Let $\alpha: V \rightarrow V$ be an endomorphism, and let $r_k = $ rank($\alpha^k$). Show that $r_k-r_{k+1} \geq r_{k+1}-r_{k+2}$.  I have already shown that Im($\alpha^{k+1}$) $\leq$ Im($\alpha^{k}$) and Ker($\alpha^{k+1}$) $\geq$ Ker($\alpha^{k}$) for all $k$, and if $\alpha$ is an automorphism then equality holds throughout; but I can't seem to show why the rank has to obey the above. Rank-nullity doesn't seem to be helping.
120708
Let $S: V \to V$ be a linear transformation. Show that $V \supseteq \operatorname{Im}(S) \supseteq \operatorname{Im}(S^2) \supseteq \cdots$ and $\{0\} \subseteq \ker(S) \subseteq \ker(S^2) \subseteq \cdots$. If $a_k = \operatorname{rank}(S^k)$, deduce that $a_k \geq a_{k+1}$ and that $a_k - a_{k+1} \geq a_{k+1} - a_{k+2}$.  I can show everything except for $a_k - a_{k+1} \geq a_{k+1} - a_{k+2}$, which eludes me. I attempted using the rank-nullity theorem, but no luck so far. Any ideas/hints?
57655
1) How can I show that any one to one endomorphism of an Artinian module is an automorphism.   2) I also want to show that any onto endomorphism of a Noetherian module is an automorphism.   I need counterexamples for modules without chain conditions.
20584
Let $R$ be a reflexive and transitive relation on $X$. Show that $R \cap R^{−1}$ is an equivalence relation on $X$.   I am stuck and feel like I am doing something wrong, any help would be much appreciated!  Here is what I have so far:  $R = \{1,2,3\}$  $X = \{2,3,4,5,6\}$  $R$ on $X = \{(1,2), (1,3),(1,4),(1,5),(1,6),(2,4),(2,6),(3,6)\}$  $R^{-1} = \{1,1/2,1/3\}$  $R \cap R^{-1} = \{1\}$  $R^{-1}$ on $X = \{(1,2),(1,3),(1,4),(1,5)(1,6)\}$  TRUE  Am I completely off-base or am I close?  Thanks!
103007
Question. Let $M = \left\{ \begin{bmatrix} a &amp; b \\ -b &amp; a \end{bmatrix} \,\middle\vert\,  a, b \in \mathbb R \right\}$.  Show that $(M,+)$ and $(\mathbb C,+)$ are isomorphic.  Show that $(M^{*},*)$ and $(\mathbb C^{*},*)$ are isomorphic.    (Here, $M^{*}$ is defined as excluding the zero-element) I understand that there exists an isomorphism $\phi:M \rightarrow\mathbb C$ defined by $\begin{bmatrix} \alpha &amp; \beta \\ -\beta &amp; \alpha \end{bmatrix}\mapsto \alpha+i\beta$, ie. I can show that $\phi(M_{1}+M_{2}) = \phi(M_{1})+\phi(M_{2})$ and $\phi(M_{1}*M_{2}) = \phi(M_{1})*\phi(M_{2})$. However, I'm unsure how to go about the two subquestions seperately.
95508
The following is from Hungerford's graduate algebra book, Chapter 1, Section 4, Proposition 8.    The proposition states that: If $H$ and $K$ are subgroups of a group $G$, then $[H:H \cap K] \leq [G:K]$. If $[G:K]$ is finite, then $[H:H \cap K] = [G:K]$ if and only if $G = HK$.  I am trying to work my way through the proof (the book only provides an outline). I have show that $[H:H \cap K] \leq [G:K]$. I am now trying to show the second part, that $[H:H \cap K] = [G:K]$ if and only if $G = HK$. The outline says that to show this we prove two things:  1.) $[H:H \cap K] = [G:K]$ if and only if $f$ is surjective, where $f$ is the map given by $(H \cap K)h \rightarrow Kh$ (I have shown this part).  2.) $f$ is surjective if and only if $G = HK$.   I am stuck on showing the second part.   Thank you in advance for any help you can give.
