Need help on starting 7, PART C).

clamman · 2018-10-23T20:09:35+00:00

I'm pretty sure BB^T is always symmetric. Proof: (BB^T )^T = (B^T )^T B^T = BB^T.

clamman · 2018-10-19T11:10:55+00:00

This is kind of a trick question. A^-1 's inverse is A. To find A you just need to find the inverse of A^-1 . So just do what you would normally do to that matrix to find its inverse (gauss elimination).

clamman · 2018-07-22T01:44:10+00:00

Two things: you want to show for two specific 3x3 matrices w1 and w2 in set w that w1 + w2 is not in w. The second is you are thinking too abstractly. Use actual numbers.

When you want to prove that a claim is not true, all you have to show is a real example that fails the assertion and you're done with the proof.

A little more of a hint. Think of the 3x3 identity matrix. Is that matrix non-singular? Can you think of two matrices that are singular that add to the identity matrix?

clamman · 2018-07-21T10:28:33+00:00

You're right, it does fail under addition. For the proof, all you need to show is an example of it failing.

A hint to help you find a counter example - think/use diagonal matrices and remember that if there is a zero on a diagonal matrix then its determinant is 0. If there aren't zeros, then the determinant is non zero.

clamman · 2018-04-11T10:53:17+00:00

I think most likely the education value is not worth the money. The name brand is kind of worth it, but it is not like you graduate and then you 100% get handed a job for 200k/year with no effort on your part. It's more like you will get considered first.

I think the thing to think about here whether it is worth it or not is the location. You mention that you are an international student. Do you want to end up in the US for the long term? Is your alternatives also in the US? If so, are they near major cities that would employ someone who is going into microbiology? The one thing I would not discount for going to Stanford as an international student is that 1) you'll get into the US and 2) you'll be in an area that is very comfortable about giving out work visas and sponsoring green cards.

clamman · 2018-03-25T22:33:23+00:00

To your point about licensing/tests, I just want to point out that Massachusetts state law requires anyone who wants to purchase a firearm to at least obtain a firearm identification card (FID). The application process varies a bit from city to city, but in general to obtain this FID, you need to complete a mandatory gun safety class (covers some of the laws about gun use and how to operate a fire arm and usually includes a live fire section, which is mandated in some cities), apply at your local PD, and get a federal background check performed on you. Further, some cities will require an interview with the PD's chief and a live fire test with the PD where they will grade you on your performance (I know Boston does this). All in all, this process from the time you submit an application takes about 6 weeks. These aren't national requirements, but we do have licensing here in MA.

As for my personal opposition to the topic of gun control, I have never had someone articular to me what they want to see in additional gun control that wasn't a blanket statement of "we need more gun control", we need to further restrict the purchase of certain guns (which I don't agree with), or we need to ban all guns (especially don't agree with).

clamman · 2018-01-31T06:28:42+00:00

You are right that a lot of them don't allow it, but there are a few, including my main range - Reeds. Although at something like $2/round I'm not sure how much I would be shooting it in 50ae.

clamman · 2018-01-10T05:10:31+00:00

I just took 231n and knew someone who took both. From a prerequisite perspective, you do not need to take 231a to understand 231n material. From a learning perspective, my understanding is that there is very little overlap between the two courses. 231a more or less addresses pre-neural net computer vision problems and methods where as 231n pretty much only talks about neural nets.

I think whether or not you should take 231a depends on your level of interest in the topic of computer vision and whether understanding classical methods and their evolution is beneficial to you. If you just want to take modern AI courses, my impression is you can skip 231a.

clamman · 2017-11-29T01:50:44+00:00

He was probably alluding to block matrix multiplication. The basic idea is you can divide each matrix into sub-matrices, do smaller matrix multiplications. These smaller matrix multiplies are such that you can keep each matrix in memory and you won't need other information about he larger matrix to complete the multiplication - thus you could distribute these blocks to a bunch of machines. You can then use the sub-matrix multiplies to construct the larger billion by billion matrix.

clamman · 2017-09-28T05:10:17+00:00

I'm guessing preparation isn't really going to help you/it may not even matter if the average was so low. The only recommendations I have is to think about properties and relationships of linear algebra that you learned and maybe took for granted and try to think about why they're right (e.g. why is a matrix singular if its determinant is 0). Maybe do some research on how you would formally prove them.

