00:01
And n by n matrix, a is positive definite, then there exists a positive definite matrix b, such that a equals b transpose times b.
00:11
Okay, to solve this problem, and the first thing we need to notice that we can write a as the product of, as a product of p times d times p transpose.
00:27
So this is just from our textbook that we can write down symmetric, sorry, write down the matrix as the p times the diagonal matrix and where p is a orthogonal matrix and times p is an orthogonal matrix and times p transpose with such p transpose is exactly the inverse.
01:08
All right.
01:09
And the next thing, we can observe that if we consider the matrix, a diagonal matrix c, such that the square of c is d, that is d.
01:28
That is just like taking the square root of the diagonal matrix d.
01:34
Because, excuse me, because we, for the diagonal matrix if we are just taking the square root of each entry on the diagonal, then if we, and that is c and we find the matrix multiplication of such as c, that is the same as the diagonal matrix d.
01:55
So that is same to say since c is also a diagonal matrix, hence that implies the transpose of c is the same as c.
02:08
So we have c, we have c, transpose times c equals.
02:16
All right.
02:17
And here we need to make a note that is matrix c.
02:26
C is also diagonal...