MATH 245 midterm (follow-up)

Hello everyone. Tomorrow I will be working on the document I mentioned in my previous post: an overview of all the material that will appear on the MATH 245 midterm. This document will differ from my notes, in that as I said it will be conversational and highly conceptual. That is, I will be omitting as much explicit computation as possible, focusing instead on the main ideas behind the theory.

Anyways, the main purpose of this post is to ask a question. To anyone currently in MATH 245 who might be reading this: if you have something to contribute, please leave a comment on this post! Currently I’m looking primarily for someone else to give an explanation of a solution to questions 3 and 5 on the first assignment (I’ve included the link for anyone outside the class that is interested). If you have anything else to say that might be insightful or helpful, also feel free to do so.

Of course, do not feel that in order to contribute you must go through the tedious work of typing up something formally and so on; you can just post an informal comment on this post. I’ll read it, and include an adapted version in the document (giving credit to you, of course).

The first version of the document should be posted by tomorrow night (this will include at least all of my own efforts), although if any contributions come in from now until Monday, I will revise the document and include them.

It will appear both on this blog and on the front page of

Thanks a lot, and good luck to everyone.


About mlbaker

just another guy trying to make the diagrams commute.
This entry was posted in linear algebra and tagged , , , , , , , . Bookmark the permalink.

21 Responses to MATH 245 midterm (follow-up)

  1. Would you like a full solution to number 3 and 5 or just something informal? By the way, the solution for number 3b should be pi/6.

  2. dx7hymcxnpq says:

    Thanks for the reply! It’s up to you how detailed you want to be, but ideally I’m looking for just a clear description, so that writing the formal proof(s) out afterwards would be more or less straightforward. To use \LaTeX, you can just surround math in (DOLLAR)latex … (DOLLAR)

  3. Hmm… the let me try the latex again:

    I’ll just put a copy/paste answer in and let me know if there are any questions.

    3 a) Suppose that there exists such u \in U and v \in V such that their angle is the smallest angle that can be formed with vectors in U and V. Thus, we have \theta(u,v)=\theta=\cos^{-1}\frac{u\cdot v}{|u||v|}, hence |u||v|\cos\theta=u\cdot v. Now observe that since \frac{u}{|u|} \in \mathbb{R}^{n}, it can be expressed as a combination of a vector in V and a vector in V^{\perp}. So \frac{u}{|u|}=v+w for some v in V, and w in V^{\perp}, and apply the dot product of v to both sides to yield (\frac{u}{|u|})\cdot v=(v+w)\cdot v\implies(\frac{u\cdot v}{|u|})=v\cdot v+w\cdot v, which gives (\frac{|u||v|\cos\theta}{|u|})=|v|^{2}+0\implies|v|\cos\theta=|v|^{2}\implies\theta=\cos^{-1}|v|. Now since \frac{u}{|u|} is a combination of vectors in V and V^{\perp}, by the definition of orthogonal projection, we have v=\mbox{Proj}_{V}\left(\frac{u}{|u|}\right). Thus, we have \theta(u,v)=\theta=\cos^{-1}|v|=\cos^{-1}\Bigl|\mbox{Proj}_{V}\left(\frac{u}{|u|}\right)\Bigr| .

    • Is there any way to preview the code before I post?

    • dx7hymcxnpq says:

      I don’t think there is a preview feature when it comes to comments. But you could always type a draft post on your blog and use that preview feature, then copy it all into a comment.

    • dx7hymcxnpq says:

      I have a question: it seems when you decompose \frac{u}{|u|}, you are taking it as a given that the orthogonal projection of \frac{u}{|u|} onto V is indeed the vector v, and not perhaps some other vector v' \neq v \in V… how do you know this?

      It seems that nowhere in the proof do you use the hypothesis that u and v are these “special” vectors with minimal angle…

    • Actually, I’m not saying that the orthogonal projection of \frac{u}{|u} onto V is v but rather that since its projection is a point in \mathbb{R}^{n}, then it can be expressed as a combination of vectors in V and V^{\perp}.

    • “It seems that nowhere in the proof do you use the hypothesis that and are these “special” vectors with minimal angle…”

      I’m just assuming that such vectors exists and working from there.

    • dx7hymcxnpq says:

      You’re re-using the name v for two different things, and what’s the point in assuming 2 vectors with property P exist if you never exploit that property anywhere?

    • Okay, reading your question again, I guess you’re referring to the last couple lines of my proof. I realize I might have gotten my notation a bit mixed up, so I’ll post a new version of the proof later this afternoon.

