Math Is Fun Forum

No. Consider

and

They are both invertible (they are their own inverses), but I + (-I) = O, and we know the zero matrix is not invertible. Kind of a simple example, but it works.

What does the noncommutativity of matrix multiplication have to do with the associativity? BA doesn't show up in (AB)C = A(BC).

The definition of associativity is that the order of operations does not matter, as long as the order of operands doesn't change. From this it follows easily that (AB)CD = A(BC)D = AB(CD) = (ABC)D = A(BCD); we can multiply in any order (A and B first, or B and C first, etc.), but we cannot reorder the operands (A, B, C, and D).

If you really don't believe it, take the product of r matrices A[sub]k[/sub] (where each A[sub]k[/sub] is of dimension n[sub]k[/sub] × n[sub]k + 1[/sub]), denoting the elements of A[sub]k[/sub] as

If we take, for example,

by the associativity of scalar multiplication (which I am sure you do not dispute), and similarly for other placings of parentheses (they will all reduce to parentheses around the scalar elements in the sum, which can be moved as removed as you please).

Haha, I looked at that kind of funny when you wrote it, and wondered if I had helped at all when you arrived at a trivially obvious conclusion. Good work, team.

Hessian is like the Jacobian (without further qualification they are determinants, we can also consider the Hessian or Jacobian matrix, which are useful as the matrix of certain transformations, such as change of coordinates), but its elements are second order derivatives. The Hessian is defined as

Suppose a = (a[sub]1[/sub], ..., a[sub]n[/sub]) is a critical point of f. Then the following hold:

In more advanced terms, if the Hessian matrix is positive definite at a, then a is a local minimum for f, if the Hessian matrix is negative definite at a, then a is a local maximum for f, and if the Hessian matrix has both positive and negative eigenvalues at a, then a is a saddle point of f. This is basically the second derivative test for functions of more than one variable. Note that this second derivative test may sometimes be inconclusive.

Actually there are some unsolvable games (although they are a "set of measure zero"). Here is a proof.

http://www.cs.uu.nl/~hansb/d.freecell/freecellhtml.html

Yeah, it's probably somewhat intense. But in our class we've talked about topics mentioned in the news article, and professors love to talk about their research interests, so I'm sure I can make an office visit to explain anything I don't understand. But that's only if I get to actually see the paper. Oh well. The concept of extra time-like dimensions is certainly interesting.

You're almost right... but don't forget your constant of integration!

Depending on your teacher you might lose a point or two for leaving that out. Do you know how to do partial fractions?

I saw on Facebook that you are ok, Ricky. That is good news. Hopefully all of your friends are fine as well.

Are you exaggerating on that 6 to 7 minutes? It would pause noticeably for some seconds for me, but definitely not over a minute or even half a minute. That's outrageous.

No, this is cheating!

Look at all those pieces he put in before me!

Earlier we discussed subforums, for forums such as the exercises, which would help organize them a little. Is this a possibility yet?

To explain Ricky's statement a little more, note that

We cannot rearrange these terms into a[sup]k[/sup]b[sup]k[/sup] if a and b do not commute. Do you see this?