47249
Let $f:F_n\to F_m$ be an epimorphism ($n\geq m$). Then it is true that there is a basis $X=X_1\sqcup X_2$ in $F_n$ such that $f$ maps $\langle X_1\rangle$ isomorphically onto $F_m$, and maps $X_2$ to identity. One can probably deduce this statement from the Grushko-Neumann theorem, but somehow I cannot get an elegant proof of it.   I can see that by the Grushko-Neumann we can assume that $m=1$, and hence our epimorphism factors through the abelianization. And then an ugly linear algebra appears... Any suggestions how to make the proof slick?
116946
Suppose that:  $0 \rightarrow M_{1} \rightarrow M \rightarrow M_{2} \rightarrow 0$  Is an exact sequence of $R$-modules (for $R$ commutative integral domain). Show that $\mathrm{rk}(M) = \mathrm{rk}(M_{1})+\mathrm{rk}(M_{2})$    I have shown the inequality $\mathrm{rk}(M) \geq \mathrm{rk}(M_{1})+\mathrm{rk}(M_{2})$, however for the opposite direction I'm lost. I tried to take $N$ as a linear independent subset and tried to decompose it into $N_{1}$ (the elements in $\mathrm{Ker}(g)$) and $N_{2}$ (the elements not in $\mathrm{Ker}(g)$), but I haven't got far, since $N_{1}$ gives us a linearly independent set in $M_{1}$, however $N_{2}$ can give us a linearly dependent set in $M_{2}$.   We define $\mathrm{rk}(M)$ in this case as the supremum over all cardinalities of the linearly independent subsets of $M$ (where the only finite linear combination equaling $0$ implies trivial coefficients for $r \in R$).  Would this be the proper way to show the result? I think I'm missing something important.
31554
Let $n&gt;1$ , consider $\mathbb R^n$ with standard volume form $\omega _0 =dx_1 \wedge ...\wedge dx_n$ ; let $r,R$ be two positive real numbers with $R&gt;r$ . Then is it true that for every $a,b \in B(0;r)$ , $\exists$ a diffeomorphism $f:\mathbb R^n \to \mathbb R^n $ such that $f(a)=b ; f(x)=x , \forall x \in \mathbb R^n $ with $||x||&gt;R$ and $f^*\omega _0=\omega _0$ ?   As $f^*(dx_1 \wedge ...\wedge dx_n)=(det f' )(dx_1 \wedge ...\wedge dx_n)$ , so basically given $R&gt;r&gt;0$ and $||a||,||b||&lt;r$ , we have to find a diffeomorphism $f : \mathbb R^n \to \mathbb R^n  $ with $f(a)=b$ ; $f|_{\mathbb R^n \setminus B[0;R]}=Id$ and $det f'(p)=1 , \forall p \in \mathbb R^n $ .  Definitely this cannot work out in $\mathbb R$ , because the only differentiable function satisfying $det f'(p)=1 , \forall p$ and $f(a)=b$ is $f(x)=x+b-a$ and this function never gives identity unless $a =b$
33147
I'm sure I've done this before in abstract algebra. Regardless it's escaped me now.   I have proved that for $T:U\rightarrow V$, with $dim(U)=m$ and $dim(V)=n$ that $rank(T)\le m$ which is obvious, but I have proved it none the less.  I want to.... this is where I get stuck. I'm not quite sure how to say it.  Suppose we have a $T$ for example that takes $U\subset\mathbb{R}^3\rightarrow V\subset\mathbb{R}^3$, we could still have a $T$ that maps a plane to something (a plane, line or point) in this case rather than $rank(T)\le 3$ I can state $rank(T)\le 2$.  (Correct my notation here, I don't like writing subset, I mean number of dimensions of the space!)  To show this I can say $F:P\subset\mathbb{R}^2\rightarrow U\subset\mathbb{R}^3$.  Then let $G:P\rightarrow V$ and that $G=TF$, now $rank(G)\le dim(P) = 2$  Now, if $U,P,Q,R$ are spaces of dimension $\le 2$ and P,Q,R are subspaces of U  $A:U\rightarrow P$  $B=A:P\rightarrow Q$  $C=A:Q\rightarrow R$  We can now say $A^2=BA$ then $A^3=CBA$  We know that for $A$ rank(A)$\le 2$, this thus provides an upper-bound for rank($A^3$)  I want to show now that if I keep applying a linear transformation that the rank is a monotonically decreasing ($\le$) sequence. I am unsure on a proof that the rank of a transformation cannot be greater than the domain's dimensions. Although this is trivial.  (NOTE:  I have proven that if a set of n vectors, R, span a vector space V, and you take a set W of m linearly independent vectors in V that m$\le$n)  If rank(A)=2 then rank($A^2$)=2 and so rank($A^3$)=2  if rank(A)=1 then rank($A^2$)=0 and rank($A^3$)=0  if rank(A)=0 then rank($A^2$)=0 and rank($A^3$)=0  But again, I can't prove this, or at least I am unsure of how to write it.  something to clarify:  For a map $T:U\rightarrow V$ how do we distinguish (using notation) whether or not we actually use all dim(U)?  For example we might have a T that maps a plane in R^3 to a line in R^3, this can be expressed as a composition of maps, one that takes a 2 dimensional vector space (coordinates of the plane in 3 space) to a 3 dimensional point in T's domain, which T then maps to a line.  We have rank - the dimensions of the image - to make this distinction on the map's target (the rank of T is 1 if it maps something to a line, but the dimensions of the target is 3)  How do we write this? What do I say to describe this?   How do I distinguish between the number of vectors in a basis and the number of dimensions the space that basis is in has?  Sorry for the length of this! I tried to talk though my problem in the hope of seeing it myself, no such luck, thanks.
91745
Why the finite order endomorphisms are diagonalizable over the complex numbers (or any algebraically closed field where the characteristic of the field does not divide the order of the endomorphism) with roots of unity on the diagonal?
23648
I am trying to prove a series of propositions:  Given any homomorphism p from $\mathbb{R}$[X] to $\mathbb{R}$[X], show that it is equal to $\phi_g$ for a unique g in $\mathbb{R}$[X], with $\phi_g$(f) = f(g(X)). I've expanded the expression $p(f) = p(\sum_{i=0}^n a_iX^i) =\sum_{i=0}^n p(a_i)p(X)^i$ but I'm not sure how to show $p(f) = \sum_{i=0}^n p(a_i)p(X)^i = \sum_{i=0}^n a_ig(X)^i = \phi_g(f)$.  Show that if h,g $\in \mathbb{R}$[X] are such that h(g(X)) = X, then g(X) = aX+b for a $\in$ R$^x$ and b$\in \mathbb{R}$.
86528
How can I verify the following statement?     The image of a morphism $f : A \longrightarrow B$ is defined as $\text{Im}(f) = \text{Ke}r(\text{coke}r f)$ whenever it exists (e.g., in every abelian category). The morphism $f : A \longrightarrow B$ factors   uniquely through $\text{Im} f \longrightarrow B$ whenever $\text{Im} f$ exists, and $A \longrightarrow \text{Im} f$ is an epimorphism and a cokernel of ker $f \longrightarrow A$ in every abelian category.
111350
Let $V$ be a vector space and $\alpha \in \operatorname{End}(V)$   (i) If $V$ is finite dimensional, then $\alpha$ is injective iff $\alpha$ is surjective.  (ii) Give example showing (i) is false if $V$ is not finite dimensional.  So on (i), since $V$ is finite dimensional, then $V$ has a basis with finite cardinality and hence $\dim(V)=n$. Also, the following holds, $\dim(V)=\dim(\ker(\alpha) + \dim(\operatorname{im}(\alpha))$ Since $\alpha$ is injective, then the $\ker(\alpha)$ is $0$ which implies that the $\dim(\ker(\alpha)$ is $0$ which implies that $\dim(\operatorname{im}(\alpha))=\dim(V)$ and that shows that $\alpha$ is surjective. (Is that right?)   On the second one, (ii), I just have no idea. I mean shouldn't Axiom of Choice be able to work here?
