bit.ly/PavelPatreon lem.ma/LA - Linear Algebra on Lemma bit.ly/ITCYTNew - Dr. Grinfeld's Tensor Calculus textbook lem.ma/prep - Complete SAT Math Prep
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@MathTheBeautiful3 жыл бұрын
Go to LEM.MA/LA for videos, exercises, and to ask us questions directly.
@jaimelima24202 жыл бұрын
The necessity for having positive definiteness comes the need decompose A into a product A=M^T * M so se could do using Dirac’ notation = = and have valid inner products in the new vectors u and v. In other words, matrix A must be Gramian. Thanks for teaching us this!
@kabeerjamal9518 Жыл бұрын
Thank you, this was helpful
@ennevudoppioo97636 жыл бұрын
For the matrix at 10:15 wouldn't picking one of the roots of the polynomial suffice to disprove posirive definiteness? for example if instead of (-3,1) we chose (-4,1) the result would be zero but (-4,1) is not the zero vector therefore the matrix isn't positive definite.
@ahming1234 жыл бұрын
you again?? got to subscribe. Short and sweet explanations
@antonellomascarello46982 жыл бұрын
Awesome lesson 🙂
@menturinai13877 жыл бұрын
For the 2x2 matrix A = [a,b;b,c], is it also enough to say det([c]) > 0 and det(A) > 0 if and only if A is positive definite? Does this slightly different check for positive definiteness generalize in a nice way for larger matrices so we don't have to choose sub-matrices growing from the top left?
@MathTheBeautiful7 жыл бұрын
Yes. Because an identical and simultaneous switch of rows and columns is a similarity transformation.
@josephndagijimana66104 жыл бұрын
Thank you
@eddiechen63892 жыл бұрын
very good stuff thanks!
@MathTheBeautiful2 жыл бұрын
Glad you liked it!
@wolftribe666 жыл бұрын
at 7:40, how does he go straight to the answer without first multiplying the matrix with [a,1] then the result with [a,1]?
@MathTheBeautiful6 жыл бұрын
Hi Edward, Yes, this can definitely be done in one step. Take a look at this video: kzbin.info/www/bejne/bnvMfoinl6esfaM It doesn't answer your question directly, but the answer is in there. Let me know if it's helpful.
@jeremykua85254 жыл бұрын
Is it true positive definiteness should be a term used only for symmetric (quadratic) matrices? And this does not apply to non-symmetric matrices?
@MathTheBeautiful4 жыл бұрын
Yes, that correct.
@studentcommenter58584 жыл бұрын
But here you have assumed y=1. Isn't there a possibility for inner product to be negative for other y value (say y = -16)?
@fanyfan74663 жыл бұрын
I know this is late but I may as well throw this out there. The idea is that for any vector, say, [a b] you can multiply it by a scalar so that one of the entries is 1. For example, if I had vector [a b] I can multiply by the scalar 1/b and get [a/b 1] and I'll just say a/b=x for simplicity. Then hopefully you can see that a scalar multiple of any vector can be written as [x 1] But why is it okay to just scale a vector like that? Remember here he's talking about the inner product of a vector with itself and when that's positive. If I scale the vector by some scalar, c, I will be multiplying the scalar in the inner product twice basically. This means that the inner product will be scaled by c^2 which is positive for all real number scalars. Hence, if the inner product is positive for a vector with itself, it will also be positive for a scalar multiple of a vector with itself. So all we need to do is show a scalar multiple of every vector satisfies this and then we can multiply scalars without worry. And as I showed above, you can get any vector (this is all for R^2 of course) in the form [x 1]. Putting all this together, showing this for a vector shows it for every scalar multiple of that vector and since every vector can be written as a scalar multiple of a vector in the form [x 1], showing it for all vectors in that form shows it for every vector. Hope this helps
@ares122657 жыл бұрын
And we have an information slice, the point at which information begins to multiply. Well this is so, not especially really thinking, just what would be added to the trend of volotility.