 Sure unlike the Kelly Hamilton unlike the short triangularization theorem you have seen or in your undergraduate curriculum. So this is the Kelly Hamilton theorem. Has anybody not heard of the Kelly Hamilton theorem? So let PA of t be the characteristic polynomial of A then What does the Kelly Hamilton theorem say? It satisfies its character equation so PA of A is zero Yeah, it's the zero where the zero is actually a zero matrix, right? This is a polynomial, but I'm evaluating the polynomial at t taking the value equal to an n cross n matrix a So I'm replacing t with an n cross n matrix and I'm evaluating the characteristic polynomial and I get All zero matrix. So this is a zero of size n cross n So that is what the Kelly Hamilton theorem says that That any matrix satisfies its characteristic polynomial. So again, it's a Yeah, this is again one of those Very cool results of linear algebra that is completely non-intuitive To me at least I am I cannot give you an intuitive reason Why a matrix would satisfy its own characteristic polynomial? Okay, so let's So have you guys seen a proof of this theorem? so PA of t is by definition determinant of t i minus a so that implies that if I wanted to find PA of a That is equal to the determinant of a i minus a Which is equal to the determinant of the all zero matrix Which is equal to zero Is that correct? Yes, sir. Yes, sir. This is an incorrect proof You can see that Real numbers are it is not a matrix Yeah, it's a that's a very very good point. So Obviously, I already observed that PA of a is actually a matrix of size n cross n But if I do it this way, you know, this thing is giving me a scalar value. This is one cross one So some I mean there is no meaning to saying that PA of a is equal to determinant of a i minus a So what went wrong was I should have actually First evaluated this and I get a polynomial out of this and then substitute A instead of t in that polynomial. I cannot directly substitute replace t with a inside the Expression for the determinant. Okay, so this is an incorrect proof. Yeah, here. Let me make another attempt. So We know that PA of lambda equals zero for every Eigen value lambda of a Okay, that is that we know because by definition an Eigen value is a zero of a characteristic of the characteristic polynomial so we also know that given any Polynomial q of x the Eigen values of q of a are What? Q of lambda. Okay, so I just evaluate that polynomial at the Eigen values and I get the Eigen values of q of a Okay, so that means that if I wanted to find the Eigen values of PA of a Okay, that would be PA of lambda which is equal to zero because this is true for every Eigen value of a Okay are all equal to zero So that means that if whatever this PA of a is it's a matrix whose Eigen values are all equal to zero. So that implies PA of a equals zero Okay, so this is so so this is the proof So none of you object So the fallacy of this proof is that any Yeah, even if the Eigen values are zero, it doesn't mean that the matrix should be equal to zero We already seen this example zero one zero zero. There are many other examples where The matrix is not zero, but it's Eigen values are all equal to zero Okay, so this is also incorrect Okay, so now we'll actually Now this is the correct proof So basically PA of t is a polynomial of degree n with zeros lambda one through lambda n Okay, and also the leading coefficient of this PA of t is one that is the coefficient of T power n is one So that implies I can write PA of t in The form t minus lambda one into t minus lambda two T minus lambda n. So this is the form of The Characteristic polynomial now from short theorem So I said we'll see some uses of short theorem. So this is the use. We are using the theorem We can write a equal to u t u Hermitian I'm just taking the u to the other side where t is upper triangular With lambda I on diagonal. So then if I evaluate PA of a Because I've written this polynomial out in the expanded form like this I can now actually substitute For a in this in this polynomial equation. So I'll write it as PA of u t u Hermitian Which is equal to u t u Hermitian minus lambda one I Times u t u Hermitian minus lambda 2 I Etc up to u t u Hermitian minus lambda n I Okay, so that is equal to I Can I can write I as u u Hermitian and pull out a u on the left and a u Hermitian on the right and write it as u t minus lambda one I u Hermitian Times u t minus lambda 2 I u Hermitian u t minus lambda n I u Hermitian and This u Hermitian u is all the identity matrix So I'll be left with a u on the left and a u Hermitian on the right and all the middle use will actually Disappear that just the identity matrix. So I can write this as u t minus lambda one I T minus lambda n I u Hermitian Which is equal to u times PA of t So this this thing is actually the characteristic polynomial of t as well because t is similar to a and u Hermitian and so clearly PA of a will be zero if And only if this PA of t is equal to zero now PA of t Is equal to this matrix if I look at t minus lambda one I lambda one is the one comma one element of t and so this will have a zero on the top and It may have something here and this is upper triangular and next matrix You know has something here the one comma one element, but that's the second element will be zero and Then this will have zeros all down the first column and over here. There could be arbitrary things and then there are zeros down here and This the rest of this row can be arbitrary, but this part will be upper triangular and So on and the last one will have some structure like this and then zero at the bottom and then zeros down here But now this form here is exactly the form we studied in the property where this element is zero And so when I multiply these two together the first two elements will become zero So if I multiply these two the top To cross to top left to cross to sub matrix is zero came by the Properties we just discussed and if I take the first three matrices together Okay, after that then you will see that the top left Three cross three sub matrix Equals zero and so on so if you multiply all of these together The top left n cross n Is zero or P a of t is the zero matrix and so that proves the result So I'm a bit over time so we'll stop here