46056
Let $V$ be a vector space , $T:V \to V$ be a linear operator , then is it true that   $\ker (T) \cap R(T) \cong R(T)/R(T^2) $ ( where $R(T)$ denotes the range of $T$ ) ?    I know that the statement is true when $V$ is finite dimensional as I can show that if $V$ is finite   dimensional , then $\dim (\ker (T) \cap R(T))=rank (T) - rank (T^2)=\dim R(T)/R(T^2)$ ,   but I don't know what happens in general . Please help . Thanks in advance
71710
I am doing a self-study using Artin's algebra. There is a problem which asks us to prove that adjoining an element to a ring doesn't do anything if that element was already in the ring.  More formally, say that $\alpha=a$ where $a\in R$. We want to show that $R\approx R[\alpha]$. (Where $\approx$ means "isomorphic to".)   My proof: We can consider $R[\alpha]\approx R[x]/(x-\alpha)$, which intuitively means "killing" $x-a$ in the ring of polynomials. Any function $f\in R[x]$ can be written as $(x-\alpha)g(x)+r(x)$ for some g,r by the division algorithm. Since the degree of r must be less than the degree of $x-a$, r must be a constant polynomial. Since $(x-\alpha)=0$, the residue of $f$ in $R[x]/(x-\alpha)$ is just r, which is some element of R. We can see that every $s\in R$ has a corresponding constant polynomial $f(x)=s$ in $R[x]$, which means the simple inclusion map $s\mapsto f(x)=s$ is our desired isomorphism.  So, this proof seems fine to me, except that I seem to have proven that adjoining any element keeps the ring the same. This sounds implausible to me. Where did I go wrong?
27599
This question is an extension of .  Let $F_X$ denote the free group on the set $X$. For any group $G$ and subset $S\!\subseteq\!G$, $\langle S\rangle$ denotes the subgroup generated by $S$ and $\mathrm{rank}(G) :=\min\{|S|;\:S\!\subseteq\!G, \langle S\rangle\!=\!G\}$.  PROPOSITION: a) $F_X\cong F_Y\:\Leftrightarrow\:|X|=|Y|$ b) $\mathrm{rank}(F_X)=|X|$  Thus for every cardinal number $c$, there is (up to isomorphism) exactly one free group of rank $c$.  Proof:    a) $(\Leftarrow)$: If $f\!:X\rightarrow Y$ is the bijection, then $\varphi(x_1\ldots x_k):=f(x_1)\ldots f(x_k)$ is the isomorphism.  $(\Rightarrow)$: $F_X\!\cong\!F_Y$ $\Rightarrow$ $\mathrm{Ab} F_X\!\cong\!\mathrm{Ab} F_Y$ $\Rightarrow$ $\oplus_{x\in X}\mathbb{Z}\!\cong\!\oplus_{y\in Y}\mathbb{Z}$, so $|X|\!=\!|Y|$, since rank is known to be an invariant of free modules.   Alternatively, $\big(\oplus_{x\in X}\mathbb{Z}\big)\otimes_\mathbb{Z}\mathbb{Q}$ $\cong$ $\oplus_{x\in X}\big(\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Q}\big)$ $\cong$ $\oplus_{x\in X}\mathbb{Q}$, so $\oplus_{x\in X}\mathbb{Q}$ $\cong$ $\oplus_{y\in Y}\mathbb{Q}$, even as $\mathbb{Q}$-modules, but isomorphic vector spaces are known to have equipollent bases.  b) Since $\langle X\rangle\!=\!F_X$, $\mathrm{rank}(F_X)\leq|X|$. Suppose we have $Y\!\subseteq\!F_X$, $\langle Y\rangle\!=\!F_X$, $|Y|\!&lt;\!|X|$.  QUESTION: how can I finish the proof of b), i.e. prove that $F_X$ can't be generated by a subset with smaller cardinality than $|X|$?
95959
If $A\in\mathfrak{M}_{m,n}(F)$ and $B\in\mathfrak{M}_{n,m}(F)$, prove that $AB=I_m$ implies $rk(A)=rk(B)$. Here $A\in\mathfrak{M}_{m,n}(F)$ means $A$ is a ($m \times n$)-matrix with entries in $F$. My approach: I tried to consider $A$, $B$ as linear maps, $L_A:F^n\to F^m$, $L_B:F^m\to F^n$. And we know $rk(A)=\dim{\mathrm{Im}~L_A}$, $rk(B)=\dim{\mathrm{Im}~L_B}$. Then by the Dimension Theorem, it suffices to show that $n-\dim{\ker{L_A}}=m-\dim{\ker{L_B}}$. However, I don't know how to prove this.