00:01
Let a be a column vector in rn.
00:05
Satisfying a transpose times a equal 1, define a matrix a equal in minus 2 a times a transpose.
00:17
Show that matrix a is an orthogonal matrix.
00:22
So we're going to say vector a to denote the column vector small a and matrix a to denote the capital a, meaning matrix a defined this way.
00:36
So you see, when we define, first, when we say that a transpose a equal 1, this is the inner product of a by itself, that is, the inner product of a.
00:50
A by a is 1.
00:53
And that inner product, we know, is exactly the same as a transpose times n, because a is a column vector in rn.
01:05
So the transpose of a is a row vector, that is, it's a matrix with only one row, and a is a matrix with only one column.
01:17
When we do the matrix product here, you get the only row of the matrix a transpose, inner product with the only column of a, to give us then the inner product of a by a.
01:32
A.
01:34
So we have that.
01:36
And when we define matrix a, we use the identity matrix of order n, n by n, and this operation now, you see, it's the other way around as this one, that is, here we have a transpose times a, which is the inner product of a and a, because a is a column vector, and for the same reason, a being a column vector means that this product in this order give us a matrix n by n that we can see it by noting the following a is being a column vector is n by 1 that is n rows one column so it's transpose is one column one row sorry and one in columns because we obtain the transform the the trans transpose of a by changing rows and columns was changing rows into columns so if we have in a n rows and one column then the transpose we have one row and n columns so the product can be done a is transpose can be done because the numbers of columns of the first factor a is equal to the number of rows and second factor a transpose and the resulting matrix is n by n so the product of a column vector times its transpose is a matrix of order n by n while the product of the transpose of the vector by the vector itself is the the inner product of the vector by itself.
03:24
Okay, so that's why here it makes sense, so let me get rid of this for a moment.
03:32
And so it makes sense here to do the identity matrix of order n minus this matrix here, which is scalar negative two times the matrix a times a transpose.
03:45
Okay, so knowing that, then we can proceed to calculate.
03:50
So what we're going to prove is that a times x transpose is the identity matrix because the matrix q is orthogonal if and only if the product of q and x transpose and the product of q transpose and q is the identity matrix.
04:14
In other words, the inverse of matrix q is x transpose.
04:20
That is, that's the definition of orthogonal matrix.
04:24
So what we'll get to prove is that the product of this matrix here by its transpose is an entity matrix of order n.
04:34
And we are going to calculate this.
04:37
And while we are doing the calculations, we are going to verify that the product the other way around, a transpose times a is the same as this one.
04:49
We will see that while we are doing the calculations.
04:51
So let's see, a then is i n minus 2 times a times a transpose, and that times the transpose of this matrix.
05:06
So that's equal to i n minus 2 a times a transpose times, and now the transpose of the subtraction or sum is equivalent of two matrices is the sum of the subtraction of the transpose of the matrices.
05:23
That is, the transpose of in minus 2aa transpose is equal to in transpose, the transpose of the first matrix here, minus the transpose of the second matrix 2aa transpose.
05:40
Transpose.
05:42
Now this is i n minus 2 a a transpose times the transpose of the identity matrix is the same identity matrix because that matrix is symmetric.
05:56
Then the transpose of the scalar times the matrix is the scalar times the transpose of the matrix.
06:04
So we have this...