Let's do an example of diagonalizing a symmetric matrix. A nice little property about orthogonal columns comes in later when we're dealing with the symmetric matrix and we're going to see how that plays out to make the computation and the calculation a bit easier for us. Here we're given A and we'll notice that this matrix has a diagonal and about the diagonal everything is the same, so negative 2, negative 2 and negative 1, negative 1, negative 1, negative 1. This would be a symmetric matrix. You can confirm this by simply taking the rows and making them columns in a new matrix and checking if they're identical to this original matrix. We do have a symmetric matrix. Now if you take this matrix and you get the characteristic polynomial, you will end up with 0, of course, equals negative Lambda minus 8, Lambda minus 6, Lambda minus 3, so we have eigenvalues of 8, 6, and 3. We could take this right now and make this into our P matrix. P is composed of the eigenvectors of my original matrix, which would be these, so I'd be okay and I have the eigenvalues which turn into my diagonal matrix. I do know that in my A equals P times D times P inverse, the D will equal 8, 6, 3 and the diagonal matrix will equal eigenvalues so that won't change, but we can use a nice little trick here, which is we can realize that these eigenvectors compose a matrix. Instead of calling it P, we'll just call it, a place holder right now, let's call it Z. That marker is not as good as I thought. Z which equals, as we would normally do and as we learned last time to do, we can take our eigenvectors and we can make this P matrix but we're going to do a nice little trick to make it a better P matrix out of it. For now we have a little place holder. We'll take this, write down our eigenvectors. Now we'll realize that the columns of this Z matrix are orthogonal to each other. It's a nice little property. Now, we also have another nice little property, which is that for a matrix with orthogonal columns, if we normalize them, if we make the length of each column 1, then we'll have an orthonormal set of columns and when that's the case, we'll realize that the inverse of that matrix is actually its transpose and that will be much nicer for us here because instead of calculating P inverse, we'll be able to just calculate P transpose, which is a much nicer thing to calculate than an inverse of a matrix. Instead of taking Z here, we want to take an orthonormal set of columns instead of an orthogonal set of columns. This is orthogonal, each column is orthogonal to each other. We also want them to be normal, so we just want to divide by their norm, which is their length, which we learned in course 2 and if we divide by each column's length, then the column will be unit size and we'll have an orthonormal set of columns. To do that, negative 1 squared plus 1 squared plus 0 squared is 2. Negative 1 squared is 1 plus 1 squared is 1. When you add them up, you get 2 but when you take the square root of that, which is square root of 2. So if we divide each of these entries by their norm, we'll get a normal unit vector of length 1. We have negative 1 divided by square root of 2,1 divided by square root of 2 and 0. Now we just go along and we do this for each column. We have negative 1, squared is 1 plus negative 1 squared is 1, so it's 2, plus 2 squared is 4, 2 plus 4 is 6. So we'll have negative 1 over square root of 6, negative 1 over square root of 6, 2 over square root of 6, and then when we do the last one, 1 squared plus 1 squared plus 1 squared is 3, so we'll have 1 over square root of 3, 1 over square root of 3, 1 over square root of 3, and we'll call this P. Instead of just taking the original eigenvectors of our original A matrix, we normalize them. We made them unit length and that gave us something nice, which is that when we write A equals PDP inverse, we know P. It's this right here. We know D it's right here and now, since the columns of our P matrix are orthonormal, instead of using P inverse, we know P inverse will be equal to P transpose and that's a nice calculation, so when we deal with diagonalizing a symmetric matrix, we should consider that we can make this calculation even easier by using this trick.
