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# Proof of formula for determining eigenvalues

## Video transcript

I've got a transformation, m that's a mapping from Rn to Rn, and it can be represented by the matrix A. So the transformation of x is equal to A times x. We saw in the last video it's interesting to find the vectors that only get scaled up or down by the transformation. So we're interested in the vectors where I take the transformation of some special vector v. It equals of course, A times v. And we say it only gets scaled up by some factor, lambda times v. And these are interesting because they make for interesting basis vectors. You know, the transformation matrix in the alternate basis-- this is one of the basis vectors. It might be easier to compute. Might make for good coordinate systems. But they're in general, interesting. And we call vectors v that satisfy this, we call them eigenvectors. And we call their scaling factors the eigenvalues associated with this transformation and that eigenvector. You know, hopefully from that last video, we have a little bit of appreciation of why they're useful. But now in this video let's at least try to determine what some of them are. You know, based on what we know so far, if you show me an eigenvector I can verify that it definitely is the case, or an eigenvalue. I could verify the case. But I don't know a systematic way of solving for either of them. So let's see if we can come up with something. So in general, we're looking for solutions to the equation A times v is equal to lambda v. It's equal to lambda times the vector. Now one solution might immediately pop out at you, and that's just v is equal to the 0 vector. And that definitely is a solution, although it's not normally considered to be an eigenvector just because one, it's not a useful basis vector. It doesn't add anything to a basis. It doesn't add really the amount of vectors that you can span when you throw the basis vector in there. And also, it's not clear what is your eigenvalue that's associated with it. Because if v is equal to 0, any eigenvalue will work for that. So normally when we're looking for eigenvectors, we start with the assumption that we're looking for non-zero vectors. So we're looking for vectors that are not equal to the 0 vector. So given that, let's see if we can play around with this equation a little bit and see if we can at least come up with eigenvalues maybe in this video. So we subtract Av from both sides, we get the 0 vector is equal to lambda v minus A times v. Now, we can rewrite v as-- v is just the same thing as the identity matrix times v, right? v is a member of Rn. The identity matrix n by n. You just multiply and we're just going to get v again. So if I rewrite v this way, at least on this part of the expression-- and let me swap sides-- so then I'll get lambda times-- instead of v I'll write the identity matrix, the n by n identity matrix times v minus A times v is equal to the 0 vector. Now I have one matrix times v minus another matrix times v. Matrix vector products, they have the distributive property. So this is equivalent to the matrix lambda times the identity matrix minus A times the vector v. And that's going to be equal to 0, right? This is just some matrix right here. And the whole reason why I made this substitution is so I can write this as a matrix vector product instead of just a scalar vector product. And that way I was able to essentially factor out the v and just write this whole equation as essentially, some matrix vector product is equal to 0. Now, in order-- if we assume that this is the case, and we're assuming-- remember, we're assuming that v does not equal 0. So what does this mean? So we know that v is a member of the null space of this matrix right here. Let me write this down. v is a member of the null space of lambda I sub n minus A. I know that might look a little convoluted to you right now, but just imagine this is just some matrix B. It might make it simpler. This is just some matrix here, right? That's B. Let's make that substitution. Then this equation just becomes Bv is equal to 0. Now, if we want to look at the null space of this, the null space of B is all of the vectors x that are a member of Rn such that B times x is equal to 0. Well, v is clearly one of those guys, right? Because B times v is equal to 0. We're assuming B solves this equation and that gets all the way to the assumption that B must solve this equation. And v is not equal to 0. So v is a member of the null space and this is a nontrivial member of the null space. We already said the 0 vector is always going to be a member of the null space, and it would make this true. But we're assuming v is non-zero. We're only interested in non-zero eigenvectors. And that means that this guy's null space has to be nontrivial. So this means that the null space of lambda In minus A is nontrivial. The 0 vector is not the only member. And you might remember before, that the only time-- let me write this in general. If I have some matrix-- I don't know. I've used A and B. Let's say I have some matrix D. D's columns are linearly independent if and only if the null space of D only contains the 0 vector. Right? So if we have some matrix here whose null space does not only contain the 0 vector, then it has linearly dependent columns. And I just wrote that there to kind of show you what we do know and the fact that this one doesn't have a trivial null space tells us that we're dealing with linearly dependent columns. So lambda In minus A-- it looks all fancy, but this is just a matrix-- must have linearly dependent columns. Or another way to say that is, if you have linearly dependent columns, you're not invertible, which also means that your determinate must be equal to 0. All of these are true. If your determinate is equal to 0, you're not going to be invertible. You're going to have linearly dependent columns. If your determinate is equal to 0, then that also means that you have nontrivial members in your null space. And so, if your determinate is equal to 0 that means there's some lambdas for which this is true, for non-zero vectors v. So, if there are some solutions, if there are some non-zero vector v's that satisfy this equation, then this matrix right here must have a determinate of 0. And it goes the other way. If this guy has a determinate of 0, then there must be-- or if there's some the lambdas that make this guy have a determinate of 0, then those lambdas are going to satisfy this equation. And you could go the other way. If there's some lambdas that satisfy this, then those lambdas are going to make this matrix have a 0 determinate. Let me write this. Av is equal to lambda v for non-zero v's if and only if the determinate of lambda In minus A is equal to the 0 vector. No, not the 0 vector. Sorry, it's just equal to 0. The determinate is just a scalar factor. And so that's our big takeaway. And I know what you're saying now, how is that useful for me, Sal? You know, we did all of this manipulation. I talked a little bit about the null spaces. And my big takeaway is, is that in order for this to be true for some non-zero vectors v, then lambda has to be some value. So if I take the determinate of lambda times the identity matrix minus A, it has got to be equal to 0. And the reason why this is useful is that you can actually set this equation up for your matrices, and then solve for your lambdas. And we're going to do that in the next video.