 With this, now we can state the next theorem. If u and c to the n cross n, then the following are equivalent. A, u is unitary. So basically this is telling you various properties of unitary matrices and also various conditions which you can state which are equivalent to saying that u is unitary. So u is non-singular, u Hermitian equals u inverse, u Hermitian equals the identity matrix, u Hermitian is unitary, e, the columns of u form an orthonormal set, the rows of u form an orthonormal set and finally for all x, the Euclidean length y, u, x is the same as that of x, which is y Hermitian y, x Hermitian x. So it is making many statements about unitary matrices and equivalent ways of stating that a matrix is unitary. So let us see how to show this. The first few parts are very simple. Later, just for the last part, we will need to do a little bit more work. So first of all, for equivalence, we have to show that all these properties imply each other. Now, so if I take the first two properties, if u is unitary, then by definition, u Hermitian u is i and so u Hermitian is an inverse of u. So u Hermitian u equals i, I will write it this way, u Hermitian is equals u inverse. So u Hermitian is an inverse of u and so these are reversible statements and so saying that a is unitary is the same as saying u Hermitian equals u inverse. So this is actually establishing this both ways and similarly by the property I just mentioned, ba equals i if and only if ab equals i. So which means that u u Hermitian equals i and again this is a reversible statement. So if I say u Hermitian u equals i, okay let me put it, write it a little more clearly, since ba equals i if and only if ab equals i, u Hermitian u equals i is the same as saying u u Hermitian equals i. So a and c are also equivalent statements. Now u Hermitian is unitary that immediately follows from, so if u u Hermitian equals i that is the same as saying u Hermitian whole Hermitian times u Hermitian equals i. So u Hermitian satisfies the definition of unitary matrix. So essentially this means that d is true and in fact d is equivalent to a. So that establishes the equivalence of abcd. Now if ui is the ith column of u then u Hermitian u equals the identity matrix is equivalent to saying ui Hermitian uj equals 1 if i equals j and 0 otherwise. And so that means that columns of u are orthonormal and so this same as saying a is equivalent to e and so e said that columns of u form an orthonormal set by just doing the same thing with u Hermitian the rows of u form an orthonormal set. So similarly e is equivalent to f. So now the last part is this g which says that for all x the euclidean length of y equal to ux is the same as that of x that is y Hermitian y equals x Hermitian x. One way is very easy if a holds that is u is unitary and y is equal to ux then y Hermitian y is equal to x Hermitian u Hermitian ux just substituting y equals ux which is equal to u Hermitian u is the identity matrix. So this is equal to x Hermitian x. So this proves one way so a implies g and so what remains is to show the other way that is g implies a. So for g implies a so let us first consider the 1 cross 1 case. So n is equal to 1 then what happens here is that if I take y equal to ax this is the scalar thing. So everything here is a scalar then if I look at y Hermitian which is the same as a conjugate y this is equal to the condition says that this is equal to x Hermitian x for every x. Now we need to show that this implies the matrix a the 1 cross 1 matrix a is unitary. So if this is true this implies that mod a square equals so y Hermitian y is equal to mod a squared times x star x. So if I take x star x to the other side I get mod a squared equal to 1 which implies a conjugate a equals 1 which is the same as saying a is unitary. This is the definition of a unitary matrix. So the 1 cross 1 case it is easy. Now for the 2 cross 2 case now let us consider the 2 cross 2 case and then we will generalize it to the n cross n case. So consider then here let us let u equal to the matrix u 1 1, u 1 2, u 2 1, u 2 2 and u Hermitian u. So let us say this is a matrix a 1 1, a 1 2, a 2 1, a 2 2. It is just some notation I am defining. So now here by definition this is a Hermitian symmetric matrix. So here we have that a 1 2 equals a 2 1 star and let us call this both equal to a same. Okay now we will take some specific choices for this vector. So if x equal to this vector 1 0 then so once again just to recall what we are trying to show here is that if y Hermitian y equals x Hermitian x for all