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**A Criterion for Positive Definiteness of a Symmetric Matrix**. Following along are instructions in the video below:

x and y now these are not vectors so well start by talking about this function and Ive changed my notation a little bit since we all have only one vector Im just caught Im calling it entries X Y Z if we go to three dimensions so were dealing with only one vector a and when youre down to one vector a so in other words youre considering lowercase a transpose a capital a lowercase a then its no longer called a linear by linear form it is now called a quadratic form because theres only one input so this is a quadratic form you can think of it as a quadratic function its a perfect multi-dimensional generalization of quadratic functions quadratic functions are simple ax squared thats all you can have in one dimension Im not talking about linear shifts or constants just the quadratic part and thats the most general you can get in n dimensions in this case two dimensions so were looking at a quadratic form and were questioning its positive definiteness in other words not in other words its a definition were questioning the positive definiteness of this matrix its the this matrix is positive definite if the Associated quadratic form is positive definite its just the synonym and the question of positive definiteness can only be asked of a mid of a matrix thats symmetric and this matrix is and so were asking is it positive definite in other words is this expression always positive for any x and y as long as theyre not both a 0 so to connect this notation to the notation we had before Ill just write that our vector a is this thats what our vector is Im now focusing on its entries a little bit more and skipping the subscripts in three dimensions itll be X Y Z so that was that vector they cannot be simultaneously zero because thats what it means but thats what it would mean for the vector to be zero zero one would be a good example so here I started with a matrix that has one dead giveaway and negative number on the diagonal which would correspond to a negative quadratic term and the reason why its an instant giveaway for lack of positive definiteness is that you can easily extrude I go to extract that entry if I set all of these components of the vector to zero except the one that thats in the same position as the negative number on the diagonal it will very simply extract that number and so the example that was that came from the audience says 0 1 of course 0 1 because if you try 0 1 or if this was in more dimensions all entries are 0 except for the one where the negative number is that that right there is a counter example that will produce minus 8 and so this is clearly not positive definiteness which means that this matrix cannot be used for the classical in a product satisfying these three properties so this was easy how about this is this positive definite yes this can only be decided by a vote if you say that its not positive definite you have to give me

a counter example you have to say here is a vector for which its the result will not be positive thats how this game is played zero negative one lets try zero negative one remember I know what youre going after you want to make this eight negative but if Y is negative one the way eight shows up in the expression if you were to get it back to this form would be eight y squared right so your minus 1 will get paired up with another minus one and it will be positive very strongly positive its nice to have large positive numbers on the diagonal so zero minus one does not break it another guess no strictly real discussion maybe if you while we imaginary numbers then you can have 1000 10 and you put an eye and there you go so as I mentioned before when you allow imaginary numbers these will change more specifically surprisingly this one will change commutativity changes complex numbers are very funny in the way they affect change and these two will be perfectly intact yes this one will change – 3 1 lets try – 3 1 and then well talk about how they may have come up with it – 3 1 so from here well have minus 3 you know Im sorry from here well have nyah from here well have H so were its 17 and these will give us minus 9 and minus 9 my goodness 17 – 18 sir isnt that beautiful three and one how did they come up with that excuse me negative three and one so let us come up with this ourselves and once we do it once and then we do it again for the matrix a B BC well just see something super cool and that will give us our first hint of a criterion at positive definiteness something thats very natural and that youre very much familiar with will come out but heres how I would think about it so let me copy down the matrix heres what Ill do Ill say maybe there is a vector like this so Im going to try and solve for it I will try alpha beta which would have been perfectly fine alpha beta here alpha beta here multiplied out see what this expression tells me thats a perfectly fine way of doing it Ill simplified by one step instead of alpha beta I realize that I can scale this vector whatever way I want so I would first try alpha one right thats sort of like alpha betas scaled by beta 1 over beta so I will just drive alpha 1 just to have less complexity so we have no nothing to do but to multiply this out so lets see what we will get and now I look at this expression and I remarked to myself well I want to find out for for which its negative but Im looking at a quadratic equation with a positive coefficient for the square so I know its a parabola that looks like this and does it ever get negative well it would get negative if it if this is a quadratic equation had a zero right because it has

