Combinatorial proof of the identity ${n+1\choose k+1}={k\choose k}+{k+1\choose k}+...+{n-1\choose k}+{n\choose k}$
Combinatorics identity question
How to intuitively understand eigenvalue and eigenvector? I'm learning multivariate analysis and I have learnt linear algebra for two semester when I was a freshman.  Eigenvalue and eigenvector is easy to calculate and the concept is not difficult to understand.I found that there are many application of eigenvalue and eigenvector in multivariate analysis. For example     In principal components, proportion of total population variance due   to kth principal component equal    $$\frac{\lambda_k}{\lambda_1+\lambda_2+...\lambda_k}$$   I think eigenvalue product corresponding eigenvector has same effect as the matrix product eigenvector geometrically.   I think my former understanding may be too naive so that I cannot find the link between eigenvalue and its application in principal components and others.   I know how to induce almost every step form the assumption to the result mathematically. I'd like to know how to intuitively or geometrically understand eigenvalue and eigenvector in the context of multivariate analysis(in linear algebra is also good).  Thank you! This is from Fraleigh's First Course in Abstract Algebra (page 82, Theorem 8.16) and I keep having hard time understanding its proof. I understand only until they mention the map $\lambda_x (g) = xg$. Can someone explain this proof step by step? Thank you so much! Here is the proof:  Let $G$ be a group. We show that $G$ is isomorphic to a subgroup of $S_G$. Define a one-to-one function $\phi: G \to S_G$ such that $\phi(xy)=\phi(x) \phi(y)$ for all $x,y \in G$. For $x \in G$, let $\lambda_x: G \to G$ be defined by $\lambda_x (g) = xg$ for all $g \in G$. The equation $\lambda_x (x^{-1} c) = x(x^{-1} c) = c$ for all $c \in G$ shows that $\lambda_x$ maps $G$ onto $G$. If $\lambda_x (a) = \lambda_x (b)$, then $xa=xb$ so $a=b$ by cancellation. Thus $\lambda_x$ is also one to one, and is a permutation of $G$. We now define $\phi: G \to S_G$ by defining $\phi(x) = \lambda_x$ for all $x \in G$. To show that $\phi$ is one to one, suppose that $\phi(x) = \phi(y)$. Then $\lambda_x = \lambda_y$ as functions mapping $G$ into $G$. In particular $\lambda_x (e) = \lambda_y (e)$, so $xe=ye$ and $x=y$. Thus $\phi$ is one to one. It only remains to show that $\phi(xy) = \phi(x) \phi(y)$, that is, that $\phi_{xy} = \lambda_x \lambda_y$. Now for any $g \in G$, we have $\lambda_{xy} (g) = (xy)g$. Permutation multiplication is function composition, so $(\lambda_x \lambda_y)(g) = \lambda_x (\lambda_y (g)) = \lambda_x (yg) = x(yg)$. Thus by associativity, $\lambda_{xy} = \lambda_x \lambda_y$. If $l$ and $n$ are any positive integers, is there a  proof of the identity     $$\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(l-k)^n=n!\;$$   which uses the Inclusion-Exclusion Principle?  (If necessary, restrict to the case where $l\ge n$.)    This question is closely related to   and also "PuTTY key format too new" when using ppk file for PuTTY SSH key authentication I wanted to connect to my Unix server using SSH keys. For that I had converted my id_rsa file from ./ssh directory to a ppk file using PuTTYgen, but after loading my .ppk key I get this warning:  PuTTY key format too new   If I use my default id_rsa file, then I get error:  OpenSSH SSH-2 private key (old PEM format)  Either way I am getting stuck. Can anyone please help me? I know that I can use  to make cross-references and hyperlinks clickable.  That makes the clickable areas outlined in fluorescent green, however.  How can I make the green boxes go away? Limit $ a_{n+1} = \frac{1}{3}(2a_n + \frac{5}{a_n^{2}}), a_1 > 0 $ Let $ a_{n+1} = \frac{1}{3}(2a_n + \frac{5}{a_n^{2}}), a_1 &gt; 0 $. Show the sequence converge and find it's limit.    I did the following:  We see that $a_n$ is positive for every n obviously (considering $a_1 &gt; 0$). Then:  $$ a_{n+1} - a_n = \frac{2a_n^{3} + 5}{3a_n^{2}} - a_n = \frac{5 - a_n^{3}}{3a_n^{2}}$$  We see that for each $a_n &gt; ^3\sqrt{5}$ the sequence is decreasing. That means the proof collapses to prove  $a_n &gt; ^3\sqrt{5}$ in order for $a_n$ to be decreasing. Suppose $ a_n &gt; ^3\sqrt{5}$ then obviously $a_{n+1} = \frac{2a_n^{3} + 5}{3a_n^{2}} &gt; \frac{2(\sqrt[3]{5})^{3} + 5}{3(\sqrt[3]{5})^{2}} = \sqrt[3]{5}$ so we deduced $a_{n+1} &gt; ^\sqrt[3]{5} $.  (It's important to note $a_1$ might be less then $\sqrt[3]{5}$, hence the sequence will decrease from the second element.) So we observed $a_n$ is always positive and decreasig hence it's bounded below so it's bounded. As for it's limit:   $$\begin{align} L = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \frac{1}{3}(2a_n + \frac{5}{a_n^{2}}) = \frac{1}{3}(2L + \frac{5}{L^{2}}) =  \sqrt[3]{5}\end{align}$$  I'm not sure as of the correctness of the solution. I'm aware that it's not a contradiction because it only implies it's in the epsilon-environment of $\sqrt[3]{5}$, or rather in this case $(\sqrt[3]{5}, \sqrt[3]{5}+\epsilon)$.  Is it correct?  EDIT: If anyone encounter a question in the likes, to sum it up you obviously should show the sequence is montonic and bounded. In this case for example we can show it using AM-GM inequality to see the sequence is bounded by $\sqrt[3]{5}$ and then in the difference of two consequective elements we see the difference is negative due to every element being greater then the third root of 5. From here we know the sequence is bounded below and decreasing hence it's bounded. If we plug in the limit of the sequence to itself we will see it's the third root of five. And that's it. Why does English spelling use silent letters? Why have a letter in a word when it’s silent in pronunciation, like the b in debt?  Can anyone please clarify my uncertainty here? How to style text in hyperref \url? 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I installed Python 3.3 from the source file and it was compiled into /usr/local/bin. Then I created a symbolic link between /usr/bin/python and /usr/local/bin/python3, so that whenever I invoke python from the command line it uses the latest version.  Now I am trying to install MySQL Python and I got the following output:  apt-get install python-mysqldb  Reading package lists... Done Building dependency tree Reading state information... Done python-mysqldb is already the newest version. The following extra packages will be installed:   apt-listchanges python-apt Suggested packages:   python-glade2 python-gtk2 python-apt-dbg python-vte python-apt-doc The following packages will be upgraded:   apt-listchanges python-apt 2 upgraded, 0 newly installed, 0 to remove and 142 not upgraded. 3 not fully installed or removed. Need to get 0 B/394 kB of archives. After this operation, 250 kB of additional disk space will be used. Do you want to continue [Y/n]? 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