Proof: Orthogonal Matrices Satisfy A^TA=I

  Рет қаралды 12,517

Mu Prime Math

Mu Prime Math

2 жыл бұрын

One way to characterize orthogonal matrices is to say that a matrix orthogonal if and only if A transpose times A is the identity matrix. In this video, we prove this result using basic matrix calculations and the definition of orthonormal vectors.
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Music: C418 - Pr Department

Пікірлер: 37
@BuddyNovinski
@BuddyNovinski Жыл бұрын
My professor never explained these concepts. I blame the curriculum, but it's hindsight. There was too much time on proofs, and not enough time on problems. Recently, I discovered that it helps to put oneself into the subject, something our distinguished young man does well.
@pipertripp
@pipertripp 4 күн бұрын
AA^T = I is also true since the the rows of A also form an orthonormal set. The whole thing is a pretty brilliant result. Not have to compute the inverse of A using row reduction is a huge win computationally as it's much faster to transpose A than perform all of that Gauss-Jordan bollocks.
@ashveet420
@ashveet420 Жыл бұрын
New to the channel and I love the content. Keep it up with the amazing uploads .
@jotadiolynedicci3361
@jotadiolynedicci3361 2 жыл бұрын
I love this video ! I think you could also also add that A has always a determinant equals to 1 or -1. It would be interesting because there is the group SO_n(R) (orthogonal and det = 1). With a little work, you can find SO_2(R) and discover the rotation matrix !
@MuctaruKabba
@MuctaruKabba 11 ай бұрын
This is a great explanation of the algorithm of orthogonality. It helped me a lot in trying to understand the concept and its use. Thank you.
@claudefazio
@claudefazio Жыл бұрын
Your explanations are very clear!
@alejrandom6592
@alejrandom6592 Жыл бұрын
First time I read this theorem it didn't make any sense. Turns out it's really intuitive, thanks for making this video ♡
@MikeT10101
@MikeT10101 Жыл бұрын
Thank you. Great job! I just subscribed.
@nehalkalita
@nehalkalita 2 ай бұрын
This explanation is even better than the ones from Chat-GPT.
@CarsonKennedyArias
@CarsonKennedyArias 8 ай бұрын
Awesome dude you cleared everything up for me! Thank you
@lancelot6201
@lancelot6201 Жыл бұрын
Thanks a lot, it really helped😃
@domination2720
@domination2720 4 ай бұрын
instant subscribe the best lesson on this topic thank you sir
@krishnaks4622
@krishnaks4622 7 ай бұрын
Thankyou for such a good video
@user-wu8yq1rb9t
@user-wu8yq1rb9t 2 жыл бұрын
Of course , I'm here! *Just Great* . Thank you so much. I have two requests: At first, please be more active. Second, please more Matrix (actually more anything, especially math for physics). Again, thank you so much ❤️
@luisfortuna7893
@luisfortuna7893 2 жыл бұрын
New subscriber. Great content and explanations! I wonder if you are a math major. I am just curious because I am a math major myself, and I find the few videos I’ve watched on your channel to be well-constructed and straightforward. Congrats on your channel! Luis F
@MuPrimeMath
@MuPrimeMath 2 жыл бұрын
Yes, I'm a math major!
@luisfortuna7893
@luisfortuna7893 2 жыл бұрын
@@MuPrimeMath Awesome. I kind of figured that from how to present the topics. If you do not mind me asking, where did you go (or are going) to school?
@MuPrimeMath
@MuPrimeMath 2 жыл бұрын
I'm currently an undergraduate at Caltech
@luisfortuna7893
@luisfortuna7893 2 жыл бұрын
​@@MuPrimeMath I have never looked at the math curriculum from your school, but from your videos, it is obvious that it reaches great depth. You are very knowledgeable, indeed; kudos to you! I will be graduating from a small university in NJ (FDU) this May, but I am sure I only got to scratch the surface of many interesting topics in mathematics. Anyway, I wish you good luck in all your future endeavors! I like to connect with mathematicians from across the globe. Is there any way to connect with you (like LinkedIn) for future reference? P.S. Sorry for the long comments 😅
@BuddyNovinski
@BuddyNovinski Жыл бұрын
@@luisfortuna7893 Yes, I found him on LinkedIn and sent him birthday greetings and praise for how he has clarified much of the math I took at Penn State. I don't think he wants me to reveal his name here, so I've now called him "our distinguished young man".
