1.If a is a column vector, then how many non-zero eigenvalues does the matrix aa' have? what are the eigenvalues? What are the corresponding eigenvectors? What are the eigenvectors corresponding to the zero eigen values?
Since aa' is a rank 1 matrix, there is only one nontrivial eigenvalue. If the eigenvalue is denoted c, then aa'v=cv. Also, taking the transpose of both sides, (aa'v)'=(cv)' meaning v'a'a=cv'. However, this time, a'a is a constant, so eigenvalue c=a'a. Now to find the associated eigenvector. Recall that a(a'v)=cv. Since a'v and c are both scalars, the equation may be rewritten (a'v)a=cv, and a is a multiple of v. Also, replacing v with a gives (a'a)a=(a'a)a, an equality. So a must be the eigenvector.
As for the eigenvectors, corresponding to 0, if a=[a_1 a_2 ... a_n], then the eigenvalues are of the form
[a_2+...+a_n ] [-a_1]
[-a_2 ] [a_1+a_3+...+a_n]
[-a_3] [-a_3]
. .
. and . and so forth.
. .
[-a_n ] [-a_n]
Keep in mind there are n of these.
12. given the matrix A=(5 -3;-3 5), find a matrix M, such that A=M*M. Now find a matrix M such that A=M'*M
M=[ a b ] where a=-3sqrt(2)/2,b=sqrt(1/2) (took long time, though)
[ b a ]
Represent M as [a b] multiply M^2=A out and do algebra. You'll find a=d and c=d. After
[c d]
plugging this back into M^2=A, multiply out again. a^2+b^2=5 and 2ab=-3. Substitute and solve.
For the second part, M is symmetric, so M=M'. Thus, the same M applies. (sad part is it took me almost as much time to figure out 2nd than first.)
13. Suppose x_1, x_2...x_n are IID from [0,1] uniform interval. What is the expected value of the maximum. What is the expected value (max-min).
\[E[max_i {X_i}]= int_0^1 int_0^1 ... \int_0^1 max_i {X_i} dx_1 dx_2 ... dx_n\]
We have to split up the n-dimensional space [0,1]x[0,1]x...x[0,1] into sections. One section will have x_1 be the maximum, another section with x_2 the maximum, and so forth. By symmetry, we can just consider the space where x_1 is the maximum Without loss of generality and multiply by n at the end. So assuming x_1 is the max, we can use the conditional probability
\[E[max_i {X_i}| X_1 is max]= int_0^1 int_0^x_1 ... \int_0^x_1 [x_1] dx_1 dx_2 ... dx_n\]
\(= [int_0^1 ... [x_1] [int_0^x_1 dx_2] ... [\int_0^x_1 dx_n] dx_1]
= [int_0^1 ... [x_1] [x_1] ... [x_1] dx_1]\)
\( = {[1]^(n+1)}/(n+1)\)
\( =1/(n+1)\)
There are n of these, so E[max_i {X_i}]=n/(n+1). Off the bat, you can see that at least for n=1, this formula is true.
As for the max-min, I'm gonna make the assumption that E[max-min]=E[max]-E[min] since the integral can be parsed. Through the same argument as above,
\[E[min_i X_i| X_1 is the min ]= \int_0^1 x_1(1-x_1)^(n-1)dx_1=1/[n(n+1)]. \]Multiplying by n yields
E[min]=1/n+1, which again makes sense at lease for n=1. Thus, E[max-min]=(n-1)/(n+1).
20. Given a fair coin, what is the expected number of trials you need to go to get 2 consecutive heads. 3 consecutive heads. generalize to N.
\[2(1/4)+3(1/8)+4(2/2^4)+5(3/2^5)+6(5/2^6)+7(8/2^7)+8(13/2^8)+9(21/2^8)+…\]
I don't have a closed form, but the formula is \(\sum_{i=1}^{infinity} (i+1)(F_{i-1}/2^{i+1})\), where F_i are the Fibonacci numbers.
Explanation: The first two are easy. There is a ¼ chance of getting two heads on the first two flips. There is a 1/8 chance you will get THH on three rolls. From henceforth, it's just a matter of finding the different ways you can end the sequence in THH, where the previous terms have no HH's. This is where the Fibonacci terms come in to play. I'm too tired to generalize to N (or even 3 for that matter). Variance, forget it.
