 This is the second of two videos that looks at eigenvalues and eigenvectors. In the first video we have seen how to find eigenvalues and we write these as lambda. For each lambda, how do we find the eigenvector, an eigenvector that goes with it? We know that our fundamental equation that we're working with here is that when matrix M multiplies an eigenvector v, it just gives us back that v scaled by lambda. And another way to write that is that M minus lambda times the identity multiplied by v is equal to vector zero. This is the same equation written two different ways. What we need to know now that we have obtained our lambda values, we just need to look at one of these equations and figure out an acceptable vector. I find that it's more useful to use the form on the right hand side. Okay, let's look at a particular example. We'll have the matrix 2, 4, 1, minus 1. We looked at this before and we found already that its eigenvalues are equal to 3 and minus 2. What we're going to do now is we're going to take those values one at a time and figure out an acceptable eigenvector. We're going to write our vector that we need to find as just x and y where we need to find these x, y values. Now take a look at this green underlined equation and in particular the matrix which is a difference of two different matrices M and lambda times the identity. Now that we have our lambda value of 3, we could write out that difference, that difference matrix. It's going to be 2 minus 3 and then just 4 and then just 1 and minus 1 minus 3. There it is. We're saying that when that multiplies our vector x, y it gives us 0, 0. So let's go ahead and clean this equation up. We have minus 1, 4, 1 minus 4 onto x and y. If we want to be explicit about that we can multiply out. It means minus x plus 4 y and x minus 4 y and that we know is equal to 0, 0. Now what we immediately notice here is that whilst this this equation between two columns, two column vectors, is telling us two things, it's actually telling us the same equation twice. So we can see here that we're saying minus x plus 4 y is equal to 0. We're also saying that x minus 4 y is equal to 0. That's telling us the same thing. Is that a problem? No. That's exactly what we want to see at this stage. We should find that when we work on eigenvalue and eigenvector problems based on a two by two matrix then really only one of these rows in the final expression constrains us and the other one doesn't add any new constraint. So this is exactly what we want. So now how do we go ahead and solve it? We're saying that minus x plus 4 y is equal to 0. Of course we can just rearrange this to say instead that 4 y is equal to x and that's the only constraint we have. What we're allowed to do is choose, we can choose, the simplest values of x and that will make this work. So I'm going to choose y is equal to 1 and then I'll find that x is equal to 4 and that is a perfectly acceptable eigenvector for 1 to go with my eigenvalue. We will always have this freedom in choosing the elements of our eigenvector. Really this freedom simply corresponds to choosing how long the eigenvector is. In other words it's magnitude because if a particular eigenvector satisfies our equations a scaled version of that same eigenvector will still satisfy with the same eigenvalue. Now while the eigenvector can have any length we might specifically have been asked for a normalized eigenvector. That simply means we need to take the one that we found and scale it to have unit length. So in this case since it's 4 1 we need to divide by root 17 to scale to unit length. Simple as that. So there we are that's our eigenvector and a normalized version of it. Now we still haven't found the eigenvector for the other eigenvalue which was minus 2. Let me just move this up on the screen to make space to do that at the bottom. So here we go we do exactly the same procedure. We subtract minus 2 on the diagonal 2 minus minus 2 and 4 and 1 minus 1 minus minus 2. Lots of minuses there. So let's tidy that up. That's going to be 4 4 1 x y is equal to 0 0. As before we see that really this is the same equation twice. There's only one constraint and we can read it off simply as x is equal to minus y. So if I choose x is equal to 1 for example then I'm going to write down an eigenvector 1 minus 1 or if I'd chosen y is equal to 1 then it would have been minus 1 1. It doesn't matter they're both correct eigenvectors to go with our eigenvalue but if we want to normalize we'll need to divide by the magnitude 1 over root 2. Okay so there are acceptable eigenvectors to go with the eigenvalue minus 2. Okay so now let's find the eigenvectors that go with the eigenvalues for our 3 by 3 matrix M which was minus 2 1 3 1 minus 1 0 minus 1 1 2. We looked at that before in the previous video and we found the eigenvalues which were minus 1 root 2 subscripts on our lambdas here so we know which one we're dealing with. Let's deal with lambda 1 first which is the one that has value minus 1. So I'll write over here the little equation that we're using over and over again which is that M minus lambda times the identity multiplied by our vector is 0. Okay we need this difference matrix so we subtract off the diagonal 1 minus minus 1 and then 1 3 1 and minus 1 minus minus 1 and 0 minus 1 1 and 2 minus minus 1 and that's on x y and z because we now need an eigenvector with three elements and it's going to be equal to uh we simplify the matrix to minus 1 1 3 1 0 0 minus 1 that again is on our x y z eigenvector is equal to 0 0 0. Now what we immediately notice is that as before we don't really have three different equations captured by our matrix equation we only have two. In fact this is very obvious in this case because the bottom row is the same as the top row. That's not always the case it's not always the case that the rows are actually identical but we will always find if we check that there are only really two independent equations when we're dealing with three by three eigenvalue problems we only have two equations really. Now I'm going to highlight this row here 1 0 0 that's just saying in fact that x is equal to 0 now if we take either the top row or the bottom run we have minus x plus y plus 3 z is equal to 0 or y is equal to minus 3 z. Okay so now we simply choose any values of y and z x has been dictated to us but any values of i y and z that satisfy these rows. So if i choose z is equal to 1 that's going to give me y is equal to minus 3 and I can straight away then write down a satisfactory eigenvector it will be 0 minus 3 1 as simple as that. It doesn't matter where the minus sign is I could equivalently have chosen z is equal to minus 1 and then I'd have 0 3 minus 1. If I normalize then I'll need 1 over root 10 that being 3 squared plus 1 squared and so that is a complete solution for our first eigenvector. We found it in simple form and in normalized form. This is the eigenvector that goes with eigenvalue minus 1. We can go ahead however and check this eigenvector to make sure that it works. So for that we'll simply need to write out our matrix m the original matrix which was minus 2 1 3 1 minus 1 0 minus 1 1 2. We have our vector 0 3 minus 1 we just need to do this sum. So the first element is going to be a minus 