FAST eigenvalues

Are you tired of finding eigenvalues? Then check out this video for a 2-minute way of finding eigenvalues of any 2x2 matrix! This is based on a beautiful quadratic trick introduced by Po-Shen Loh at Carnegie Mellon University, but known since Babylonian times.

[littledragonwolf]

Its funny cuz this is the one thing in math i actually figured out by myself.

[francaisdeuxbaguetteiii]

Another brilliant video by master Peyam! This is god-tier maths

[insouciantFox]

If only I didn’t take my Linear Algebra final LITERALLY A WEEK AGO.

[lourdjonsalen]

Watching these kind of Math techniques not just for fun or Math satisfactions but bcoz I know I will use these soon! Thank you for this, Dr. Peyam!

[Thankmelr]

“Introduced by Loh but known since Babylonian times”. That’s an interesting notion of “introduced” right there.

[jamesbentonticer]

I like the all black shirt and tie but it does look even better when you have chalk dust all over it.

[tomatrix]

This is cool, especially the extension to solving quadratics

[SonGoku-tymg]

Your way makes me genius in front of my teacher and class mates

[VerSalieri]

I enjoyed it, but honestly, for a 2 by 2 matrix...just finding them directly is quick enough.

And...I dont’t think it can be “simply” generalized to higher order matrices... it can, just not simply.

[starter]

Another powerful tool that has made its way through a lot of ODE, PDE, Numerical discovery is Geshgorins Circle Theorem where we can bound the eigen values by the diagonals of the matrix and the sum of the magnitude of each entry on the row of the diagonal entry.

In your example, without even computing determinant, we can already say one eigen value is within the interval [5, 9] ( 7+ |2| and 7- |2|) and the other eigenvalue is within the interval [-3, 5] (1+ |4|, and 1 - |4| ) . Though this has issues with determining distinct eigenvalues in intersecting intervals Geshgorin's Circle theorem is still a powerful tool for bounding the eigenvalues of a matrix. Its also a good way to check if your eigenvalues are incorrect if they dont even lie in the intervals in the first place.

[Android-]

Another fascinating solution by mid-point method. Great content. Thanks.

[hungryplate]

Please tell us something about 3*3 eigen values.

[shivaudaiyar]

Thanks for such a great content sir with love from India

[yasinozkan]

If all the rows or all the columns of the matrix A is the same value s, which is 3 in that case, then s is an eigenvalue of A. Then by the trace of A 8-3=5 is the other eigenvalue. Best regards.

[nestam]

You really upload this the day after my linear algebra test

[miro.s]

And the fastest method for exactly this matrix would be considering eigenvalues as two points in complex plain at possibly intersection of boundary with real axis, the two circles: (M, R) = ([7, 0], 2) and ([1, 0], 4). If we search for real eigenvalues on boundary of each circle, we see immediately that they can not be anything else than 5 and 3. More precisely, we get following conditions: |7-x|<=2 and |1-x|<=4

[matteovissani]

Amazing. Is there a similar trick for matrix bigger than 2x2?

[StefanReich]

Are you tired of finding eigenvalues? Looking for some excitement in your life? Then learn this one weird trick and become the center of any mathematicans' party.

[kevingrant]

Hey Dr peyam does this method work for 3x3 matrices?

[damiane]

Excellent!!!
Is the same method to solve a quadratic?