I think you're heading in the right direction - b is orthogonal to every column vector in A, so they are going to do a pretty poor job when used to approximate b
this is not quite true
## \vec{b} \in Nul(A^T) ##
is more accurate
i think you've pretty much got it, but you need to outline how...
For the k component of curl, imagine a loop in the xy plane around the point in question, now imagine a line integral around the loop, what is the net result? Consider this for each axis and you should get close. You'll need to be careful with convention, to decide on +-.
Try to put some reasoning behind your arguments, rather than just throwing expressions. I haven't done the work so can't just tick a box, just trying to guide your thinking..
Another good way to start that may help is always to try a simple example, pick say n=5 and k=2. Now consider the ways...
I think you're onto it, but i'd go back to your definition of a spanning set. If it only requires that the set spans V and uniqueness is not required, then there may exist more than one way to form a given vector v, e.g. consider if your set had both (1,0,0) and (2,0,0) in it..
well deciding on a set with a1≤ a2≤ ... ≤ ak≤ n effectively partitions n integers into k+1 groups, preserving order, so maybe you could see if you can figure how many ways there is to form k+1 partitions from n objects
with these questions you need to be careful with terminology and some definitions... In particular you need to be clear which "universe" you're working in, so I assume you're asking which of these sets is closed in the set R? (correct me if I'm wrong..)
a) the fact the natural numbers are...
though equivalent, the discrete veiw point for the probability mass function may be simpler to envisage here:
f(x) = p, if x=1
f(x) = (1-p), if x=0
f(x) = 0, otherwise
Now the expectation of a function of x, say g(x) will be:
E[g(x)] = \sum_{x_i} g(x_i)f(x_i) = pg(1)+(1-p)g(0)
If x...
Your E(Y) is not correct. Rather than inputtting the coniditional distributions to start, try writing the fromula for E(Y) and work from that to see where the conditional distributions can be used.
yeah so I would probably start by trying to find the characteristic equation of the matrix
as they're only 2 non-zero rows in the first column, hopefully it shoudl simplify a fair bit