Q: How prove that variance-covariance matrix is nonnegative definite?

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Let A be a matrix which is both symmetric and skew symmetric. so AT=A and AT= -A so A =- A that implies 2A =zero matrix that implies A is a zero matrix

Let's prove that rho(A)=2-norm(A) for A symmetrical and then prove the relation between 1-norm and 2-norm. Both are easy.

If x is a null matrix then Ax = Bx for any matrices A and B including when A not equal to B. So the proposition in the question is false and therefore cannot be proven.

There is no definite way to prove how boys (or more specifically, mankind) were made, and I doubt that there was a reason why men were made too.

Automated proofs are a complicated subject. If you are not an expert on the subject, all you can hope for is to write a program where you can input a sample matrix (or that randomly generates one), and verifies the proposition for this particular case. If the proposition is confirmed in several cases, this makes the proposition plausible, but is by no means a formal proof.Better try to prove it without writing any program.Note: it is not even true; it is the inverse of the matrix which gives identity when is multiplied with the original matrix.

What is "a 3b"? Is it a3b? or a+3b? 3ab? I think "a3b" is the following: A is an invertible matrix as is B, we also have that the matrices AB, A2B, A3B and A4B are all invertible, prove A5B is invertible. The problem is the sum of invertible matrices may not be invertible. Consider using the characteristic poly?

I think so. Copy and paste method could be used to prove this. But this is only my opinion.

The trace of an nxn matrix is usually thought of as the sum of the diagonal entries in the matrix. However, it is also the sum of the eigenvalues. This may help to understand why the proof works. So to answer your question, let's say A and B are matrices and A is similar to B. You want to prove that Trace A=Trace B If A is similar to B, there exists an invertible matrix P such that A=(P^-1 B P) Now we use the fact that Trace (AB)= Trace(BA) for any nxn matrices A and B.This is easy to prove directly from the definition of trace. (ask me if you need to know) So using this we have the following: Trace(A)=Trace(P^-1 B P)=Trace (BPP^-1)=Trace(B) and we are done! Dr. Chuck

A - B = B - AThis statement is very difficult to prove.Mainly because it's not true . . . unless 'A' happens to equal 'B'.

Recall that if a matrix is singular, it's determinant is zero. Let our nxn matrix be called A and let k stand for the eigenvalue. To find eigenvalues we solve the equation det(A-kI)=0for k, where I is the nxn identity matrix. (<==) Assume that k=0 is an eigenvalue. Notice that if we plug zero into this equation for k, we just get det(A)=0. This means the matrix is singluar. (==>) Assume that det(A)=0. Then as stated above we need to find solutions of the equation det(A-kI)=0. Notice that k=0 is a solution since det(A-(0)I) = det(A) which we already know is zero. Thus zero is an eigenvalue.

I am finding it difficult to find a definite answer for you. From the information that I did find it seems that it is the purchaser that will bring the bill of sale to the DMV to prove the purchase and new ownership of the said vehicle.

Nobody can make that kind of definite statement. All we can say is that none have yet been observed in orbit around Mercury or Venus. It's very difficult to prove a negative.