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.
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.
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.
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.
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?
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.
Go to LEM.MA/LA for videos, exercises, and to ask us questions directly.
This dude here just calculated eigenvalues on the fly and here I'm wasting my ink.
This all of the sudden became so much more simple, thank you!
eigenvalues by inspection? what's the secret shortcut? which video has the pre requisite??
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.
@@rishabhgarg9217 ruclips.net/video/2mPl3qKMFL4/видео.html&list=PLlXfTHzgMRUIqYrutsFXCOmiqKUgOgGJ5&index=40
@@MathTheBeautiful thanks a lot
This was breathtaking!
Thank you
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.
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 ?
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.
This video was f**ked up
Which is the video for the determinant trick? Thank you!
ruclips.net/video/O6jWE9E00mU/видео.html
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.
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?
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.
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)
From 2014 you become more "cool", i like it.
What even?