Let $A$ be a $5\times5$ matrix with real entries such that the sum of the entries in each row of $A$ is $1$. Then the sum of all entries in $A^3$ is   (a) $3$   (b) $15$  (c) $5$  (d) $125$  $1$ is an eigenvalue of $A$, that's all I know. How could we go to the sum of all entries of $A^3$ ?
Let $A\neq I$ be a $5\times5$ matrix with real entries such that the sum of the entries in each row of $A$ is $1$. Then the sum of all the entries in $A^3$ is  1)$\space 3$ $\qquad $2)$\space 15$ $\qquad$ 3)$\space 5$ $\qquad$ 4)$\space 125$  Solution:  The answer is $5$ if $A = I$, but for $A \neq I$,  I found that vectors $\begin{pmatrix} 1\\ 1\\ 1\\ 1\\ 1\\ \end{pmatrix}$ and  $\begin{pmatrix} -1\\ -1\\ -1\\ -1\\ -1\\ \end{pmatrix}$  are eigen vectors for the matrix $A$ corresponding to eigen value $1$. From here, I dont know how to proceed, please help me solve the question..and if there are mistakes in my argument, please tell me..
"Seen" status is earlier than the user's last post I posted what I think is a clever answer to:    But then I noticed the question was posted a while ago on Mar 2.  I wanted to see what the chance of the user logging back in and seeing my answer, but the  says the user was seen on Feb 19.  I thought this was supposed to be the date they last logged in? How could they have asked a question on Mar 2 if they haven't logged in since February?  Thanks Will wolves attack and kill ocelots in Minecraft? I know they kill sheep. I just don't want my cats being eaten. My processor is 64-bit - does that mean I need the amd64 image? I'm trying to find an alternative to the standard tombstone qed symbol for more informal papers. I initially thought of an empty square with Pub written inside, but what would actually be better would be a small beer mug symbol. (much like the one found in this set of icons )  Unfortunately there doesn't seem to such a symbol in the comprehensive latex symbol list (although many funny symbols can be found there). Has anyone thought about this? Is there any quick-and-dirty solution? Does C1 code coverage analysis exist for Ruby? I'm currently using  to get C0 code coverage analysis for a rails project that I'm working on.  However, those results are practically meaningless- I have 100% coverage according to rcov (as it only covers C0 analysis) and I've barely written half the test cases for the functionality that exists thus far.  I'm used to the useful results from the code coverage in Visual Studio 2008 Team, which has C1 coverage.  Are there any tools that provide similar coverage for ruby? L293, L298 and SN754410 H-bridge drivers on low voltage power supply This question is about the following three integrated H-bridge drivers: -    or L293D (D = protection diodes added)  (protection diodes included)  (no protection diodes)   Time after time the same question keeps coming up - someone is using one of these devices (on a low voltage, usually around 6V or less) and they are just not performing adequately. The reasons are listed further below but my question is this: -  What H-bridge drivers are preferred when controlling a low-voltage motor?   Information  The L293 and the SN754410 are nearly identical and crucially, if you try and control a 1 amp load, you are faced with dismal performance: -    The tables tell you (typical conditions) that the upper transistor drops (loses) about 1.4 volts when driving a 1A load and, the lower transistor drops (loses) about 1.2 volts when driving a 1A load. The upshot is that if you have a 6V, 1A motor and 6V battery, don't expect to see more than 3.4 volts across the motor: -  \$V_{OUT} = 6V - (1.4V + 1.2V) = 3.4V\$  Worst case scenario is you might only see 2.4 volts across it.  What about the L298? It's got a nice big heat-sink whereas the L293 and SN754410 are regular-looking chips. Here's what the volt drop (losses) look like: -    It's the same story - for a 1A load, you can expect to lose up to 3.2 volts and, what you thought might be 6V across your motor, is at best 4.2 volts and at worst only 2.8 volts.  Clearly none of the devices listed are suitable for low voltage applications where the motor might be expected to draw in excess of 0.5 amps. What is a loss function in decision theory? My notes define a loss function as the 'cost' incurred when the true value of $\theta$ is estimated by $\hat\theta$. What kind of cost is it talking about? monetary cost? or is it something related to errors? 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My question is not only about possible ways to validate / evaluate clustering performance, but is broader - on what basis do we select/prefer one clustering method/algorithm over another one. Also, is there common warnings that we should look around when we are selecting a method to cluster our data?  I know that it is very general question and very difficult to answer. I only would like to know if you have any comment or any advise or any suggestion for me to learn more about this. Export iMessages in human readable form for archival I have a long (several months' worth) iMessage conversation with a contact.  