Problem  Let $A$ and $B$ be similar matrices. Prove that the geometric multiplicities of the eigenvalues of $A$ and $B$ are the same. [Hint: show that, if $B=P^{-1}AP$, then every eigenvector of $B$ is of the form $P^{-1}\vec{v}$ for some eigenvector $\vec{v}$ of A].  Attempted Proof Write $A \vec{v} = \lambda \vec{v}$.   Since $B=P^{-1}AP$, this is equivalent to saying $$B (P^{-1}\vec{v})=P^{-1}APP^{-1}\vec{v} = P^{-1}A\vec{v} = \lambda (P^{-1} \vec{v})$$ Since $P^{-1} \vec{v}$ is an eigenvector of $B$ only for eigenvalues of $A$, the dimensions of the eigenspaces must be the same. Is this proof complete?
Suppose $A$ and $B$ are similar matrices. Show that $A$ and $B$ have the same eigenvalues with the same geometric multiplicities.      Similar matrices: Suppose $A$ and $B$ are $n\times n$ matrices over $\mathbb R$ or $\mathbb C$. We say $A$ and $B$ are similar, or that $A$ is similar to $B$, if there exists a matrix $P$ such that $B = P^{-1}AP$.
Video produced from images only plays in VLC but no other players Show $a\Bbb Z+b\Bbb Z = \gcd(a,b)\Bbb Z$ Is there a way to call process builder from Apex? I can't work out why Altium is doing this, or how to undo it, or how to make it stop doing it.  It happens when I use the 'Find Similar Objects... ' option; all of objects that it hasn't found are faded out and I can no longer select them.  Anyone know how to fix this?  Additionally, if anyone knows how to change, en masse, the components' footprint library to 'Any' through the SCH inspector, that'd be really helpful too. Back transforming regression results when modeling log(y) I'm fitting a regression on the $\log(y)$.  Is it valid to back transform point estimates (and confidence/prediction intervals) by exponentiation?  I don't believe so, since $E[f(X)] \ne f(E[X])$ but wanted other's opinions.  My example below shows conflicts with back transforming (.239 vs .219).  set.seed(123)  a=-5 b=2  x=runif(100,0,1) y=exp(a*x+b+rnorm(100,0,.2)) # plot(x,y)  ### NLS Fit f &lt;- function(x,a,b) {exp(a*x+b)}  fit &lt;- nls(y ~ exp(a*x+b),  start = c(a=-10, b=15))  co=coef(fit) # curve(f(x=x, a=co[1], b=co[2]), add = TRUE,col=2,lwd=1.2)  predict(fit,newdata=data.frame(x=.7)) [1] 0.2393773  ### LM Fit # plot(x,log(y)) # abline(lm(log(y)~x),col=2) fit=lm(log(y)~x) temp=predict(fit,newdata=data.frame(x=.7),interval='prediction') exp(temp)         fit       lwr       upr 1 0.2199471 0.1492762 0.3240752 Determine pid of terminated process Dependency injection and Libraries/Frameworks Why do we use the Lagrangian and Hamiltonian instead of other related functions? There are 4 main functions in mechanics. $L(q,\dot{q},t)$, $H(p,q,t)$, $K(\dot{p},\dot{q},t)$, $G(p,\dot{p},t)$. First two are Lagrangian and Hamiltonian. Second two are some kind of analogical to first two, but we don't use them. Every where they write that using them can cause problems. But why? Removing intercept from GLM for multiple factorial predictors only works for first factor in model I am running a binomial logistic regression with a logit link function in R. My response is factorial [0/1] and I have two multilevel factorial predictors - let's call them $a$ and $b$ where $a$ has 4 factor levels  $(a_1,a_2,a_3,a_4)$ and $b$ has 9 factor levels $(b_1,b_2,\dotsc,b_9)$. Therefore:  mod &lt;- glm(y ~ a + b, family = binomial(logit), data=pretend) summary(mod)   The model output would then show all the information about the model as well as the coefficients.  There is a factor level for both a and b missing (a1 and b1) from the summary output. I understand that it is constrained in the "intercept" of the model. I have read that if I want to remove the intercept term and see the estimates for those factor levels I can just add -1 to the model formula, i.e.:  mod2 &lt;- glm(y ~ a + b - 1, family=binomial(logit), data=pretend) summary(mod2)   In the new model (mod2) the intercept term is then gone and variable a's factor-level a1 is given amongst the list of coefficients. But, variable b's factor-level b1 is still missing and given that there is no intercept term anymore, how can I interpret the odds-ratio for that factor level then?  Could someone please explain to me how to get the coefficient for b1 too and why this is happening? How many $4$ digit even numbers have all $4$ digits distinct ?   