In this video we will learn how can we calculate eigenvectors of a matrix. We will also see what is geometric multiplicity and algebra multiplicity correspondent to an eigenvalue. So, what do you mean by eigenvectors of a matrix. Let A be an n x n real matrix. Then the values of lambda for which Ax equal to lambda x. Where x is x1, x2 to xn has a nontrivial solution. These are called eigenvalues and a vector x. This vector x corresponding to eigenvalue lambda. This is called eigenvector corresponding to the eigenvalue lambda. So next is, what do you mean by algebraic and geometric multiplicity for eigenvalue? So let A be an n x n matrix with lambda as one of the eigenvalue. Then we know that when we have a characteristic polynomial corresponding to a matrix, what is the characteristic polynomial? This is determinant of A minus lambda I equal to zero, right? So if matrix is of order n x n, so this polynomial is of degree n, right? So if lambda 1, lambda 2 up to lambda n are the routes of this characteristic polynomial. So some of the lambdas may repeat, right? All lambdas may be distinct, some lambdas may repeat also like some lambda maybe complex also, right? So the number of times the lambda repeats is the algebraic multiplicity, right. And corresponding to that lambda the number of linearly independent eigenvectors is called geometric multiplicity. In general, for an eigenvalue lambda algebraic multiplicity is always greater than equal to geometric multiplicity. Let us state one example to understand all of this. Let us take this matrix of order 2 x 2. We will find this eigenvalue and the corresponding eigenvectors, right? So how to find the eigenvalues of this? We will write a corresponding polynomial of this matrix which is determinant A minus lambda I and put it equal to zero. So this implies determinant of minus lambda 3, 2, -1 minus lambda equal to zero. Which further implies minus lambda minus 1 minus lambda minus 6 equal to zero and implies lambda square plus lambda- 6 equal to zero. So this further implies lambda plus 3 into lambda- 2 equal to 0 or lambda equal to 2 and -3. So these are the eigenvalues of this matrix. Now you can easily see that the eigenvalue lambda equal to 2, is repeating only once. So the algebraic multiplicity of lambda equal to 2 is 1. And similarly lambda equal to -3 is repeating only once. We can say that as the algebraic multiplicity for lambda equal to -3 is also 1. Now let us try to compute geometric multiplicity corresponding to lambda equal to 2, right, or eigenvector corresponding lambda equal to 2. So we will find eigenvector corresponding to lambda equal to 2 first of all. So how we can find it? See we have this equation A- lambda I times x equal to zero. For lambda equal to 2, this will be A- 2I times x equal to zero. And this implies, what is A? A is this, but you have to subtract -2I from this A, so this will give -2, 3, 2, -3. And x is what, x is x1, x2 put it equal to 0, 0. So this implies 2x1 equal to 3x2. So what will be x then? x will be x1, x2 transpose. So there are so many x1, x2 to satisfy this equation. You have to take only one, right? So you take say x2 equal to 2 for example, if you take x2 equal to 2 then x1 will be 3. So we can say that linearly independent eigenvector corresponding to lambda equal to 2 is this. So how many linearly independent eigenvectors corresponding to lambda equal to 2, only one. So we can say that geometric multiplicity corresponding to Lambda equal to 2 is 1. Similarly we can find eigenvector corresponding to lambda equal to -3. How to find that. Again we will take for lambda equal to -3. Again we will take A + 3I X equal to zero. So this will give 3, 3, 2 and 2 times X1, X2 equal to 0,0, right? This implies 3X1 + 3X2 equal to zero or 2X1 + 2X2 equal to zero, which is the same equation. So it is X1 + X2 equal to 0. So the eigenvector corresponding lambda equal to -3 will be X1, X2, X1, X2 is what minus X1 you can put. So this is basically X1 times 1 minus 1. So that means the linearly independent eigenvector corresponding to lambda equal to -3 is 1- 1. So what is the geometric multiplicity for lambda equal to -3 is 1, right? So in this way, if a matrix is given to you and you want to find out its algebraic multiplicity or geometric multiplicity, we can find it using this process. Now take another example, here you have a matrix of order 3 x 3. And eigenvalues is known to you, eigenvalues of this matrix are 0, 0 and 5. You can also verify by finding the characteristic polynomial of this matrix, right? Now you can see that the algebraic multiplicity for lambda equal to zero is 2 here, because this eigenvalue zero is repeating two times. However, the algebraic multiplicity corresponding to the lambda equal to 5 is 1. Because this is repeating only one time. Now, if you compute eigenvector corresponding to lambda equal to 5, that is this eigenvector, right? You can easily find out using this A- 5I times X equal to zero, right? Now, let us start to find out the eigenvectors corresponding to lambda equal to zero, right? So since algebraic multiplicity is always greater than equal to geometric multiplicity. So this implies for lambda equal to zero, this implies that geometric multiplicity will be less than equal to 2 because this eigenvalue is repeating two times. So it's algebraic multiplicity is 2, right? So that means the geometric multiplicity is either 1 or 2 in this case. So how to check whether geometric multiplicity for lambda is equal to zero is 1 or 2. So we will find eigenvectors, linearly independent eigenvectors in fact corresponding to lambda equal to zero. So you have this matrix. Now for lambda equal to zero, it is A minus lambda I times X equal to zero, right? So this implies the eigenvalue substitute lambda equal to zero it is A is equal to zero. This implies A is 1, 0,-2, 0, 0, 0- 2, 0, 4 and this is X1, X2, X3 is equal to 0, 0, 0. So this implies X1- 2 X3 equal to zero. And from the last equation -2x1 + 4X3 equal to 0, which is the same equation, which is the first one, same equation as the first one, right? So this implies X1 equal to 2X3. Now, what is your X? X is X1, X2, X3, right. So X1 is 2X3, X2 is X2, X3 is X3. Now this can be written as X3 times 2, 0, 1 and X2 times 0, 1, 0. So these two are the linearly independent eigenvectors corresponding to lambda equal to 0. Because all other eigenvectors are the linear combination of these two vectors, right? So we can say that for lambda equal to 0, the geometric multiplicity is 2. In this video, we have seen how we can compute eigenvector corresponding to eigenvalues. We have also seen what is algebraic and geometric multiplicity cross point. When I can really offer one important property, we have noticed that algebraic multipliicity is always greater than equal to geometric multiplicity for an eigenvector.
