 Recall that a matrix is normal if A A Hermitian equals A Hermitian A. It commutes with its Hermitian and we saw that it is a generalization of unitary real symmetric and Hermitian matrices and we saw a simple 2 cross 2 example of a matrix that is normal but not unitary Hermitian or skew Hermitian. You also defined unitary diagonalizability. So a matrix is said to be unitary diagonalizable if it is unitary equivalent to a diagonal matrix. So the similarity transform that will take the matrix to a diagonal matrix is in fact a unitary matrix. Okay, so this is the theorem which gives us some fundamental properties of normal matrices. So if the matrix A which is of size n cross n has eigenvalues lambda 1 to lambda n, then these four properties are equivalent. First is that A is a normal matrix. The second is that A is unitary diagonalizable and the third is that the sum of the squares of all the entries of A which is also the Frobenius norm squared of the matrix A is equal to the sum of the squares of its eigenvalues. And the fourth is that there is an orthonormal set of n eigenvectors of A that is n has a full set of eigenvectors which are orthonormal to each other. Okay, so we will see how to show this. So if, so the starting point is again shows unitary triangularization theorem. So let t be upper triangular of size n cross n such that u Hermitian A u equals t. So this means that t is unitary equivalent to A. Okay, that means that this when we say A is normal then that is the same as saying that t must be normal. So that is something you can immediately verify that. So A normality of A is equivalent to normality. So you can easily check that if A is normal then t is normal and this is an if and only if condition. Yes, go ahead please. Sir what is normality of A and normality of t? Normality is the property that the matrix commutes with its conjugate transpose. So normality of A is the property that A A Hermitian equals A Hermitian A. Normality of t is the property that t t Hermitian equals t Hermitian t. So if A is normal then t is normal. Okay, but t is upper triangular. Okay, so if I look at upper triangular matrix and I write out this condition t t Hermitian equals t Hermitian t that looks like this. So I will write t t Hermitian. So t Hermitian would be a lower triangular matrix and t is an upper triangular matrix t Hermitian. So this is t Hermitian, this is t and that is the same as t t Hermitian. Okay, and now if you see what happens when you equate the entries of these two products, what I am going to argue is that if t is normal and upper triangular then it must be diagonal. So if I take the one-one element of both the left hand side and the right hand side here then what we get is the one comma one element of this is going to be t11 times t11 star. And so that will be t11 t11 star and the one comma one element here would be this column times this row which is t11 t11 star plus the summation j equal to 2 to n t1j t1j star. Okay, so now this is mod t11 squared and it cancels with this mod t11 square and this is mod t1j square. So this means that summation j equal to 2 to n mod t1j square equals 0 which in turn means that these are all non-negative quantities. So if you add up all of these and you are still getting 0, every one of them must be 0. So t1j equals 0, j equals 2 through n. So basically other than the one comma one element, all other entries in the first row must be equal to 0. Similarly, if you equate the 2 comma 2th entry, what you will get is that t2j equals 0 for j equal to 3, 4 up to n and so on. So just proceeding this way, we get tij equals 0 for j greater than i and i going from 1 to n and of course tij equals 0 for j less than i and i going from 1 to n because t is upper triangular. Okay, so for j less than i, the entries are always equal to 0. So this means that t is diagonal which means that a is uniterally diagonalizable. The other way is actually much simpler. So p implies a. So what we need to show is that if a is uniterally diagonalizable, then a is a normal matrix. So basically any diagonal matrix is already normal. For any diagonal matrix D, D-homission D just contains mod di squared along the diagonals and so that is equal to D-D-homission for any diagonal. So all diagonal matrices are normal and unitary equivalence preserves normality. So basically therefore b implies a. So a and b are done. Now if you want to show b implies c, what we need to show is that if a is uniterally diagonalizable, then summation of mod aij squared equals the summation of lambda i squared. If a is uniterally diagonalizable, then the resulting diagonal matrix will have the eigenvalues lambda i on the diagonal because after all it's a similarity transform and if you are using a similarity transform to get a diagonal form, then the diagonal entries must be lambda i. So and further, I will just show this in a second, unitary equivalence is a property that preserves Frobenius norm. So that means that sigma ij equal to 1 to n mod aij square is equal to the trace of a-homission a which is equal to sigma i equal to 1 to n mod lambda i squared. So to see this, I mean this is simple. Basically if a is uniterally diagonalizable, then there exists a diagonal D containing eigenvalues of a and unitary u such that D equals u-homission a-u and so then summation of lambda i squared which is equal to the trace of d-homission d is equal to trace of and substitute for d, u-homission, a-homission, u-u-homission a-u which is equal to the trace of this is just the identity matrix. So u-homission a-homission a-u and this trace is a similarity invariant and this u-homission times u is a similarity transform and so you have trace of a-homission a which is equal to sigma ij 1 to n mod aij square. To go the other way, okay so to show the other way, so specifically that we want to show that if summation of aij squared equals summation of lambda i square, then a