@alimalik_yoga
@alimalik_yoga Жыл бұрын
Awesome thank you
@tomkerruish2982
@tomkerruish2982 2 жыл бұрын
Spring Break already? Hey, does Fleming Hovse still have that stupid bell? One of my favorite matrix identities is det(exp A) = exp(tr A). Any chance you'll cover that in an upcoming video?
@praveendwivedi635
@praveendwivedi635 Жыл бұрын
Very nice ...❤ Sir
@johnunbehaun6638
@johnunbehaun6638 Жыл бұрын
What college and what background is this lecture from?
@kingbeauregard
@kingbeauregard 2 жыл бұрын
Suppose I applied a coordinate transformation to the matrix that did not stretch the matrix, just rotate it. The magnitude of the determinant would not change but it could flip between positive and negative, right?
@MuPrimeMath
@MuPrimeMath 2 жыл бұрын
The sign of the determinant describes the orientation of the vectors relative to each other. Swapping two vectors changes the orientation, so if you switch two columns in a matrix, the determinant changes sign. Rotation matrices always have positive determinant because rotations preserve orientation. As you said, the volume of the region spanned by the vectors does not change upon rotation, so the determinant of a rotation matrix is always +1.
@kingbeauregard
@kingbeauregard 2 жыл бұрын
Here's a question: what would you say the determinant of a matrix represents? I know a few ways to use determinants, but I don't have a good sense of what it really means. I get the feeling that it's got something to do with how much volume (in however many dimensions) the vectors delineate, but that's more of a vague sense than an educated opinion.
@MuPrimeMath
@MuPrimeMath 2 жыл бұрын
I have a few videos about the determinant. Here is one: kzfaq.info/get/bejne/d5-VfceC2ZvVpas.html
@angeldude101
@angeldude101 2 жыл бұрын
That's pretty much exactly it. It's how much the matrix scales the unit n-volume. Of course it only tells you about the n-volume and not any lower dimensional volumes, so it's a little limited there. Something I can definitely say for certain is that not understanding this makes learning _anything_ about linear algebra absolute hell, which is exactly why the university course I took on LA didn't bother communicating it well. (Apologies for the minor rant. That was not a very fun class even if I've since warmed up to LA significantly since then.)
@kingbeauregard
@kingbeauregard 2 жыл бұрын
@@MuPrimeMath ... okay, there are two concepts in that video that have blown my mind: 1) that an ordinary nxn matrix describes a stretching of orthogonal unit vectors, and 2) that the determinant indicates the total amount of stretching. I'm going to have to process this for a while. Thanks!
@kingbeauregard
@kingbeauregard 2 жыл бұрын
@@angeldude101 I am frustrated by how poorly math is taught a lot of the time too. I haven't had to touch this stuff in decades, but I want to understand it even if belatedly. "It's how much the matrix scales the unit n-volume." - Mind, re-blown. I'm still processing this.
@BuddyNovinski
@BuddyNovinski Жыл бұрын
Determinates were yet another concept I missed because I could not envision them. It's nearly a half century late, but I gradually realized that I must see the concepts. Thanks to our distinguished young man and others such as Grant Sanderson (3 Blue 1 Brown), I can see the concept. It even makes sense of nullspace, which was a mystery to me when I took linear algebra at Penn State.
@milos_radovanovic
@milos_radovanovic 2 жыл бұрын
Was this proof by inspection?
@tomkerruish2982
@tomkerruish2982 2 жыл бұрын
I'd say proof by definition.
@mzbros8155
@mzbros8155 2 жыл бұрын
Yes it is actually, and I made a video on it here: kzfaq.info/get/bejne/mrenZ9pqvM28k4U.html
@miro.s
@miro.s 2 жыл бұрын
For me the video is incomplete. You should also show that in case of square matrices the row vectors create orthonormal set.
@MuPrimeMath
@MuPrimeMath 2 жыл бұрын
That follows from the fact that if square matrices A,B satisfy AB=I, then also BA=I. Hence if A^TA=I, then (A^T)^TA^T = AA^T = I, meaning that A^T is also an orthogonal matrix.
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