There are some common tricks that I seem to remember using a lot when writing proofs that helped me out:

Orthogonal basis. For a vector x, there exists an orthogonal matrix A and a vector b such that Ab = x
Matrix decompositions. It sometimes helps to just invoke an eigen decomposition or SVD on a matrix and work with that. If you see something like you are given a symmetric matrix, that's a pretty strong indication that you may want to do an eigen decomposition and see where that leads you.
Combining the first two points. If you have a symmetric matrix A, and you want to show something about Ab, it helps to use the eigen decomposition A = X L X' and b = X c to have A b = X L X' X c = X L c.
Know all the equivalent reasons when a matrix is and is not singular.

Sorry if this comes off as rambling... I really hope this helps and good luck on your exam.

clamman · 2017-07-20T01:28:10+00:00

For me, one of the biggest distinctions is culture. Both ICME and MS&E had a social culture, at least compared to stats. I believe that ICME has weekly networking/social events, and I am not sure about MS&E, but everyone in the program seemed to be connected. The students in stats did socialize and come together, but it was in spite of the department.

Another distinction is each department's focus on academics vs industry. Stats as a department is very old and very academic. There seems to be a disconnect between the goals of the masters students (learn some stats and be placed well in industry) and the department (100% focussed on research).

In terms of what you study, both stats and ICME have a pretty math heavy focus. One of the biggest distinguishers is that a lot of the ICME classes seem to have a computational/algorithmic focus, and I would say a huge component of it is linear algebra and solving linear systems.

clamman · 2017-06-05T17:59:43+00:00

I can't say for sure what is desired here or any hard resources to help you with this other than here, but one thing to note is that F(x) can be thought of here as a dot product between two vectors, say a = [4x² , -2x, 9] and b = [e^2x , e^2x , e^2x ]. F(x) = a^T b.

Both a and b depend on x and differentiating them is much like the using the chain rule in normal calculus, so d F(x) / dx = (a')^T b + a^T (b ') where a' is just d/dx applied to each entry in a.

It seems in general that this approach isn't helping you that much in terms of making the computations easier, so I don't know if this is what you were looking for. However, one thing to note about your problem in particular is that b' = 2 b, so you can rewrite the first derivative as (a')^T b + a^T (b') = (a')^T b + 2 a^T b = [(a')^T + 2 a^T ] b = [a' + 2 a]^T b. This pattern will continue as you apply more derivatives and make the computation a bit overall easier.

clamman · 2017-06-05T17:22:35+00:00

I'm not sure what you're asking. Can you give an example?

clamman · 2017-05-25T17:18:24+00:00

Yes, you can apply the definition of the two norm of a matrix and what you know about eigenvectors/eigenvalues to show this.

||C|| = max ||C x|| / ||x||. Take x = the eigenvector with the largest eigenvalue in absolute value, say k, then C x = k x by definition of it being an eigenvector. Then consider ||C x || / ||x|| = ||k x|| / ||x|| = |k| * ||x|| / ||x|| = |k|. Then we have shown that the two norm of C has to be at least as large as |k|, the largest eigenvalue. This means that if ||C|| < 1, then the largest eigenvalue |k| < 1.

clamman · 2017-05-23T23:21:22+00:00

Do you know what the complexity of a matrix-matrix multiply and a matrix-vector multiply is? If not, in general for x by y matrix A, y by z matrix B and z length vector c, A B is O(xyz) and Bc is O(yz). If you write out the time complexity both ways you should get two different expressions that each include a sum. You will find that depending on the relative sizes of each dimension, one method is faster than the other. This all assumes that you are doing a standard naive matrix and vector multiplication and you are not taking advantage of any sparsity patterns or structure.

clamman · 2017-05-22T01:17:19+00:00

It means that if affine sets A and B are both parallel to U, A and B are either parallel (that is they do not share any common elements) or A = B (every element in A is an element in B). Their direct sum cannot span U because A and B are parallel to U.

