In this video, we will learn what similar matrices are. We will also see some of the important properties of similar matrices. What are similar matrices? Let A and B be two square matrices of order n. Then A is set to be similar to a matrix B if there exists an invertible matrix P such that A is = PBP inverse. Symbolically right, A is similar to B like this. Basically, if we have a matrix P, which is invertible matrix, such that A can be expressed as product of these three matrices, then we can say that A is similar to B. Let us take one example. Let A is = this matrix, B is this matrix. Then A and B here are similar. Why are they are similar? Because there exist an invertible matrix P. You can see the determinant of this matrix is 2, such that B equal to P inverse AP or A = PBP inverse. How can we see that whether such P exist or not? You can take, this is basically AP = PB. You can take P as matrix A, B, C, D. A is given to you 2 minus 3,1 minus 2. What is P? A, B, C, D. This is A, B, C, D. What is B? 1, 0, 0, minus 1. When you equate each term from the left hand side to the right hand side, you will be getting four equations and four unknowns. You can solve them and you can find the values of A, B, C, and D, which are be 3, 1, 1, 1. If this matrix P is invertible also, then we can say that there exists a matrix P such that A = PBP. Hence, we can say that these two matrices are similar. Now let us see some important properties of similar matrices. The first property of similar matrices is that determinant of A is = determinant of B, if A is similar to B. How we can show it? The proof is easy. You can easily see that if matrix A is similar to matrix B. This implies there exist an invertible matrix P such that A is equal to PBP inverse. Now you did a determinant both the sides. Determinant of A will be = determinant of PBP inverse. Determinant of A into B is same as determinant of A into determinant of B, hence, we can write this expression as determinant of P, determinant of B, determinant of P inverse. This is determinant of P, determinant of B. Determinant of P inverse will be the reciprocal of determinant of P. This is one of determinant of P. This cancels out. So, this is = determinant of P. We can say that determinant of A is same as determinant B, the first property. The second property is that they have same characteristic polynomial. If A and B are similar matrices, they have same characteristic polynomial. Now how can we show this? In order to prove this, we have to simply show that the characteristic polynomial are same. Let us take A minus Lambda I. This is =, we can write it like this because A is = PBP inverse, PBP inverse. Minus Lambda I is what? PP inverse. B into P inverse is the identity matrix. Now this is further =, we can write it like this. Determinant of P, B minus Lambda I times P inverse, the determinant of this, which is = determinant of P, determinant of B minus Lambda I and determinant of p inverse. Since determinant of P inverse is the reciprocal of determinant of P, so this cancels out. This is nothing, but determinant of B minus Lambda I. We can see that determinant of A minus Lambda I is = determinant of B minus Lambda I. The characteristic polynomial is nothing but determinant of A minus Lambda I = 0. Since these two expression are same, that means the characteristic polynomial of both are same. Since they have the same characteristic polynomial, so we can simply say that they have the same eigenvalues also, because eigenvalues are nothing, but the roots of the characteristic polynomial. Now trace of a matrix. Of course, the test will be equal. Why the test will be equal? Because if they have the same eigenvalues and the sum of the eigenvalues is nothing but the trace of the matrix, hence the trace the similar matrix are also equal. Next is if x is an eigenvector corresponding to an eigenvalue Lambda of the matrix A, then P inverse x will be the eigenvector corresponding to Lambda for B. This proof is also easy, you can see that Ax = Lambda x. Where x is not = 0, which represent that for matrix A Lambda is an eigenvalue and the corresponding eigenvector is x. Now you can replace A by PBP inverse because A is similar to B. It is PBP inverse x = Lambda into x. This further implies now because P is an invertible matrix you can pre multiply both sides by P inverse. It is B, P inverse x is equals to Lambda into P inverse x. Now you can take this P inverse x sub Y. We can say that BY is = Lambda Y. From here, we can easily conclude that for the matrix B, the eigenvalue is Lambda and the corresponding eigenvector is Y. For the eigenvector Y, in order to prove that Y is an eigenvector, one thing is required that Y should be non zero. Y is non-zero Because what is Y? Y is P inverse x. If it is zero, then this implies x is zero. You can multiply both sides by P. This implies x is 0, which is not true. What we have concluded, so from here we can conclude that Y is not = 0 Hence, Y is an eigenvector. What is Y? Y is P inverse. We have also seen this property. In this video, we have seen that what similar matrices are. We have also learned some of the important properties of similar matrices, like they have the same characteristic polynomial and hence the same eigenvalue.
