Now that you know what an icon basis is, you may be wondering how to find it. The process actually not too difficult. It entails solving an equation with a determinant. Here it is. First take a look at the matrix with entries two, one, zero and three. And take a look at how it acts on these points around the square. This should make the entire transformation clear. As you've seen before, the two horizontal vectors get stretched by two and the diagonals get stretched by three. And for the other points, this happens. Now compare this to another matrix one that simply stretches the entire plane by a factor of three in any direction. This matrix has entries three, zero, zero and three. And it's really three times the identity matrix. Notice one thing, these two transformations are not the same transformation, but they do coincide in many points. In other words, they act the exact same way for infinitely many points all the points in this line. So to be more specific, in this diagonal, the two transformations do the exact same thing. So they match in infinitely many points. Now that is strange. The transformation should only match at one point, the point zero, zero. When they match at infinitely many points, something non singular is happening, and what's happening? Well, let's look at their difference. If these two transformations match at infinitely many points, that means the difference is zero at infinitely many points. So if you apply the difference of these two matrix to any vector in any of these diagonals, you get the vector zero, zero. In other words, this matrix times a vector x y for infinity metric vectors is 0, 0. Now that is the trait of a singular transformation. Recall that a non singular transformation has a unique solution to the equation matrix times vector equals 00, that's the vector 00. So if you have any finitely many solutions to this, means your matrix is singular and you can verify that this is indeed a singular matrix as a determinant is zero. Now let's do something similar but for another transformation on the right. Our transformation does exactly as it did before. And now let's compare it to the transformation that stretches the plane by two in every direction. These two are not the same, but they match in this entire line. So in other words, matrix on the left times xy is able to matrix at the right times xy for any vector x,y on this line, that's infinitely many points. So we can do the same procedure, take the difference and that matrix times a vector, equals 00 for infinitely many vectors. That means that the matrix 01, 01 or the difference between our matrix and two times the identity is a singular matrix. And you can check indeed that it is singular matrix because its determinant is zero. So what is special about the eigenvalues? So what happened for the eigenvalue value two and the eigenvalue three. Let's think about it in general. If lambda was a negatve value, then the transformation given by our matrix and the transformation given by scaling the plane entirely by a factor of lambda are equal for infinitely many vectors x, y. That means their difference times a vector is equal to 0, 0 for infinitely many vectors and equation when it flew many solutions. Therefore this matrix two minus lambda one, zero, three minus lambda has to be a singular matrix or it's determinant is zero. It's determinant is given by this equation when we expanded, and that's called the characteristic polynomial. So basically to find the eigenvalues lambda, all we need to do is look at the characteristic polynomial and find the roots. The place where the characteristic polynomial is zero are the eigenvalues. So in this case, they're going to be two and three. So now that you have the eigenvalues, let's try to find the eigenvectors. So we call that the eigenvector is the vector that satisfy the equation matrix times vector, equals eigenvalue times vector. So if we expand this, we get these equations over here and the solution for these are x=1, y=0 or any multiple victims. So here is one of the eigenvectors, the one corresponding to the eigenvalue two. We do the same thing with three, solved for these equations and get 1, 1. So the eigenvector 1, 1 is corresponding to the eigenvalue of three. Now you're ready for a quiz. Find the eigenvalues and eigenvectors of this matrix. And the solution is that for the eigenvalues, it's 11, 1 and for the eigenvectors is 2, 1 corresponding to the eigenvalue 11 and minus one two corresponding to the eigenvalue one. Why? Well, if you look at the characteristic polynomial, it is the matrix with entries nine minus lambda 4, 4 and three minus lambda. So that expands as lambda square minus 12, lambda plus 11, which factors as lambda minus 11 times lambda minus one. Therefore the eigenvalues are lambda equals 11 and lambda equals one. And I'll leave it as an exercise for you to solve the equations for the eigenvectors and verify that they are going to be 2, 1 and minus 1, 2 or some multiple of them, right? Because all that matters is the direction.