He lost these messages when he re-installed his OS, but I still have the archive.  There is some important information there that we'd like to keep.  Is there a simple way to export this conversation for archival in a human readable form, and also keep any images?  I can select parts of the conversation, copy it to the clipboard and paste it into a rich text document, but this discards all images.  I'm looking for a way that keeps the images.  Note: I use iMessage on OS X only, not iOS.  (Ideally I'd also like to export all non-image file attachments, but I'm not expecting that this will be possible.) I understand that people explain (in layman's terms at least) that the presence of mass "warps" space-time geometry, and this causes gravity. I have also of course heard the analogy of a blanket or trampoline bending under an object, which causes other objects to come together, but I always thought this was a hopelessly circular explanation because the blanket only bends because of "real" gravity pulling the object down and then pulling the other objects down the sloped blanket. In other words, to me, it seems that curved space wouldn't have any actual effect on objects unless there's already another force present.   So how is curved space-time itself actually capable of exerting a force (without some source of a fourth-dimensional force)?  I apologize for my ignorance in advance, and a purely mathematical explanation will probably go over my head, but if it's required I'll do my best to understand. What is numeric promotion? Why is it so hard to find the roots of polynomial equations? The question that follows was inspired by  question:  When trying to solve for the roots of a polynomial equation, the quadratic formula is much more simple than the cubic formula and the cubic formula is much more simple than the quartic formula.    That the general solutions to various polynomial equations are so complex and difficult to derive seems to suggest a fundamental limitation in the problem solving capabilities of the mathematical machinery. Does this intuition of mine make any sense? What should I make of it? 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I have googled the problem every way I can think of and have not been able to find an answer to why the layer is suddenly un-editable or how to fix this. Let's have 2 numbers, N and K, where K divides N. The number of K-combinations from a given set S of N elements is a well known formula. Let's concatenate N/K groups (resulting in N elements) such as the resulting set is N. How many possibilities are there, i.e. what's the formula?  For instance:    N=4, K=2, C(4, 2) = 6 {1,2},{1,3},{1,4},{2,3},{2,4},{3,4}   The 3 possibilities are:     {1,2},{3,4}   {1,3},{2,4}   {1,4},{2,3}   I generated these combinations and I think the number goes like this:  (4, 2): 3 (6, 3): 10 (6, 2): 15 (10, 5): 126 (9, 3): 280 (10, 2): 945 (14, 7): 1716 (12, 4): 5775 (15, 5): 126126 (15, 3): 1401400   Apparently, the result always divides with (N-1). Force applied off center on an object Proper use of the IDisposable interface I know from reading  that the "primary" use of the IDisposable interface is to clean up unmanaged resources.  To me, "unmanaged" means things like database connections, sockets, window handles, etc.  But, I've seen code where the Dispose() method is implemented to free managed resources, which seems redundant to me, since the garbage collector should take care of that for you.  For example:  public class MyCollection : IDisposable {     private List&lt;String&gt; _theList = new List&lt;String&gt;();     private Dictionary&lt;String, Point&gt; _theDict = new Dictionary&lt;String, Point&gt;();      // Die, clear it up! (free unmanaged resources)     public void Dispose()     {         _theList.clear();         _theDict.clear();         _theList = null;         _theDict = null;     }   My question is, does this make the garbage collector free memory used by MyCollection any faster than it normally would?  edit:  So far people have posted some good examples of using IDisposable to clean up unmanaged resources such as database connections and bitmaps.  But suppose that _theList in the above code contained a million strings, and you wanted to free that memory now, rather than waiting for the garbage collector.  Would the above code accomplish that? What ultimates can be damage boosted by Mercy? What ultimates can be damage boosted by Mercy?  We already know that , but are there other ultimates that can be damage boosted by Mercy? Can all ultimates be damage boosted by Mercy? So I am running on Windows 7 (64-bit) Home Premium with 12 GB RAM (Intel Core i7 920).  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On one occasion the party encountered an Elizabethan seaman who demanded to know if Elizabeth was still queen. They party replied "err... yes" because they didn't want to have to explain time travel.