Positive Definite Matrices and Minima

MIT 18.06SC Linear Algebra, Fall 2011
Instructor: Martina Balagovic

A teaching assistant works through a problem on positive definite matrices and minima.

License: Creative Commons BY-NC-SA

[mekala]

We can also calculate C for the semi-specific case by taking the correct combinations of the first two columns or rows. Matrix will be positive semidefinite when det is zero. We know matrix is singular when determinant is zero.
[2 -1 -1]
[-1 2 -1]
[-1 -1 2+C]
If we take - 1 * column1 + (-1 * column2) :
-2 - 1 = -1
1 - 2 = -1
1 + 1 = 2
So if matrix is singular 2 + c = 2 ----> c = 0

[alperaslan]

If she was my lecturer, I would never miss any of her classes.

[justpaulo]

It seems to me that the pivot test was much faster and easy.
In addition it seems that from the U matrix you can read directly the final "complete the square" equation.

[Robocat]

Overall very good explanation. But I don't see how the formulas in the upper right corner could help us with the completing the squares. It's not the same. Anyway it's very simple that you can actually do it in your head when you already know the pattern.

Another point is that in the end you talked about the null space of positive definite matrix but you didn't come up with a conclusion. It's not mentioned in the last lecture either.

[hsccbo]

You sholdn't expect determinant test to work for positive semi-definite matrices.
Consider the matrix,
[0 0]
[0 -1]
which certainly passes the ">=0" condition for all the principal submatrices yet its spectrum is {0, -1}.

[matthewjames]

is A = [ -1 0; 0 -4] negative definite?

Method 1: Determinant method
-1 < 0 (first sub matrix)
-1*-4-0*0 = 4 > 0 (second sub matrix)
Therefore this matrix is indefinite

Method 2: eigenvalues
lambda1 = -1 < 0
lambda2 = -4 < 0
Therefore this matrix is negative definite

What am I doing wrong here!?

[reactioner]

I like how teaching assistants are smiling after 5 second being out of view

[SIYA-PAL]

Very easy way अब अंपने बताया means u understand me very ejey

[radicalpotato]

Are they all necessary and sufficient tests, anyone?

[АлександрСницаренко-рд]

when c=0, the matrix is a positive semidefinite?

[SIYA-PAL]

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[SIYA-PAL]

Aanp bahoot axcha study krati hai mujhe anpka sab kuch samajh aata hai it's hindhi but wriiten in English 😄😄😄😄😄🇮🇳🇮🇳

[SIYA-PAL]

I Lives in India but i ever study by u bicouse u r a English girl so I can to talk English by hearing your voice 🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳 i lives in up loknow India 🇮🇳

[mozhdehyazdanifard]

Thanks for your quite useful video. Your explanations are pretty clear for dummies like me!
You are pretty smart, and also a good teacher. You are also beautiful. My brother was distracted by your beauty.. He could not take his eyes off you, to look at the whiteboard. He had to watch your video some couple of times.
Do not worry. I'll explain it for him.

[mozhdehyazdanifard]

I think what you said is all incorrect!!!

I'll bring another matrix that takes into question all what you have mentioned. That is why I'm dummy :(


1) Completing the Square:

Using the method "completed squared" does not seem to be compatible with other test methods! However, this is the most reliable method, so other methods cannot be used! For example, we know that the following matrix should have equal results with what you presented:
[2 -2 -2 ]
[0 2 -2 ]
[0 0 2+c]

Using "completed squared" we have:

2* x^2 + 2* y^2 + (2+c)* z^2 - 2 xy -2xz - 2yz

This is the same expression as what you have derived for your matrix. --> As you said: C > 0


2) determinant test:

What we see is that the determinant of the whole matrix is 2 * 2 * (2+c) = 8 + 4C > 0 ---> C> -2 !!!!


3) pivot test:

It's an upper triangle matrix. All that is required is to divide the rows by 2, except the third one:

2 * [1 -1 -1 ]
2* [0 1 -1 ]
(2+c) [0 0 1 ] ---> 2( x-y-z)^2 + 2 * (y-z)^2 + (2+c)* z^2 !!!! --> C > -2, but
that is not equal to what you derived for "completing the square"!

[mgk]

Very few women can do maths the way she's doing.