Oxford Linear Algebra: Spectral Theorem Proof
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- เผยแพร่เมื่อ 17 พ.ย. 2022
- University of Oxford mathematician Dr Tom Crawford goes through a full proof of the Spectral Theorem. Check out ProPrep with a 30-day free trial to see how it can help you to improve your performance in STEM-based subjects: www.proprep.uk/info/TOM-Crawford
Test your understanding of the content covered in the video with some practice exercises courtesy of ProPrep. You can download the workbooks and solutions for free here: www.proprep.uk/Academic/Downl...
You can also find several video lectures from ProPrep explaining the Spectral Theorem here: www.proprep.uk/general-module...
And further videos explaining the Gram-Schmidt process are here: www.proprep.uk/general-module...
Finally, fully worked video solutions from ProPrep instructors are here: www.proprep.uk/general-module...
Watch other videos from the Oxford Linear Algebra series at the links below.
Solving Systems of Linear Equations using Elementary Row Operations (ERO’s): • Oxford Linear Algebra:...
Calculating the inverse of 2x2, 3x3 and 4x4 matrices: • Oxford Linear Algebra:...
What is the Determinant Function: • Oxford Linear Algebra:...
The Easiest Method to Calculate Determinants: • Oxford Linear Algebra:...
Eigenvalues and Eigenvectors Explained: • Oxford Linear Algebra:...
The video goes through a full proof of the Spectral Theorem, which states that every real, symmetric matrix, has real eigenvalues, and can be diagonalised using a basis of its eigenvectors.
The first part of the proof uses the eigenvalue equation to show that any eigenvalue is in fact equal to its complex conjugate, and thus is real.
The second part of the proof shows that a matrix similarity transformation using an orthogonal matrix exists, and results in a diagonal matrix. We first construct an orthonormal basis (where the first vector is an eigenvector) using the Gram-Schmidt process, and then use these vectors as the columns of our orthogonal matrix. Next, we show that the resulting similarity matrix is also symmetric. This then allows us to conclude that the first row and first column are diagonal as required. The final step is to use induction on the size of the matrix. Assuming the result is true for a (n-1) x (n-1) matrix, we use our earlier calculation to construct the final orthogonal matrix, and show that when it is used as a change of basis matrix the result is diagonal, as we wanted.
Produced by Dr Tom Crawford at the University of Oxford. Tom is an Early-Career Teaching and Outreach Fellow at St Edmund Hall: www.seh.ox.ac.uk/people/tom-c...
For more maths content check out Tom's website tomrocksmaths.com/
You can also follow Tom on Facebook, Twitter and Instagram @tomrocksmaths.
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Get your Tom Rocks Maths merchandise here: beautifulequations.net/collec...
Check out ProPrep with a 30-day free trial to see how it can help you to improve your performance in STEM-based subjects: www.proprep.uk/info/TOM-Crawford
I just realized I had Linear Algebra 17 years ago. In 3 more years, it will be the median of my life. I'm starting to feel old.
I subscribed to your channel a couple months ago, but have not watched a single video. This video showed up on my home page.
This is the best presentation and proof of the spectral theorem I have seen. Beautiful logic and clarity of thought. Thank you.
In the statement of the Spectral Theorem, note that we really want to say "there exists an _orthonormal_ basis of R^n which consists of eigenvectors of A" (otherwise the equivalent condition you write in terms of orthogonal matrices is quite misleading!)
Thanks for the fun video though!
love how clear your explanations are, proprep seems super worth getting too 😋
Fella I was always told I was thick as muck in Maths at school. Yelled at my teachers, and sent out of the class for not understanding algebra. Would have really loved someone like you to inspire me. Instead I've been terrified my whole life of Maths.
I just went over this before thanksgiving, thank you for the clarification of this.
that theorem blew my mind when i was in college...
Thank you, that was really interesting. I've come bake to my course of linear algebra at first year of university :')
Very clear explanation!
Would it be possible to have a video explaining the proof for Cochran's theorem? Thank you!
Thanks 🙏🏻 Thanks a lot for this informative and useful video ❤️
Literaly just started doing this 5m ago. Thank you
Nice video sir Tom!
Gorgeous; thank you :)
This is extraordinary
Just one question, how do se prove that A actually has any eigenvalue? Does it come directly from its simmetry?
Do I understand correctly that _v'_ is the component-wise conjugate, i.e. _v = (a + bi, c + di) => v' = (a - bi, c - di)?_ If so, is the inner product of v with its conjugate v', i.e. _v^T * v',_ really equal to the inner product of v with itself, i.e. _v^T * v,_ as shown at ~10:55?
In case anyone wants to know why the first statement of part (II) is equivalent to saying there is an orthogonal matrix R such that R-1AR is diagonal, the intuition can be found from 3b1b's video about eigenvectors starting from here: th-cam.com/video/PFDu9oVAE-g/w-d-xo.html
Thanks a lot for this really clear proof Tom - there's loads of examples online but thanks for actually walking us through it :)
Lot's of thanks from India sir 😅
謝謝！
He's got good chalk writing
Nice
nice, but what kind of symmetry does the matrix have? Symmetry of rotation? Center of symmetry? Axis of symmetry? Any of the above?
A matrix is said to be symmetric if it's equal to its transpose.
What exactly is "v bar"? Is it the 'conjugate' of vector v? I'm confused
What does II of the thm. say if R was changed to C or some other field? Is the proof any different if it was done on linear maps between arbitrary inner-product spaces instead of Euclidean spaces? What does the thm. say if the dimension was infinite?
The theorem also works over C, but you need to change from symmetric matrices to Hermitian matrices (i.e. matrices that equal their *conjugate* transpose). The proof works in the same way for arbitrary inner product spaces. If the dimension is infinite, one essentially gets into functional analysis and there are various spectral theorems - e.g. the same statement as the basic spectral theorem holds for compact self-adjoint operators on a (real or complex) Hilbert space. Generalising beyond that, in order for the statement to remain true, you also have to generalise your notion of eigenvectors, and this rapidly gets rather complicated.
You are an example of "don't judge a book by it's cover."
Hi Tommy 😁
Isn't the proof by induction a bit of overkill her? ;-)
Just considering e_i instead of e_1 and deducing that A_{i,i} = 1, and A_{i,j} = 0 for i
e j does the trick, no?
We only know that v1 is an eigenvector. So the other columns don’t necessarily reduce to be diagonal.
@@TomRocksMaths Ha, forgot about that fact. Thanks!
This is cool and all but never forget that this is the same guy who forgot his circle theorems :)
Jk man you are awesome!
💐 ᵖʳᵒᵐᵒˢᵐ
You haven’t shown there is at least one real eigenvalue for A.
It should follow easily from the fundamental theorem of algebra. Great video nonetheless.
Uh easy
He doesn't even look like a mathematician, cuz when I saw him the first time, I thought he was some kind of musician