x then u must be a unitary matrix. Okay so if I take y Hermitian y that is equal to Hermitian 1 0 times u Hermitian u which is this matrix a 1 1 a a star a 2 2 this times 1 0. Now this this product here it just pulls out the entry a 1 1 and what we are given is that this is equal to x Hermitian x is equal to 1. So a 1 1 is equal to 1 so a 1 1 equals 1. Similarly so considering x equal to 0 1 a 2 2 equals 1. So this matrix is of the form 1 a a star and 1. So we just need to show that a equals 0 and u Hermitian u will be the identity matrix or u will be unitary. Now if I take x equal to the vector 1 1 then y Hermitian y is equal to 1 1 times the matrix 1 a a star 1 1 times the vector 1 1 which is in turn equal to so if I expand this out this is 1 1 times 1 plus a a star plus 1. So it becomes 1 plus a plus a star plus 1 which is the same as 2 plus a plus a star and this is what we are given is that this is equal to x Hermitian x which is equal to 2. So this means that a plus a star which is equal to the real part of a 2 times the real part of a is equal to 0. So which in turn implies that the real part of a equals 0. Similarly if you choose x equal to 1 i then and you do the same thing you will have 1 i this matrix times 1 i. So then this becomes 1 minus i a a star minus i and then you do this times 1. So that gives me 1 plus i a plus i times this a star plus i sorry here it will become minus i because I am taking the conjugate transpose. So it is minus i times a star plus i and so that becomes equal to so minus i squared is minus 1 which cancels with this and so I will be left with i a minus i a star and that is supposed to be equal to the norm of this vector which becomes 1 minus 1 which is 0 norm squared of this vector. So this is equal to 0 and just for the sake of clarity I will write it as x Hermitian x which is equal to 0. Did I say that correctly or have I made a mistake? So how will one cancel? Yeah thanks for asking that so the ones are not cancelling right minus i squared is actually plus 1. So this is actually 2 plus i a minus i a star and similarly x Hermitian x is not 0. If I take x Hermitian x it becomes the inner product between 1 minus i and 1 i so that becomes 1 minus i square and so this is also equal to 2. So as before I get 2 2 plus i a minus i a star is equal to 2 so that means i a plus i a star equals 0 and I can take out i and take it to the other side. So I will be left with a minus a star which is equal to 0 which implies that the imaginary part of a is also equal to 0. So we've shown that the real part of a 0 and we've shown that the imaginary part of a 0 so that implies a equals 0 so that means that u Hermitian u is equal to the identity matrix or u is unitary. So what's left is to show that this holds in the n cross n case also. So here we'll use exactly this idea that we just discussed but what we'll do is so once again let a equal to u Hermitian u and suppose x is such that okay I'll just write it like this x is this vector with zeros there's going to be two ones and zeros everywhere else where these two are in the ith and jth position okay. So then if I take if I define then y equal to ux implies that if I take y Hermitian y this will pull out this is equal to x Hermitian ax which can then be written as okay I I'll just do one thing because I'm so let's call this xi it's a vector which has zeros everywhere except xi in the ith position and xj in the jth position because I want to use xi equals 1 xj equal to 0 then xi equals 1 xi equals 0 xj equals 1 and then xi equals 1 xj equals 1 and finally xi equals 1 and xj equal to i that's what we did we do consider four cases we considered x equal to 1 0 and 0 1 x equal to 1 1 and x equal to 1 i so those are the four xi xj values that I'm considering later but for the purpose of writing this I'll just say xi x is a vector where only the xi and xj are nonzero okay so then if I do x i x Hermitian ax I will get none of the other entries matter it becomes xi star xj star times aii aij aji ajj times xi xj and so now this is exactly the 2 cross 2 case we considered earlier and so so now we can follow the previous arguments which means I will choose xi xj equal to 1 0 0 1 1 1 and 1 i to show that this matrix aii aij aji ajj is nothing but the identity matrix 2 cross 2 and so but i and j are arbitrary here every principle 2 cross 2 sub matrix of a is the identity is a 2 cross 2 identity matrix which implies a is equal to i u Hermitian u equals i and hence u is unitary okay so those six statements are equivalent seven statements okay