is uniterally diagonalizable. So when you use the Schur's triangularization theorem to find unitary transform such that u-homission a-u equals t, lambda i, i going from 1 to n are the diagonal elements of t or will be the diagonal elements of t. So that is basically how the eigenvalues are related to the matrix a. So there exists a u such that u-homission a-u equals t and the diagonal entries of t are these lambda i's. So basically if somebody told us that summation so i equal to 1, okay ij both, ij equal to 1 to n, aij square. So if we were to try to compute this, this is equal to the trace of a-homission a which is in turn equal to, I will just write it so that it is clear, trace of a-homission a and I will substitute for u and I will write this as trace of u-homission t-homission u-u-homission t-u which is equal to trace of u-homission t-homission t-u which is equal to the trace of because trace is uniterally invariant to this kind of a similarity transform. This is a similarity transform on t-homission t and the trace is invariant to similarity transformation. So this is the same as the trace of t-homission t which I can write as since the diagonal entries are lambda i so the trace of t-homission t can be written as sigma it's the sum of the mod of all the entries in t, mod square of all the entries in t and I keep the diagonal entries separate lambda i square plus sigma i less than j mod t ij square okay this these two this equality is coming because of c so we are assuming c is true and we are trying to show that the matrix must be uniterally diagonalizable so this immediately implies that sorry let me put it this way this this quantity here and this quantity here are equal because of our assumption that summation aij squared equals summation lambda i squared so that immediately implies that sigma i less than j mod tij squared equals zero or tij equals zero for i less than j and of course tij equals zero for i greater than j because t is upper triangular okay so if the row index is bigger than the column index all these entries are always zero so t is diagonal so that means that the upper triangular matrix we got by applying a unitary transformation of a through the sure triangularization theorem is in fact diagonal so which implies that a matrix a is uniterally diagonalizable which is the statement b then b implies d so if a is uniterally diagonalizable we want to show that there is an orthonormal set of n eigenvectors of a so if a is uniterally diagonalizable then it means there exists a u such that u Hermitian a u equals lambda which is equal to this diagonal matrix containing lambda 1 through lambda n that so if i take u to the other side or rather multiply by u on both sides I will get a u equals u lambda or each of these columns of u are actually so this lambda is a diagonal matrix so it means that a ui equals lambda i ui or there exists an orthonormal eigenvectors of a and the other way is also exactly the same if there exists an orthonormal eigenvectors um so if there exists an orthonormal all these steps are reversible and so basically it follows of a then a u equals u lambda where lambda is a diagonal matrix containing the eigenvectors of it so that implies u Hermitian a u since these eigenvectors are orthonormal u Hermitian equals or u Hermitian u equals equals the identity matrix so if I do u Hermitian from multiply by u Hermitian on the left then I get u Hermitian a u equals lambda which is a diagonal matrix so which means that a is uniterally diagonalizable sir yeah sir this i1 lambda i1 lambda 1 lambda to lambda n how are you saying that they are distinct always they are not that's a crucial point actually so the this is the main one of the main differences between what we said earlier and what we are saying now earlier for diagonalizability one of the conditions was that one of the sufficient conditions was that the eigenvalues need to be distinct we don't need that here so the important consequence of this is that normal matrices are non-defective okay they're always diagonalizable and the algebraic multiplicity equals the geometric multiplicity of the matrix so we've seen that Hermitian matrices are a special case of normal matrices if a is Hermitian then it means a equals a Hermitian and therefore a a Hermitian equals a Hermitian a which is both equal to a square and so Hermitian matrices are normal and for Hermitian matrices we can say one small extra thing which is known as the spectral theorem for Hermitian matrices so if a and c to the n cross n is Hermitian then all eigenvalues of a are what what can we say about the eigenvalues of real exactly and b a is uniterally diagonalizable see this is useful because all covariance matrices are Hermitian symmetric okay by definition the covariance metric is the expected value of xx Hermitian where x is a vector and so since covariance matrices are Hermitian symmetric matrices all eigenvalues of a covariance matrix are real and any covariance matrix is uniterally diagonalizable now the statement b here immediately follows from the fact that the matrix is any Hermitian matrix is normal and a normal matrix we just showed this that any normal matrix is uniterally diagonalizable now the point a that the eigenvalues are real follows because if it's if a matrix any diagonal and Hermitian symmetric matrix must be real and so and uniterally if I mean this unitary equivalence preserves Hermitian symmetric and so if I find a matrix that is uniterally equivalent to the matrix a and is diagonal then the matrix t which is uniterally equivalent to a must be Hermitian symmetric as well and if it's Hermitian symmetric and diagonal it is a real valued matrix so all the eigenvalues are real between unitary diagonalizability here and diagonalizability that we discussed earlier which was through similarity transforms is that we no longer need distinct eigenvalues in other words normality is enough