0:00
Hello there.
00:01
Okay, so for this exercise, we've got a matrix defined as follows, is qa.
00:07
It's defined as a polynomial, sort of a saying, of coefficients and the powers of the matrix a to the end up to 0, which is just the identity matrix.
00:23
So we got this matrix qa, and where a is an n -by -n matrix.
00:31
So now what we need to show is that if we find, well, actually we have here a similar matrix b, such that is equal to p inverse a p, the similarity condition, then the matrix q of b can be written as the p inverse qa p.
00:55
Okay, so we need to show this relation here.
01:00
So let's start.
01:04
So first let's consider q of b.
01:10
This is equal to a, n, bn, plus a .n minus 1, b, n minus 1 plus dot, dot, dot, a1, b plus a0, and the identity matrix.
01:40
Then we are going to replace with the condition here.
01:46
So we got that q of b will be equals to a, n, p inverse, ap, and due to here we're taking the n power, we know that b to the n is equal to we know that b to the n is going to be equals to p to the inverse a and you're going to use this property so be to the n will be just p to the minus to p inverse a to the n plus a n minus 1 p inverse is a n minus 1, p plus a1, p inverse, a p, plus a0 on the identity matrix.
02:55
And here we can take out all the p because the p inverse is distributed on all the terms here...