I want to prove $\exp : \mathfrak{gl}(n, \mathbb{C}) \longrightarrow GL(n,\mathbb{C})$ is surjective.   The textbook gives hint as  "using Jordan canonical form"  So my guess is expressing matrix in terms of similarity transformation, but how does it guarantee that the exponential map is surjective?   Can you give me some proof of above statement.
21836
Let $M_n(\mathbb{F})$ be the set of all $n\times n$ with entries in $\mathbb{F}$ and let $\exp:M_n(\mathbb{C})\to M_n(\mathbb{C})$ be defined by  $$ \exp(A)=\sum_{k=0}^{\infty}\frac{A^k}{k!},$$ for all $A\in M_n(\mathbb{C}).$  I want to prove that $\exp$ is a surjective map from $M_n(\mathbb{C})$ to $GL(n,\mathbb{C})=\left\{A\in M_n(\mathbb{C})\,\middle| \det(A)\neq0\right\}$, how do I go about that?   I mean saying that $\exp:M_n(\mathbb{R})\to GL(n,\mathbb{R})$ is an analogous to saying $\exp:\mathbb{R}\to \mathbb{R}_{&gt;0}$ and this is also pretty intuitive, since, in analogy with the case of numbers, $A^0=I\;\forall A$, so $\exp(0)=I+0+\frac{0^2}{2!}+\dots=I$, so even for $A=0$ we get $\det\left(\exp(A)\right)\neq0$ and so because of the first term we can never get a zero determinant. But I have no idea how to prove the subjectiveness.  Thanks in advance.
36711
I'm searching for a way to generate the group $\mathrm{GL}(n,\mathbb Z)$. Does anyone have an idea?  The intention of my question is that I am searching for an easy proof of the existence of the epimorphism:  $\Phi:\mathrm{Aut}(F_n )\to \mathrm{Aut}(F_n/[F_n,F_n])=\mathrm{Aut}(\mathbb {Z}^n)=\mathrm{GL}(n,\mathbb {Z})$  I know that $\Phi$ is an canonical homomorphism since the commutator subgroup is characteristic in $F_n$. So I need some nice generators of $\mathrm{GL}(n,\mathbb {Z})$ to find their preimages in $\mathrm {Aut}(F_n)$ to prove the surjectivity of $\Phi$.  Thanks for help!
47791
Consider the Killing form of the Lie algebra $\mathfrak{gl}_{m}$. Then $\{e_{ij}\}$ is a basis for this Lie algebra where $e_{ij}$ is a matrix with 1 in the $i$th row, $j$th column and 0 everywhere else. Then every $A = (a_{ij}) \in \mathfrak{gl}_{m}$ can be written as $A = \sum_{i, j = 1}^{m}a_{ij}e_{ij}$. Since $[e_{ij}, e_{kl}] = \delta_{jk}e_{il}-\delta_{li}e_{kj}$, we have $ad(a)e_{ij} = \sum_{kl}(a_{ki}\delta_{lj} -a_{jl}\delta_{ki})e_{kl}$. From this we have $$ad(a)ad(b)e_{ij} = \sum_{k, r, s}(a_{rk}b_{ki}\delta_{sj} + a_{ks}b_{jk}\delta_{ri} - a_{ri}b_{jk}\delta_{sk} - a_{js}b_{ki}\delta_{rk})e_{rs}.$$ From this how can I see that the Killing form is $2m tr(ab) - 2tr(a)tr(b)$. I know that the Killing form is just the trace of the above centered equation, but I seem to be getting confused with all the indices.
106539
How do I see that the representing matrix of the dual map $f^*$ between finite-dimensional dual spaces is given by the transpose of the representing matrix of $f$? Here I want to assume that the matrix $f^*$ is represented with respect to the dual basis.  Apparently this result is very well-known, but I would like to see proof of this.