  4. Revision of 3a)

    (I’ll be back at 3 so please tell me if you find any inconsistencies

    Suppose that there exists such u\in U and v\in V such their angle is the smallest such angle that can be formed with vectors in U and V . Notice that we have \theta(u,v)=\theta=\cos^{-1}\frac{u\cdot v}{|u||v|}\implies|u||v|\cos\theta=u\cdot v . Also note that for any given 0<k\in\mathbb{R} , we have \theta(u,kv)=\theta(u,v) since \theta(u,kv)=\cos^{-1}\frac{u\cdot kv}{|u||kv|}=\cos^{-1}\frac{k(u\cdot v)}{k|u||v|}=\cos^{-1}\frac{u\cdot v}{|u||v|}=\theta(u,v) . Now observe that since \frac{u}{|u|}\in\mathbb{R}^{n} then it can be expressed a combination of a vector in V and a vector in V^{\perp} . So \frac{u}{|u|}=v_{0}+w_{0} for some v_{0}\in V and some w_{0}\in V^{\perp} where v_{0}=kv for some 0<k\in\mathbb{R} . Apply the dot product of v_{0} to both sides to yield (\frac{u}{|u|})\cdot v_{0}=(v_{0}+w_0)\cdot v_{0}\implies(\frac{u\cdot v_{0}}{|u|})=v_{0}\cdot v_{0}+w_0\cdot v_{0}, which gives (\frac{|u||v_{0}|\cos\theta}{|u|})=|v_{0}|^{2}+0\implies|v_{0}|\cos\theta=|v_{0}|^{2}\implies\theta=\cos^{-1}|v_{0}|=\cos^{-1}|kv| . Now since \frac{u}{|u|} is a combination of vectors in V and V^{\perp} , by the definition of othorgonal projection, we have v_{0}=\mbox{Proj}_{V}\left(\frac{u}{|u|}\right) . Thus, we have \theta(u,v)=\theta(u,kv)=\theta(u,v_{0})=\theta=\cos^{-1}|v_{0}|=\cos^{-1}\Bigl|\mbox{Proj}_{V}\left(\frac{u}{|u|}\right)\Bigr| .

    • dx7hymcxnpq says:

      How do you know that v_0 = kv for some 0 < k \in \mathbb{R}? V isn’t assumed to be of dimension 1 (only U is). The proof seems to work, if you can show that indeed v_0 is a scalar multiple of v (to do this, I think you would probably have to use the hypothesis that u and v minimize \theta etc.)

  5. Saifuddin Syed says:

    Why did you say v_o = kv? it adds nothing to your proof. You would be done if you just substituted in the value of v_o to be proj_V(u/|u|).

    • dx7hymcxnpq says:

      I’m not sure I follow. We’re trying to show that the angle \cos^{-1} \left| \mathrm{Proj}_{V} \left( \frac{u}{|u|} \right) \right| is the minimal one, and I don’t see why there couldn’t exist vectors with a smaller angle than that…

      Also, what if U \cap V \neq \{ 0 \}? Then the definition of angle between them is different. Do we need to treat this as a special case?

  6. dx7hymcxnpq says:

    Here’s an update. I thought about what the problem is actually saying a bit. Note that \mathrm{angle}(U,V) is defined in many different ways depending on what the situation is. OK. Note that if U \cap V = W \neq \{ 0 \}, then we certainly have U \subseteq V, since U, being the span of one vector, is one-dimensional. So the proof is the following:

    1. If U \cap V = \{ 0 \}, then we must prove that there exist a \in U, b \in V such that \theta(a,b) = \cos^{-1} \left| \mathrm{Proj}_V( \frac{u}{|u|} ) \right|. Furthermore, we must prove that any other pairs of vectors yield a larger angle than these two (!).

    2. If U \subseteq V or V \subseteq U, then need to demonstrate that \cos^{-1} \left| \mathrm{Proj}_V( \frac{u}{|u|} ) \right| = 0. But this is trivial if U \subseteq V, since that huge projection just becomes a unit vector, and \cos^{-1}(1) = 0. Note it is impossible that V \subsetneq U, since U is one-dimensional and V is nontrivial.

    • It looks like I only tailored my proof for case 1 and on the note of why k is greater than 0. Well we should assume that the angle between u and v is acute as if it is >\frac{\pi}{2} then we may just multiply u or v by -1 to make the angle acute for all cases.

    • dx7hymcxnpq says:

      My question wasn’t really a problem with positive k, but the existence of such a k at all. Why does v_0 necessarily have to be a scalar multiple of v?

    • Let’s try that again:

      \frac{v\cdot u}{|v|^{2}}v is our projection of v onto u and \frac{v\cdot u}{|v|^{2}} is our k.

    • dx7hymcxnpq says:

      Don’t think that formula’s right. The projection is onto V, which is some arbitrary space, not onto a line…

    • But we’re projecting onto v, a vector, which is in V, since we assume that there are two vectors u and v that will minimize the angle beforehand.

  7. dx7hymcxnpq says:

    OK, so just to close this thread, a different proof to Problem 3(a) was given by Harmony which quite elegantly yielded the desired inequality, using the fact that the inverse cosine is a decreasing function. I’ll try to remember it and type it up once I have time.

    Also, the official assignment solutions are now up.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s