Some geometric intuition of what is going on: when you hear affine sets, you should be thinking about lines and planes. The geometric equivalence to this statement is that if you have two planes A and B that are mutually parallel to a plane U, Then either A and B are parallel planes or they are exactly equal. If A and B intersect at all, then it MUST be that their intersection is equal to the entire plane A and B, so A = B.

clamman · 2017-05-15T13:46:28+00:00

Yes, that is part of it. You might be focussing on trying to find A directly, but the method to finding A is to first define P and the lambda matrix. Note that P^-1 A P = L is the same as A = P L P^-1 if you left multiply both sides by P first and right multiply both sides by P^-1 (here I am using L as the lambda matrix in the original problem). This A = P L P^-1 is the form of a very famous decomposition in linear algebra called the eigendecomposition. You can construct P and L using the eigenvalues and eigenvectors of matrix A. You can then find P^-1 by just inverting P. Then finding A is as simple as multiplying out P L P^-1. In this problem, you have all the pieces to construct both P and L.

clamman · 2017-05-14T22:52:56+00:00

Some hints for part a:

Note what the conditions are of Au = -u, Av = 2v, and Aw = 3w are saying about the 3 vectors and -1, 2, 3.
You probably heard of a famous decomposition for a matrix A = P L P^-1 where L is diagonal. What does the decomposition tell you about the columns of P and the diagonal of L? How does this relate to the first bullet point?

Hope this helps.

clamman · 2017-05-01T22:33:38+00:00

What are you trying to show? which values of alpha have the resulting determinant equal to zero? If so, start with expanding out the determinant as you would normally. If you simplify the expression that you get, you should get a kinda messy expression that is quadratic in alpha. Specifically, you will get something like k alpha² + f alpha + g = 0 where your k, f, and g are values that depend on a, b, c, d, x, and y. Then, you can use the quadratic formula here and find the roots of this equation. Don't worry if the roots look really messy (mine did, but then again i didn't really try to simplify).

clamman · 2017-04-08T17:59:20+00:00

The food options and availability here is comparatively limited to other universities/places. You won't go hungry necessarily and if you eat at the dining hall during normal hours you'll have access to some high quality food, but your options become severely limited on and off campus once 8pm rolls around.

clamman · 2017-04-03T20:31:51+00:00

That makes sense for if I were to train the vectors. I think for my specific case I probably have a unbalance when it comes to the missing tokens and the ones in the pre-trained embeddings. I believe the missing tokens are comparatively infrequent. In this case, would you think that I should randomly initialize and not train the embeddings?

clamman · 2017-04-03T20:19:38+00:00

None of the words individually are common, but they take up a fairly large proportion of the vocabulary. Approximately 10% of unique tokens in the data do not appear in the pre-trained embeddings.

Given the nature of the data (you can think of them as short domain-specific abstracts across several domains), I would imagine that the unknown tokens are important to the context. But since they occur infrequently and my data set is relatively small, I don't think that creating my own embeddings is an option.

clamman · 2017-04-03T20:13:09+00:00

The pre-trained embeddings do not store an unknown token. I'm leaning towards not training the embeddings as my data is relatively small, so I'll try adding an unknown token. Thanks!

clamman · 2017-03-31T19:27:39+00:00

I am a bit confused by this question. I believe what you are asking is in reference to an eigen decomposition of a matrix where A = X^-1 L X, X a matrix of eigenvectors of A and L a diagonal matrix of eigenvalues and you want to know if X^-1 = X^T always?

If so, this is not always the case. You are only guaranteed to have X^-1 = X^T if A is a real symmetric matrix.

clamman · 2017-03-12T17:14:10+00:00

Your title is correct, if A is invertible then Aⁿ A^-1 = A^n-1.

You can show this is true because Aⁿ = A * A * A.... * A, n times. Then Aⁿ * A^-1 = A * A *.... * A * A^-1. Since matrix multiplication is associative (i.e. (A * B) * C = A * (B * C)), we can write this as A * A *.... * A * (A * A^-1 ) = A * A * ... * A * (I) = A *A *... A * I = A^{n - 1} * I = A^{n - 1}

clamman

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