71829
(1) Prove or disprove: For any sets $X,Y,Z$ and any maps $f:X \to Y$ and $g:Y \to Z$, if $f$ is injective and $g$ is surjective, then $g \circ f$ is surjective.  So i proved previously that $f$ is injective if $g \circ f$ is injective, but then i also proved that if $g \circ f$ is surjective then $g$ is surjective. then this question appears, so how do i prove it that it is true or false? I'm saying it's true, because $g \circ f$ is bijective, so how do i go around proving it? Or my idea is already wrong? I need some help thanks.  (2) Show that (it is true in general that) for any sets $A,B$, one has $P(A) \cup P(B) \subseteq P(A \cup B)$.  I know that power set has exactly all the subsets of $A$ for $P(A)$. but I dont know how to start.
37905
(reference is Lawvere/Schanuel, Session 31, Ex. 1)  I'm trying to calculate the exponential object $A^A$ and its evalution map $e \colon A \times A^A \to A$ in the category of graphs, where $A$ is the "arrow graph" (ie. one arrow and two dots).  In the following, $D$ is the graph with one dot and no arrows, $1$ is the terminal object in this category (graph with one dot and one arrow, the loop).  So far I have:   The points of $\mathbf{1}\to A^A$ correspond to the maps $A\to A$ (via two standard isomorphisms), and since $\mathbf{1}$ is the loop, and there is one map of graphs $A \to A$, there is one loop in $A^A$. The dots $D\to A^A$ correspond to the maps $A \times D \to A$, of which there are four, hence four dots in $A^A$. The arrows $A \to A^A$ correspond to the maps $A \times A \to A$, of which there are four, hence four arrows in $A^A$.   But I'm stuck on how to put these together to constitute $A^A$ and its evaluation map.
16678
Try to prove the following proposition.     $SL_n(\mathbb R)$ is a subgroup of kernel of $f\in$ Hom($GL_n(\mathbb R), A)$, where $A$ is an abelian group.   My first attempt is to show that $SL_n(\mathbb R)$ is in form of $ABA^{-1}B^{-1}$, then it's clear that $ABA^{-1}B^{-1}$ is in $ker(f)$. However, there exists an elementary row reduction matrix, i.e. $E_{ij}(m)$ adding m times $j$-th row to $i$-th row. Then $E_{ij}(m)\in SL_n(\mathbb R)$. Now, it is possible that $E_{ij}(m)$ is not in $ker(f)$. Then I try to find any relation between $\mathbb R^{\times}$ and $A$, but, except knowing they are both abelian groups, I cannot figure out what else connects those two group. Could you give me some hints?
96112
Let $A \in M_n(\mathbb R)$ and $\exp(A)=\sum_{i=0}^{\infty}\frac{A^i}{i!}$. How to prove $\exp(A)$ is an orthogonal matrix iff $A^t=-A$?  I can get that $\exp(A)$ is an orthogonal matrix iff $\exp(A^t)=\exp(-A)$, but how to get $A^t=-A$ from it?  What's more, is $\exp(A)$ injective or not?
27013
Let $T : V \longrightarrow W$ and $U : W \longrightarrow Z$ be linear maps. How do I prove that $(U\circ T)^{*} = T^{*}\circ U^{*}$? I'm used to seeing $V^{*}$ not $(U\circ T)^{*}$. Any help is appreciated.  $^{*}$ denotes the dual map (transpose).
99501
$$GL_n(\Bbb{R})/SL_n(\Bbb{R}) \cong \Bbb{R}^\times$$  This is trivial to prove with the first isomorphism theorem - by using the determinant as a endomorphism, then as $SL_n(\Bbb{R})$ is $1$ under the determinant, it is the kernel, and by FIT, the above is proved.  I was wondering how to prove this without the isomorphism theorem ?
111828
I know, fundamental group of $GL(n,\mathbb{C})$ is isomorphic to $\mathbb{Z}$. It's written . Actually, I've succeed in proving this, but my proof is two pages long and very technical. I want   to find some better proofs of this fact (in order to compare to mine);   to find some book or article, using which I can learn, how to calculate fundamental groups and, more generally, connectedness components of maps from one space to another;   to find something for the reference, which I can use in order to learn, how to write this proofs nicely, using standard terminology.
