00:01
We are going to prove two things here.
00:04
We are going to prove first that if two matrices of order n times n are similar, a and b, then they have the same determinant.
00:20
And then in part b, using that what we showed in part a, we will prove that if a and b are similar, then they have the same characteristic polynomial, which then imply in its turn that they have the same eigenvalues.
00:37
So we're going to prove bar a first and for that we say that let s be invertible matrix n times n such that b equal s inverse times a times s and that's the definition that a and b are similar.
01:22
That is, one of the matrices is equal to the inverse of the similarity matrix s times the first, the other matrix times the similarity matrix s.
01:36
So that's the definition of similarity between the matrices, similarity.
01:51
Then the determinant of b is equal to the determinant of s inverse times n times s times s times a times s but now we apply a property of determinants which say that the determinant of the product is equal to the product of the determinant so this is the determinant of s inverse times the determinant of a times the determinant of s which is the same as the determinant of s inverse times the determinant of s times the determinant of a because this now this is an equation where we have real numbers and there is the properties of real numbers which can be applied here the commutative and associative properties and that's equal to now we apply the property of the part of the termination the termina of the product of two matrices to on the other sense that is the is equal to the determinant of the product of the inverse of s and s so we are applying the same property we applied in the first line so now this is the determinant of the identity matrix because as inverse times s is the identity matrix by definition of inverse and so the determinant of then identity matrix is one, so we have that this is the determinant of a.
03:25
So the determinant of b equal the determinant of a.
03:30
So we have proof part a.
03:37
Two similar matrices then have the same determinant, and in part b, we start by again saying that b is s inverse times a times s.
03:51
Again, the definition of similarity.
03:53
And because we want to prove that they have the same characteristic polynomial, we are going to start by writing the characteristic polynomial of b is p of landa equal b minus, sorry, the determinant of b minus lama, lambda times the identity matrix of order n.
04:43
P of lambda the and let's put it as subindex b to say that is the characteristic polynomial of b so we have this b minus lambda the identity and now we replace b by s inverse times a times s minus lambda i because b is equal to that product here now we notice that this is the same as s inverse times a minus lambda identity times s because if we distribute this product here from the right and from the left for example we will get s minus to the s inverse times s times a times s which is this product and for the second term we will have s inverse negative sign lambda identity matrix times s and that is negative lambda is significant so we have lambda here s inverse times identity times s sorry and i will be negative lambda s inverse times s which is identity matrix that will be negative lambda identity which is what we have here so it is clear that this product here is exactly equal this product here, exactly equal to this expression here...