A 3x3 Symmetric Eigenvalue Decomposition in under 3 Minutes!

[@MathTheBeautiful]

Go to LEM.MA/LA for videos, exercises, and to ask us questions directly.

[@vaibhavsingh1049]

This dude here just calculated eigenvalues on the fly and here I'm wasting my ink.

[@elachichai]

eigenvalues by inspection? what's the secret shortcut? which video has the pre requisite??

[@rishabhgarg9217]

He had explained this procedure in other playlist.Just go the playlist by name Linear algebra part3 : transformations which has 112 videos and as far as i remember you will find those particular videos by video number 40-50.

[@MathTheBeautiful]

@@rishabhgarg9217 kzbin.info/www/bejne/aJ6znWangLJ5gpY&list=PLlXfTHzgMRUIqYrutsFXCOmiqKUgOgGJ5&index=40

[@eluz7903]

@@MathTheBeautiful thanks a lot

[@joshuasorell8107]

This all of the sudden became so much more simple, thank you!

[@adarshkishore6666]

This was breathtaking!

[@RahulThakkar1]

Which is the video for the determinant trick? Thank you!

[@MathTheBeautiful]

kzbin.info/www/bejne/hWfNiHhvepVjo7c

[@RahulThakkar1]

Thank you for the prompt reply@@MathTheBeautiful! I should have given a proper context, I meant the trick being talked at 1:17 - finding the 3rd eigenvector orthogonal without the cross product rule or checking for the null space. It's being mentioned as a Determinant trick.

[@vinkrishx]

When lamba equals 2 , then the eigen vector is -2 1 1 how did you get 2 -1 -1 ? Can you pls explain that sir ?

[@MathTheBeautiful]

The term "eigenvector" is really a shortcut for the actual concept of an eigenspace. The eigenspace here is a*[-2 1 1] . If you take a=1 you get [-2 1 1]. If you take a=10, you get {-20 10 10]. If you take a = -1, you get [2 -1 -1]. So the two vectors in your question are equivalent from the point of view of representing the eigenspace.

[@Momo-idky]

Honestly, if you would have made the video 2 minutes longer and would've explained where the 2/sqrt(6) came from I wouldn't have wasted my time and you would've taught me something. Sadly that's not really the case now.

[@Peters-lx7lt]

How does he know that the eigenvector corresponding to 0 is (0 1 -1)/sqrt(2) ? I see that this vector is orthogonal to the other one but why not choose (1 -1 0)/sqrt(2) instead?

[@eren-ajanitshimanga8759]

The eigendecomposition is not unique. Therefore as long as we have an orthonormal vector to the other two eigenvectors, it satisfies since the eigenvalue is 0.

[@MegaArti2000]

well, (1 -1 0) is not an eigenvector of this matrix, if you do M * (1 -1 0) you get (2 -1 -1), which is not paralell to (1 -1 0)

[@shivamjalotra7919]

This video was f**ked up

[@Michaelmaggimee]

Thank you

[@sundusshiraz3566]

What even?

[@13TrafalgarLaw]

From 2014 you become more "cool", i like it.