with translation if and only if there exists a bounded measurable function $m(y)$ so that $(T\hat{f})(y) = m(y) \hat{f}(y)$ for all $f \in L^2(\mathbb{R}^n)$. It is also true that $\|T\|=\|m\|_\infty$.    I was wondering if anyone had a reference to a proof of these two results or could explain why they are true.
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 $N$"?
40319
Let $V,W$ be normed vector spaces, and $L(V,W)$ be the space of bounded linear operators. Usually I would only see the statement "If $W$ is Banach, then $L(V,W)$ is Banach.". But  writes that there is a converse: "If $L(V,W)$ is Banach, and if $V$ is non-trivial, then $W$ is Banach". This is pretty interesting since I never seen a converse before. I was wondering if anyone has a nice proof.  I tried reversing the proof for the usual direction, but the inequalities can't be reversed. I also tried to start with a Cauchy sequence in $W$, and construct linear operators (using Hahn-Banach to control the operator norms), but alas I cannot say much about the distance between these operators (much less say that it is Cauchy)
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 $\alpha \in V$.
74690
I need help to answer the folowing problem:  Let $F$ be a field, $n\ge 1$ and $G=F^n\bigoplus F^n\bigoplus Mat(n,F).$ That is an element of $G$ has the form $(v,w,X)$ where $v$ and $w$ are n-by-1 vectors and $X$ is $n$-by$-n$ matrix, all with entires in $F$. Define an operator on $G$ by: $$(v_1,w_1,X_1)(v_2,w_2,X_2)=(v_1+v_2,w_1+w_2,X_1+X_2+v_1w_2^t)$$ Show that $G'=\left\{(0,0,X), X\in Mat(n,F)\right\}$. Where $G'$ is the derived group of $G$ i.e $G′=\left\{[a,b],a,b∈G\right\}$
83999
Let  $S:  V \to W$ and $T: V \to W$ be linear transformations. Given $a$ in $\mathbb{R}$, define functions $(S+T): V \to W$ and $(aT): V \to W$ by $(S+T)(v)=S(v) + T(v)$ and $(aT)(v)= aT(v)$ for all $v$ in $V$. Show that $S+T$ and $aT$ are linear transformations.
15674
Let $V$ be a real n dimensional vector space &amp; let $T:V→V$ be a linear transformation satisfying $T^2(v)=-v$ for all $v\in V$. Then how can we show that $n$ is even?  I am completely stuck on it. Can anybody help me please?
113362
I need some help to show that:  If  $| \langle u,v \rangle | = \|u\| \|v\|$ then $u=\lambda v$ for some scalar $\lambda$.  We have to consider this over an arbitrary field $\mathbb{F}$.  I appreciate your suggestions.  Thanks.
53195
Let $V$ be a finite-dimensional vector space over $F$. Let $T:V \rightarrow V$ be a linear transformation such that $ST=TS$ for all linear transformations $S:V \rightarrow V$. Show that $T = cI_v$ for some $c \in F$.
112741
If $V_1$ and $V_2$ are vector spaces over the field of rational numbers and $f \colon V_1 \to V_2$ is a map such that $f(x+y)=f(x)+f(y)$ for all $x, y \in V_1$, show that $f$ is a linear transformation.   It is easy to show that $f(0)=0$ and $f(-x)=-x$. Here scalars are from rationals. If scalar is an integer then clearly $f(\alpha x)=\alpha f(x)$. Hence $f$ is a linear transformation.  Now I have to show that when $\alpha$ is not an integer then $f(\alpha x)=\alpha f(x).$