2 times 0 and then as a 3 and I see there's a minus 3 so that does give us 0 and our second element is the only non-zero element will be minus 3 and our third element there gives us 1 and we can write that as simply minus 1 onto 0 3 minus 1 and so indeed we found that this vector works with the eigenvalue of minus 1. Now we can continue to look at to find the other eigenvectors but first let's take a pause and review the steps involved. So we're looking at rules for solving eigenvector problems. Eigenvector problem is where we have a square matrix m and we say that m multiplied by some special eigenvector gives us back that eigenvector times just by a value the eigenvalue. We find the possible eigenvalues using this equation involving a determinant of a difference of two matrices. In general there are going to be n solutions for an n by n matrix so two solutions for a two by two three solutions for a three by three matrix that's because when we write the determinant it will have lambda to the power of n as its highest order so for example we have cubed to deal with when we're working out for three by three matrices. Now having found those eigenvalues we then for each value need to figure out an acceptable eigenvector. What we've noticed is that generally we only have to use n minus one of the rows in the equation that we're working to satisfy and that meant just one row in the case of two by two problems and two of the rows in the three by three problems. We had some freedom as to what values to choose for our eigenvector and in fact that freedom corresponded to just scaling the entire eigenvector to a greater or smaller magnitude and if we were asked to normalize we would simply work it out using whatever values we like the simplest values and scale it at the last step so that it has unit length. Okay so we've covered a lot of ground for one video and this would be a good place to just stop watching if you like but I would like to carry on and solve the remaining two eigenvectors for our three by three example. Because they involve a square root two they're actually a bit more messy and tricky to do and in a way I think that makes for a good interesting example to see. So let me go ahead and cut back to the screen that we had before with our matrix m spelled out and our possible eigenvalues and we'll now take the value lambda subscript two which is square root two. So then as usual we need to subtract that down the diagonal so we'll have minus two minus square root two one three one minus one minus square root two zero minus one one two minus square root two and that is the thing which when multiplied by our unknown eigenvector xyz should give us zero zero zero. Now one thing we notice here is the rows look all different. It looks like we've got three different equations captured in this matrix equation but they are not. If we examined them carefully enough we'd find that we could generate one of these rows from the other two and in fact we're only therefore going to need to use two of them. You could pause the video and play with it and see if you can show this but it must always be the case unless we've made a slip earlier. Okay so I see that the middle row has a zero so I'm going to start with that one. It says x plus minus two minus root two times y is equal to zero. That means that if I choose a simple value for y of one then I can immediately say that x moving across is going to be one plus root two. Good so now I'll use the top line which is minus two minus root two x plus y plus three z is equal to zero and I'll substitute in the values that I've already picked and inferred. So I'm going to get one plus root two onto minus two minus root two that's the x term plus the y is one plus three z yet to be found is equal to zero. Rearrange so put z on one side divided by third. Expand this thing out minus two minus root two minus two root two minus two plus one. All right oh and there's a minus sign because we've moved it all to the other side from the z of course. Now we need to tide this up but what I notice is that inside the brackets I have a minus three and a minus three root two and that will cancel cancel with a factor of a minus and third of front and just give us a very simple expression of one plus root two. So that's our z term. Okay we've found a compatible set of x y and z values so we can now write down an acceptable eigenvector one plus root two one one plus root two there we are. That is an acceptable eigenvector and here's where we found those numbers that goes with the eigenvalue lambda two is equal to square root two. Note that I use the same subscript two on my vector so that I make it clear that lambda subscript two goes along with vector subscript two. So now our only remaining task is to look at the third eigenvalue which was negative root two and find a compatible eigenvector for that one. So as always what we need to do is take the vector m and subtract that the lambda value we found off down the diagonal and because we're subtracting minus a minus number we can just add it instead of course so that will be minus two plus root two and then one and then three and then one and minus one plus root two and zero and minus one and one and two plus root two and that matrix when multiplied by our unknown eigenvector x y z will give us zero zero zero. Now as before our middle row looks nicest here it's just telling us that x plus root two minus one put it that way around y times y is equal to zero that means if I chose y is equal to one obvious choice then x is equal to one minus root two watching for signs. Now if I take the let's say the bottom row I can have minus x plus y plus two plus two plus root two times z is equal to zero but I can substitute in the values I found so that will say that square root two minus one plus one plus two plus root two z is equal to zero. Okay I've got some work to do to find out the value of z here I'll start by rearranging just to put two plus root two z is equal to minus root two on the other side but I still need to do a bit more work divide both sides I notice I can simplify simplify by a factor of root two I can write this as z is minus one over root two plus one pause the video and check you agree with me and then I'm not happy with that because I don't want to leave z as a fraction I could do but that would be a very messy looking eigenvector I noticed there's a trick in up I have up my sleeve I know that if I multiply the top and bottom over fraction like that by root two minus one it will simplify I will then find that the top of course is one minus root two but the bottom will be two plus root two minus root two minus one and that whole expression just comes down to one finally then z is equal to one minus root two we've now found our x y and z values that are acceptable so we're seeing saying that vector three that goes with the lambda three value is one minus root two one one minus root two that is an acceptable eigenvector so we're done for our three by three matrix m we found that three eigenvalues and for each of them an eigenvector the last two of these which involve the root two were more tricky just because there was more to keep track of more messy expressions but the basic maths is the same every time