This question has been asked if we replace even with positive.   I was solving this Problem like: For first position $1-9$ (No $0$) for next $0-9$ (leaving out the digit which has been choose) for next $0-9$ (leaving out $2$ digits), but on last digit I get stuck. From what I should choose. Will it be $0, 2, 4, 6, 8$, because number needs to be even? How should I solve it? After finishing The Cayo Perico Heist, how long will it be set in hard mode? After I finished The Cayo Perico Heist, I tried setting up another heist in the Kosatka, and got this message :  The setup cost for The Cayo Perico Heist is $25000. Setting up now will set the heist to Hard Mode. Do you wish to proceed?  It seems that there is a cooldown period before the heist reverts to normal mode from hard mode. How long will it take before it reverts to normal mode? Also, does the cooldown timer run even if you're logged out of GTA Online? E.g., if I quit GTA Online just after finishing the heist, the next time I log back in, will I have to wait for the cooldown period to finish before it reverts to normal mode? Integrating $\int \sin^n{x} \ dx$ I am working on trying to solve this problem:     Prove: $\int \sin^n{x} \ dx = -\frac{1}{n} \cos{x} \cdot \sin^{n - 1}{x} + \frac{n - 1}{n} \int \sin^{n - 2}{x} \ dx$   Here are the steps that I follow in the example that I am reading:     $u = \sin^{n - 1}{x}$      $du = (n - 1) \cdot \sin^{n - 2}{x} \cdot \cos{x} \ dx$      $v = -\cos{x}$      $dv = \sin{x} \ dx$            $\int \sin^n{x} \ dx =  \sin^{n - 1}{x} \cdot \sin{x} \ dx$      $\int \sin^n{x} \ dx = \underbrace{\sin^{n - 1}{x}}_{u} \cdot \underbrace{-\cos{x}}_{v} - \int \underbrace{-\cos{x}}_{v} \cdot \underbrace{(n - 1) \cdot \sin^{n - 2}{x} \cdot \cos{x} \ dx}_{du}$      $\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \cdot \cos^{2}{x} \ dx$      $\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \cdot \left(1 - \sin^{2}{x}\right) \ dx$   Here is where I get lost. How did we go from $\int \sin^{n - 2}{x} \cdot \left(1 - \sin^{2}{x}\right) \ dx$ to $\int \sin^{n - 2}{x} \ dx - (n - 1) \int \sin^{n}{x} \ dx$? Even more specifically, where did $\sin^{n}{x}$ come from?     $\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \ dx - (n - 1) \int \sin^{n}{x} \ dx$   I get this part.     $n\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \ dx$      $\int \sin^n{x} \ dx = -\frac{1}{n} \cos{x} \cdot x \ \sin^{n - 1}{x} + \frac{n - 1}{n} \int \sin^{n - 2}{x} \ dx$   Could someone please explain what I am missing?  Thank you for your time. "who" vs. "whom" and when to omit the relative pronoun This is the boy who was asked by the teacher to stay back after school. This is the boy whom the teacher asked to stay back after school. This is the boy the teacher asked to stay back after school.   Are the sentences above correct ? I am using cURL to insert data into salesforce. It works with single record but I need to insert multiple records at a time. Is it possible to do so and is it possible via a REST interface? I have a div 200 x 200 px. I want to place a 50 x 50 px image right in the middle of the div.   How can it be done?  I am able to get it centered horizontally by using text-align: center for the div. But vertical alignment is the issue.. How do I restore the default repositories? There are now errors when updating and I cannot install most software due to a corrupted  /etc/apt/sources.list   file. Is there a copy I can download to replace it?  The file would be for Ubuntu 12.04 (Final Beta) in the United States. How do I increase the render distance? I have created a simple model here   When rendering it the end of it is cut off   How can I get the camera to render a longer distance? How do I remove a summoned Ender Dragon with no AI? So I used the command /summon EnderDragon ~ ~ ~ {NoAI:1} so that I would have an Ender Dragon that would just sit there.    I then wanted to get rid of it, but doing /killall EnderDragon didn't work, neither did /killallnpc, and I can't seem to do any damage, even with a diamond sword and maximum strength using commands.    I'm not sure why I can't get rid of it; I think it's because the dragon has no AI, but I need to get rid of it somehow, any ideas? Backup of clients' directory to Windows Server 2008 Delete specific line number(s) from a text file using sed?