96857
Prove that  $GL(n, \mathbb{C})$ is isomorphic to a subgroup of $GL(2n, \mathbb{R})$.  My Proof: For an $A \in GL(2, \mathbb{C})$, $$ A = \begin{bmatrix} a+bi &amp;c+di \\ e+fi &amp; g+hi \end{bmatrix} =  \begin{bmatrix} a &amp;c \\ e &amp; g \end{bmatrix} + i \begin{bmatrix} b &amp; d \\ f &amp; h \end{bmatrix} .$$ It follows that $ GL(n, \mathbb{C})$ is a subset of $GL(2, \mathbb{R}) \oplus GL(2, \mathbb{R})$. (it is a subset because the condition that $\det A \neq 0$ puts more restrictions on the elements of $A$ than the condition that the two little matrices are invertible.) Thus the result holds for $n=2$.  Can this argument be generalized to any $n$? The hint was to use group actions, of $GL(n, \mathbb{C})$ on $GL(2n, \mathbb{R})$, using the fact that $\mathbb{C}^n \cong \mathbb{R}^{2n}$. I don't understand why a group action would be helpful?
31461
Suppose that $l_1 = \{ (x_n)_{n \in \mathbb{N}} | \sum|x_n| &lt; \infty \}$ and $l_{\infty} = \{ (x_n)_{n \in \mathbb{N}}| \sup|x_n| &lt; \infty \}$.  Show that $(l_1)^* \cong l_{\infty}$, where $(l_1)^*$ is a dual space of $l_1$.  My attempt: Define a map $L: l_{\infty} \rightarrow (l_1)^*$ given by  $$L(x)(y) = \sum_{n \in \mathbb{N}}{x_ny_n}$$  where $y = (y_n)_{n \in \mathbb{N}} \in l_1$.   My aim is to show that $L$ is an isometric isomorphism.   Clearly $L$ is linear and and injective (choose $y = e_n$ for all $n \in \mathbb{N})$.  How to show that $L$ is a surjection and isometry?
93710
Statement: There are uncountably many surjections from $\mathbb N$ to $\mathbb N$.  Is the following proof valid:  The real numbers are uncountable.   There exists a bijection between the set of real numbers and the set of all functions from $\mathbb N$ to $\{0,1\}$.   Given a function, $f$, from $\mathbb N$ to $\{0,1\}$, construct a surjection from $\mathbb N$ to $\mathbb N$, denoted $g$, as follows:  We map the $n$th composite number to $n$, and the nth prime, to $f(n)$.   Clearly this is a surjection since the composite numbers are countable. Clearly different functions $f$ lead to different surjections $g$. Since there are uncountably many $f$s, there are uncountably many surjections $g$.
44187
Suppose $f: A \to B$ and $g: B \to C$ are both surjections.  Since $f$ is surjective, then for every $b \in B$ there exists $a \in A$ such that $f(a)= b$.  Since $g$ is surjective, for every $c \in C$, there's  $b \in B$ such that $g(b) = c$. Then $g(f(a)) = g(b) = c$.   Would this work?  edit; I didn't realize there are similar questions on this site. Don't mind if this post is deleted.
93443
$$(\mathbb{Z}/(n))[x] \simeq \mathbb{Z}[x] / (n)$$  I've shown this by showing that the map that sends $\overline{1} \mapsto [1+(n)]$ (where the bar denotes the congruence class mod $n$) and $x \mapsto [x+(n)]$ is a homomorphism that is injective and bijective, which is a bit cumbersome. Is there an easier way to show this? I can't tell if this is trivial or not.
16177
On a finite dimensional vector space, the answer is yes (because surjective linear map must be an isomorphism). Does this extend to infinite dimensional vector space? In other words, for any linear surjection $T:V\rightarrow V$, AC guarantees the existence of right inverse $R:V\rightarrow V$. Must $R$ be linear?  How about $T:V\rightarrow W$ linear surjection in general?