 just one more definition to keep going. So a linear transformation t which maps from c to the n to c to the n is called a Euclidean isometry if x Hermitian x equals tx Hermitian tx for all x in c to the n. So basically an isometry is something that preserves length the length of a vector. So it's a transformation any transformation that preserves the length of a vector is called an isometry and if it preserves the Euclidean length or the length measured through the Euclidean norm then we call it a Euclidean isometry. So basically from the previous result we then have that a complex square matrix u in c to the n cross n is a Euclidean isometry if and only if it is unitary. This is exactly the same as the statement a is equivalent to g. One other remark about these unitary matrices is that if u and v are unitary matrices u, their product is unitary. That's simply because u v Hermitian u v is equal to v Hermitian u Hermitian u v and u Hermitian u is the identity matrix. So this is the same as v Hermitian v which is equal to the identity matrix. So u v is an identity. So take products of unity matrices how many of our products you take they will all be unitary. Another remark related remark is that if x1 to xk is an orthonormal set and u is a unitary matrix then u x1 through u xk is also an orthonormal set. These are all immediate from what we have seen about unitary vectors and orthonormal matrices x1 through xk are orthonormal and u is unitary, u x1, u xk orthonormal set. Okay, so now let's continue on with unitary equivalence. So I mean all this was kind of the prelude to get to unitary equivalence. So basically for a unitary matrix we know now that u Hermitian equals u inverse for a unitary matrix. So that means that if I take the mapping from A to u Hermitian Au this is a similarity transform, u is unitary. Okay, so by similarity transform what I mean is our definition S inverse AS where S is an invertible matrix. So it satisfies that definition. So this mapping from A to u Hermitian Au is in fact a similarity transform, but it's a special similarity transform. It's a transform where the matrix involved in the transform that the matrix S in our previous definition is a unitary matrix and that's why we call it a unitary similarity or a unitary equivalence. So the similarity A to u Hermitian Au for unitary u is, so now the leads to the definition that B in C to the n cross n is unitary equivalent to A in C to the n cross n. If there exists a unitary, we're going in circles a bit here, but matrix u in C to the n cross n such that B equals u Hermitian Au. Okay, so I mean going by the fact that similarity transform is an equivalence relation, you can also show that unitary equivalence is also an equivalence relation. So it's not a misnomer, it is called that for a reason. So that means that it is reflexive, any A is unitary equivalent to itself, it is symmetric, A unitary equivalent to B implies B unitary equivalent to A and it's transitive, that is A unitary equivalent to B and B unitary equivalent to C implies A is unitary equivalent to C. Okay, so you should show that just to convince yourself it's true, we have the following theorem. If A and B are unitary equivalent, the sum of the squares of all the entries of A is the same as the sum of all the squares of all the entries of B. So intuitively, this should be obvious, but just to see why this is true, summation ij equal to 1 to n mod aij square is the same as trace of A Hermitian A. If you look at the entries that arise when you're trying to compute trace of A Hermitian A, it's exactly the same as this double summation ij equal to 1 to n mod of aij square. Now, if B is U Hermitian Au, then trace of B Hermitian B is equal to trace of, I'm just going to substitute for B. So that becomes U Hermitian, B Hermitian B, B Hermitian, sorry, U Hermitian A Hermitian U, U Hermitian Au. So I'm just substituting for B as U Hermitian Au and U U Hermitian is the identity matrix. So I'm left with trace of U Hermitian trace of U Hermitian A Hermitian Au, which is equal to trace of A Hermitian E. Why is this last step true? In the trace, you can exchange the terms in the product. You can't exchange. So you're saying that trace of U Hermitian A Hermitian Au is the same as trace of U Hermitian Au, A Hermitian A. Is that what you're saying? This is not correct. So one of the things about the trace is that you can, you can cyclically permute the terms. Equal to trace of B A, we can use that. And trace of A B equals trace of B A, which means that you are actually cyclically permuting. You can't just pick two matrices and exchange the order. So you can't just shift this matrix out here and write it as U Hermitian U A Hermitian A. That's not correct. But you can shift this matrix all the way out here. That is okay. So I can write this as U U Hermitian A Hermitian A and that is equal to, this is equal to the identity matrix. So this is equal to the trace of A Hermitian A. So now you have one more exercise to show, which is that if I have trace of A B C, then it is equal to the trace of C A B, which is equal to trace of B C A. So you need to show this. So instead, a simple, simpler thing to do is to recognize that this U Hermitian A Hermitian AU is actually a similarity transform on A Hermitian A, because U Hermitian equals U inverse. So since it's a similarity transform, we know that the eigenvalues of a matrix remain invariant to similarity transforms. And the trace is nothing but the sum of the eigenvalues. So basically, trace is similarity invariant. And so this must be not true. So that's a slightly simpler way to maybe think about it, given what we know so far. So trace is similarity invariant. U Hermitian A Hermitian AU is a similarity transform on A Hermitian A. Okay, so that's about this. I think we are almost out of time today.