if it did we know almost everywhere its positive so if its negative somewhere that needs to have a couple roots so this equation has a couple roots then somewhere for some value of alpha itll be negative well lets see if it has a couple roots lets find its discriminant I think I have enough space here the discriminant equals 36 minus 32 am I right 4 I do have roots in fact my roots are well now Im just going to guess them there they the multiplication the product is for excuse me the product is 8 and the sum is minus 6 minus 2 and minus 4 so my two roots are minus 2 and minus 4 so this expression right here is the parabola that looks like this minus 2 minus 4 thats pretty good drawing right so for most all alphas its positive but between minus 2 and minus 4 its negative thats why these guys chose minus 3 smack in the middle thats the smallest this value can be minus 3 so when you take alpha equals minus 3 youre right here but they could have taken any number between minus 2 and minus 4 and this would be a negative value so the mole important takeaway from this is that the question of positive definiteness is subtle you look here and you say to yourself boy theyre all positive numbers thats good the values on the diagonal are pretty large right we want large numbers on the diagonal to have positive definiteness because thats thats the quadratic terms thats x squared + y squared the stronger they are the more likely it is to be positive definiteness and here we have eight great the off diagonals are positive as well actually that means nothing the sign of these guys means nothing if I went from 3 to minus 3 it would change this and it would simply flip everything to the other side it just flips things to the other side so the sign of the off diagonal entries doesnt matter what matters is their relative size for a relative to the diagonal terms and in this case there are just a little bit too big so if this was larger for example 9 thats an interesting case what if this was 9 then this would be 9 and then there would be a single root right at minus 3 its a double root but only one value and that would mean that at 4 alpha equals minus 3 this value is 0 does that break positive definiteness yes it does it needs to be strictly positive we talked about this so this would still be not a positive definite quadratic form it would be positive semi-definite it would be on the brink but if this was 10 then there are no roots it would be a negative discriminant no roots and therefore the parabola is high up you can kind of see that all you have to do is lift it by one and then buy a little bit more and then the form is positive definite so for 8 no for 9 still no for 10 yes positive definiteness so its this very fine balance between the diagonal and

the off diagonal entries and try to discover exactly what that balance is and its beautiful the way it works out is beautiful so what Im going to do is instead of having specific numbers lets have a BBC and see what happens so why dont you guys copy this problem over with a BBC continue with an analysis find the discriminant and then have a Eureka moment where you say I know what that expression is enjoy you do you guys know how to do the discriminant when the middle course when the linear coefficient is even theres a simplified formula well Ill just do discriminant over 4 so I dont have to divide it to do it later discriminant over 4 equals so so you repeat your analysis for this more general matrix you arrive once again at a quadratic expression with respect to alpha and then you once again say well we know for a fact that we want a to be positive so we have a parabola thats facing up and we wanted to have no zeros we want it to be safely above the x-axis so the discriminant and in this case I wrote down discriminant divided by 4 because either otherwise youd have a factor of 4 on the right hand side is B squared minus AC so this would be a positive quadratic form if two conditions are satisfied if a is greater than zero and AC excuse me and B squared minus AC is less than zero we want the discriminant to be negative we want there to be no roots so in other words AC is greater than B squared thats what we want we want this I believe this still fits in the form to be less than zero let me say it in another way and then youll say hey I recognized something so we want a to be positive and AC minus B squared to be positive and what is AC minus B squared everybody knows what it is the determinant so we need the top right number to be positive and we want the determinant of this matrix to be positive or I can actually put it in another way itll sound funny but I need this one by one sub matrix to be to have a positive determinant so I need this determinant to be positive and this determinant to be positive so the remarkable fact that Im about to tell you that generalizes is that an n-dimensional quadratic form is positive definite if its both unnecessary and a sufficient condition if you identify the following and determinants just the top left entry by itself the top left 2×2 matrix the top left 3×3 matrix the top left 4×4 matrix I guess this is a five by five matrix and the five by five matrix if all of these determinants are positive then the quadratic form is positive definite and vice versa if the quadratic form is positive definite then all of these determinants are positive its a necessary and sufficient condition for positive definiteness so thats one criterion thats our first criterion weve just stated it we havent proven it we actually will prove it because its a lot of lovely linear algebra

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