Recent content by sessomw5098

Suppose I get the eigenvalues of A, which are \lambda_{1},\lambda_{2},\dots \lambda_{n}. Also, given any polynomial f(x), I get the eigenvalues of f(A). I'm trying to show that the eigenvalues of f(A) are f(\lambda_{1}),f(\lambda_{2}),\dots f(\lambda_{n}). Is this possible? How would I go about...

I am trying to relate eigenvalues with singular values. In particular, I'm trying to show that for any eigenvalue of A, it is within range of the singular values of A. In other words, smallestSingularValue(A) <= |anyEigenValue(A)| <= largestSingularValue(A). I've tried using Schur...

Hey, A virus test is 98% accurate and 1 in 10000 people have the virus. Given that the test is positive, what are the chances that you have the virus? This is what I've got: Since keyword “given,” I’m assuming Bayes Theorem. So, let A be the event that you have the virus. Let B be the...

Well, I think the proof is complete because we hold one constant while "inducting" the other. We can assume the converse is true due to "without loss of generality." Am I right?

I understand it now. My problem was that I wasn't using the definition of the Cartesian product. Thanks!

If S1 and S2 are finite sets, show that |S1 x S2| = |S1||S2|. Here is what I've tried: Let |S1| = m and |S2| = n. Let P(k) be true. That is, P(k) = |S1 x S2| = km. P(1) is true since, if |S1| = 1 and |S2| = 1, |S1 x S2| = 1. Now, let |S1| = k+1 and |S2| = m. Then, P(k+1) = |S1 x...