Eigenvalues of a 3x3 matrix | Alternate coordinate systems (bases) | Linear Algebra | Khan Academy

Determining the eigenvalues of a 3x3 matrix

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[kamam]

Not only did I learn how to find Eigenvalues, I also learned how to factor a cubic function! Thank you so much for all the good videos Sal, they help me out a lot!

[mohemmedansari]

since this video was created in 2009, indeed the explanation is beautiful, easy to be recalled even after 11 years.

[samus]

I've always learned as "A - lambda times I", so all you do is substract a lambda in each of the matrix diagonal.

[mrlevylev]

Ive spent a good 3 hours on a question trying to calculate eigenvalues and then I found this video and worked it out straight away! Thumbs up and thank you

[KraussHelmut]

Sal, can you PLEASE make a multi variable calc playlist? And thank you so much for everything! I think it's hard to comprehend the help you are giving us all!

[Unknwn]

First of all, thank you for the great tutorial. Though, I prefer (A- lamda l) but offcourse both work. I suggest everyone to use horner's method to reduce the equation. Way faster and in my opinion easier ;).

[justinballew]

My teacher just taught us (a-lambaI) and now I'm confused.

[MaruTheGreat]

I'm so glad this video is online, I'm currently taking a class on Quantum Mechanics (and I'm only 17), and the matrix algebra is incredibly complex, so this made it easier to understand...

[Hamzrs]

Why couldn't Patrick jmt make a video on this :(

[halorulesyourface]

so.. much... writing... but still clearer than my prof. thanks man!

[ChippyBlack]

yeah i also think its det(A-lambda), all my uni text books say that for finding eigenvalues and also later on when diagonlising matrices

[MyAfricanCats]

If you use the row echelon for of Ix-A to make the second row and the first column zero, then you can easily factorize x-3. The way in this video may not work if the solutions are not integers (so we cant guess the roots).

[Memelord_]

12:07 I just came over from professor Dave's video on this. Really helpful

[petecdun]

@TerminatorSe7en You can do it either way; given that you derive the equation from the definition of an eigenvalue, where A(v)=lambda(v), lambda(v) - A(v) = 0, but also, A(v) - lambda(v) = 0. So, you can use either det(A-lambda In), or det(lambda In - A), they're effectively the same thing.

[williamwright]

Even in the year 2021, Sal Khan is still a lifesaver.

[AlexKrippner]

No he is right, I thought the same thing but when he circled the +lambda^2 term it looked like it was negative but it wasnt

[guube]

thnks for your helps it makes me feel good and iam happy now after i have watched this vedio . please make it more for benefit

This guy is good. He proves that Math is not a secret society .

[jralocalsonly]

WOW! I missed it too. Everyone ignore this comment like the original commentor said to. We all thought it said minus lambda^2 but it actually is "+ lambda^2"

[Cipher]

You are my hero!

Also:
12:39
That's what she said!