In this video, we will learn that if a square matrix is given to you, and you want to determine whether it is diagonalizable or not. So, we have some properties by which we can easily see. So diagonalizable matrix means that A is similar to a matrix A is said to be diagonalizable if it is similar to a diagonal matrix, that means there exist invertible matrix P such that AP is equal to PD, all right, or A is equal to PDP inverse. All right, now if a matrix A is known to you, then how you can see that whether that matrix is diagonalizable or not. So this can easily be visualized using this property one. What is that property? If you have an n cross n matrix A, then it is diagonalizable if and only if it possesses n linearly independent eigenvectors. What does it mean? It means that suppose you have a matrix A of order n cross n, right? And if it has an eigen value lambda who's algebraic multiplicity is suppose K, all right? Then for diagonalizability, it's geometric multiplicity must be K, all right? If it happens for every eigenvalues, then only it will be diagonalizable. Next property is if eigenvalues of a matrix are all distinct, then of course it is diagonalizable. Why is it diagonalizable? Because algebraic multiplicity cross point to each eigenvalue is one. And algebraic multiplicity is always written including geometric multiplicity. That means geometric multiplicity will be less than equal to one. So it cannot be zero, so hence it is one only, all right? So in this case, if all eigenvalues are distinct algebraic multiplicity is always equal to geometric multiplicity. Hence it is diagonalizable. So now the question arises that how can we find that matrix P, right, because you see that we have to find a matrix P such that A is diagonalizable or A similar to a diagonal matrix. How to find that P. Suppose so you have matrix A of n cross n, right? It's eigenvalue are suppose lambda one, lambda two up to lambda n. The corresponding eigenvectors are x1, x2 up to xn, because it must have n linearly independent eigenvectors for diagonalizability, right? We can construct a matrix P as the Vector x1, which is eigenvector corresponding to lambda one. We can write it as a column vector here. X2 we can write it here and similarly xn we can write here, right? So suppose the determinant of this matrix P will be non-zero because all eigenvectors are linearly independent. So the first condition is verified that this matrix is an invertible matrix. Now, next we have to see that the construction of this type of matrix. Does it satisfy AP equal to PD condition? So let us see that. What is AP now? A into P. A is your A matrix and P is this matrix X1, this column X2, the second column and Xn in the nth column, right? Then you multiply this entire matrix the first column. Similarly for the second column, what you will obtain? This will give A into X1 as a first column, right? A into X2 a second column and similarly A into Xn as a nth column of this matrix. Now, since X1, X2 up to Xn are eigenvectors correspond to the eigenvalues lambda 1 lambda 2 up to lambda n. So we can easily say that AXj is equal to lambda j Xj for J from 1 to n, right? So AX1 will be what, lambda 1 X1. Because Xj's are the eigenvectors and corresponding eigen values are lambda j's, right? This is lambda 2 X2 and this is lambda n Xn, right? Now this can be written as X1 as a first column, X2 as a second column, Xn as the nth column. And here it is lambda 1, 0,0,0, 0, lambda 2, 0 and 0 up to lambda n, the diagonal of matrix which is nothing but PD, alright? So, we have seen that if we construct this type of matrix, of course this matrix is not unique, right? But if you construct this matrix like this, which is as a column, you can write the eigenvectors corresponding to lambda j's, right, then AP equal to PD will be satisfied and P is an invertible matrix. So this is how one can find the matrix P for which AP equal to PD. Now take this example. Now one can easily verify that eigenvalues of this matrix are 2, 2 and 1. So what is the algebraic multiplicity equals to lambda equal to 2 is 2. And the algebraic multiplicity corresponding to lambda equal to 1 is 1. For this matrix to be diagonalizable geometric multiplicity should be equal to an algebraic multiplicity corresponding to each lambda, all right? We must have geometric multiplicity corresponding to lambda equal to 2 should be 2. For 1 it is always 1, becauseâ€™it's algebraic multiplicity is 1, all right? So let us try to construct the matrix P such that AP equal to PD. What will be your D? D of course will be diagonal matrix which is 2, 0, 0, 0, 2, 0, 0, 0, 1, alright? Now let us find eigen vectors corresponding to lambda equal to 2. For this it is (A -2I) times X=0, right? This implies 0, 0, 0, 1, 0, 1, -1, 0, -1. And this is your X1, X2, X3. This should be equal to 0, 0, 0, right? So this implies x1+x3 should be 0 from the second equation. And from the third equation also you got the same equation, right? So what will be your X? X is x1, x2, x3. x1 is -x3 you can write x2, x3, and this can be written as x3 times -1, 0, 1 and x2 times 0, 1, 0. So any eigenvector corresponding to lambda equal to 2 can be expressed as linear combination of these two vectors. So we can say that the linearly independent eigenvectors corresponding to lambda equal to 2 are these two vectors -1, 0, 1 and 0, 1, 0. Similarly we can find eigenvector corresponding to lambda equal to 1. So for lambda equal to 1, we can write A-I times X=0, which implies 1, 0, 0, 1, 1, 1, -1, 0, 0, x1, x2, x3 it is 0, 0, 0. So this implies x1 equal to 0 from the first equation, then x1+x2+x3 equal to 0 from the second equation. And the third equation also gives x1=0. So this implies x2 +x3=0, right? So the eigenvector corresponding lambda equal to 1 will be x1, x2, x3. x1 is 0, all right? x2 is -x3 and it is x3. So this can be written as x3 times 0, -1, 1. So the linearly independent eigenvector corresponding to lambda equal to 1 is this vector. So what will be your P? Your P will be the eigenvectors corresponding to lambda equal to 2 and the eigenvector corresponding to lambda equal to 1. So for lambda equal to 1 it is 0, -1, 1, right? And for 2 it is -1, 0, 1 and 0, 1, 0, -1, 0, 1 and 0, 1, 0. So this will be your P matrix such that AP equal to PD, all right? So in this way, one can easily find the matrix P if the matrix is a diagonalizable matrix. In this video, we have seen the necessary and sufficient condition for the matrix to be diagonalizable. We have seen that, if a matrix is of order n cross n and it is having n linearly independent eigenvectors, it will be diagonalizable. We have also seen that how we can find the matrix P if the matrix is diagonalizable. [MUSIC]