Easiest way | Similarity of Linear Transformation with GEOMETRY & EXAMPLE | Linear Algebra

Hello friends, in this video we are going to discuss similarity of linear transformation. The similarity of linear transformation is the extension of change of basis of linear transformation. So first we recall the concept of change of basis of linear transformation. So we considered the transformation I V2V where I is a identity linear transformation Domain has basis B and the co-domain has basis B'. So, I is identity linear transformation. So, I of V equal to V. The difference is that the image vector V as basis B dash and the domain vector V as basis B. So, this identity linear transformation will give the change of basis of vector space. Now, the similarity is extension of change of By changing I to the generalized linear transformation T. V2V Where domain has basis B and the co-domain has basis B'. Suppose T of V equal to W that means image of V under the linear transformation T is vector W. V is an element of domain and W is an element of co-domain. Therefore, T V B equal to W B'. W is equal to T V. Therefore, T V B equal to T V B'. So, in the similarity, We have two steps to get the image, first step is change of bases and the second step is find the image or first step is find the image and then change the bases. We are going to understand this process using one example. Suppose, T is defined from R2 to R2, the domain is basis B equal to 1 0 0 1 and the co-domain is In its basis, B' is equal to 1 by root 2, 1 by root 2. minus 1 by root 2 plus 1 by root 2 and the transformation is defined as Tx is equal to 2x that means every vector double in its image. You can rewrite with respect to basis so TxB is equal to 2 times XB dash. So to get this xb dash you have to follow two steps. First step is change of basis and then take the image or first take the image and then change the basis. We are going to understand the geometry. Suppose we have an ellipse in the domain B. Now we are going to find the image of this ellipse using this transformation Tx is equal to 2x. So first we take the change of basis. So by changing the basis B to B', this ellipse is now transferred to the ellipse with the 45 degree rotation. We are going to take the image of ellipse under this transformation Tx is equal to 2x then we have this ellipse will double original ellipse after the transformation. So therefore to get the final result of ellipse we have to do two transformation one is identity transformation from B to B dash and the second one is transformation T from B dash to B dash. Now we are going to discuss the This is the second path to get the image. So we have ellipse in the domain. Now first we take the image. So image is double the original image. Therefore it double the image. So it is a transformation T B to B. Now we change the basis and it is rotated by the 45 degree. So here we have transformation is a composition map of two transformation T B to B and I B to B dash. So the both the process we have the same result. So this processes are called the similar linear transformation. So there's two linear transformations. First composition T B to B dash with I B to B dash v equal to I B to B dash composed with T B to B v. So the matrix associated with this linear transformation if P is a transition matrix from B to B dash tb'b'p equal to P T B B. Therefore you can rewrite T B dash P equal to P T B dash. The notation T B' B' can be written as T B' and T BB written in the form of T B. Now if we take left hand side P on the right hand side then we have P inverse. So final result T B dash P equal to T B P inverse. So therefore This T B dash and T B are called the similar matrices via the transition matrix P from B to B dash. If we consider the transition matrix P from B dash to B, then this result can be written in the form of T B dash is equal to P inverse T B. Now we are going to discuss one example if B equal to 1 0 0 1 and B dash is equal to 1 1 1 minus 1 are bases for the vector space R2 and the matrix associated with the linear transformation T R2 to R2 is T B equal to 1 minus 1 2 3 then find T B dash. Now to solve this example, We will use the result which T B dash is equal to P T B P inverse. So for that we require transition matrix v1 equal to 1 0 v2 equal to 0 and belongs to B w1 equal to 1 1 and w2 equal to 1 minus 1 belongs to B dash. Now transition matrix P from B to B dash step 1 construction of augmented matrix B dash B and apply row operations such that this first part convert into the identity So v1 v2 convert into the transition matrix. So w1 equal to 1 1 first column, w2 equal to 1 minus 1 second column, v1 equal to third column 1 0 and the fourth column 0 1. Now we try to convert first part into the identity matrix. For that we will use the row operation. So first row operation will make this entry 0. So multiply first row by minus, add it to the second row. Then we have Minus 1 added to the plus 1, 0. Minus 1 added to the minus 1, minus 2. Minus 1 added to the 0, minus 1. And 0 added to the 1, 1. Now make this entry 1 by multiplying minus 1 by 2. Then we have 0, 1, 1 by 2, minus 1 by 2. Now make this entry 0. So multiply second row by minus 1 and add it to the corresponding entries of first row. Then we have Identity in the first part and the second part 1 by 2, 1 by 2, 1 by 2 and minus 1 by 2. Therefore, the second part will give the transition matrix which is 1 by 2, 1 by 2, 1 by 2, minus 1 by 2. Now, we have to calculate TB dash. So, use the result TB dash is equal to PTB P inverse. So, this is PTB and P inverse. Then, we have final result. 5 by 2, 1 by 2, minus 5 by 2, 3 by 2. This way we can calculate TB dash using TB through this result. TB dash is equal to PTB, P inverse. Thanks for watching. Like, share, subscribe. Press the bell icon of NextGen.