Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. A number Lambda is said to be an eigenvalue of the north by north matrix A, provided there exists a non 0 vector V such that the matrix times V equal Lambda times VV is called an eigenvector of A. The characteristic equation is the absolute value of a minus Lambda times I equaling 0. What we're actually going to do, because it's a matrix, is we're going to find its determinant and set it equal to 0. So the number number Lambda is an eigenvalue of the north by north matrix A if and only if Lambda satisfies the characteristic equation. Now I went out on Wikipedia and I found kind of a cool picture that shows in the shear mapping, the red arrow changes direction, but the blue arrow does not change direction. So the blue arrow was called an eigenvector of the shear mapping and since its length is unchanged, its eigenvalue is 1. So we're looking for vectors that are going to make the original unchanged for a value. So starting with a matrices five -6 three -4, we're going to do 5 minus Lambda -6 three and -4 minus Lambda. When we find our determinant, remember we just take the major diagonal minus the minor diagonal, so our 5 minus Lambda times -4 minus Lambda -3 * -6, and that all needs to equal 0. So we get Lambda squared minus Lambda -2, and when we factor it, we can see Lambda equal 2 and -1. So now we're actually going to put that back in for our Lambda. So we're going to have two different cases. If Lambda equal 2, we get a matrix three -6, three -6. We want to figure out what the matrix AB would be that when we multiply the two together, we get 00. So we can see three a -, 6 B equals 0 and that occurs twice. And for me, the easiest way that I can tell you to do it is I actually divide by any common multiple so I can get a reduced equation. In this case, a -, 2 B equals 0. So now I can see if A equal 2B would equal 1 and that would be a true statement because 2 -, 2 would equal 0. Now this is true for any multiple of the vector 21, so it could be 4. Two it could be -2 negative one. We usually use the most simplistic and we usually use the positive. SO21 would be considered the most simplistic vector here. But remember, we can do any scalar times that V and still have it be true. So when Lambda equal -1 we're doing the same thing. We get six -6 three -3 we're going to multiply by AB and that's got equal 00. So six a - 6 B equals 0 three a - 3 B equals zero. I personally simplify it up as much as I can. So a - b = 0. So I can see that when A is 1B is 1, any scalar that also works, but our answer is 11. Let's look at a three by three. So if we have 100 negative 68212 negative 15 negative three, the first thing is we subtract Lambda along that diagonal because it's the identity matrix and the identity matrix says that it's only along the diagonal that we have our ones. So we're subtracting I times Lambda. So we get one minus Lambda, 00 negative 6-8 minus Lambda, 212 negative 15 negative 3 minus Lambda. We're going to do our determinant here. So our major minus our minors, now our two of our majors are actually going to be 0. So the only real major we're looking at is the one minus Lambda, 8 minus Lambda -3 minus Lambda. Our minors, two of our minors are going to be 0. So all we're really looking at is the -15 * 2 * 1 minus Lambda. Now, it's going to be always easiest if we can see a factor that can factor out of both terms. In this case, we have 1 minus Lambda and our 2 terms. So we're going to pull it out. That's going to leave us 8 minus Lambda times -3 minus Lambda minus the -15 * 2. When we simplify that up and factor it, we can see that Lambda is going to equal 1 and two and three. So when Lambda equaling one, we get 000 negative 67212, negative 15 negative 4. We want to find the vector ABC or the matrix ABC such that when they multiply we get 000. Well, the first equation everything cancels and we get 0 = 0, the second negative 6A plus 7B plus 2C equals 0 and the third twelve A - 15 B -4 C equals 0. So we're going to combine those two and figure out what AB and C relationship exist. If I take the first one and multiply it by two and add it with the last one, I can see that the A's and the C's are going to cancel and we get a negative B = 0. Well, if b = 0, now we can go back and we can see that first equation is -6 A plus 2C equals 0. If we look at the second equation, putting AB in zero twelve, A -, 4 C equals 0 is actually just a multiple of that -6 A plus 2C equals 0. So I'm going to divide by two because I wanted as simplistic as I can. So when a is one, we can see that C would have to be 3 and we found that B was 0. So our vector is going to be 103 when Lambda equal 2. If we do the same thing, we can see that we need to combine the second and third equation because the first equations giving us A is 0. When we put in A as zero for the second third equation, we have 6B plus 2C equals 0 and -15 B -5 C equals 0. Those are really just multiples of each other, so if I thought about dividing everything through by A2, I'd get 3B plus C = 0. I can see if B is 1C has to be -3 so 01 negative 3. Now remember I can multiply that by any scalar, so zero -1 positive 3 is also a perfectly acceptable answer. We do want to try to use the smallest numbers possible. When lambda's three, we get negative 200. We do the same steps and we can see that when A is 0, we would have our B having to be two and our C would then be -5. We can multiply by a scalar, so we could have had zero -2 positive 5 on a 4x4 matrix. 1040014000300003. We're going to subtract the Lambda from the main diagonal, and then we're going to realize that in a form by 4, we have to actually get it down to a three by three to evaluate the determinant. So when I look at this, I'm going to choose this first column because I have lots of zeros in this first column. So I'm going to take that one minus Lambda. So it's in the 1st row, first column. So I have -1 to the 1st row plus first column times the entry in that first row, first column one minus Lambda, and then I ignore the row in the column to multiply it by the three by three determinant that's left. So 1 minus Lambda, 4003 minus Lambda, 0003 minus Lambda. We can see that because we have zeros in the bottom portion of this matricy that the only real diagonal that's going to matter is the main diagonal, because all of the other major diagonals are going to be 0, and all of the minor diagonals will also be 0. So we get one minus Lambda times the quantity 1 minus Lambda, 3 minus Lambda, 3 minus Lambda, or 1 minus Lambda squared 3 minus Lambda squared equaling 0. So when we stick in Lambda equaling one, we're going to find ABC and D This time we can see that 4C equals zero and two C = 0 and two d = 0. So what that really means is we know C&D are always 0, but we don't know what A&B are. So A&B can basically be anything as long as they're anything independently. So our two different answers are going to be A is one when all the other three are 0, or our B is one when all the others are zero. And once again, we have the scalar. So that's really saying A can be any value, A could be 5A could be 20, A could be -1,000,000, or B could be any value, and we would still get valid statements. When we look at Lambda equaling 3, we're doing the same procedure as before. We can see that -2 A plus 4C equals 0, and -2 B plus 4C equals zero. I once again divide and get into everything as simple as I can. So I'm going to take out a 2 from all of that, and I get negative a + 2, C equals 0, and negative b + 2 C equals 0. So if A was 2C would be 1. Well if C is 1 by the second equation, then B has to be two. So I can see that we'd have 221 and D we didn't have any information on. So D would have to be 0 for that matrix. Or when AB and C are 0D could be one. And then we also have any scalar of that. So that second vector is really saying D can be any value when AB and C are all 0. The last type is looking for complex conjugate eigenvalues. So we start with the given two by two matrix and we can see that our diagonal is going to turn into negative Lambda. So we have negative Lambda -436 negative Lambda. Doing the determinant we get Lambda squared plus 144 equaling 0. So Lambda is going to equal positive or -12 I. So when I have Lambda equaling -12 I I get the matrix -12 I -436 twelve I. When lambda's -12 I I get the matrix 12 I -436 twelve I times AB equaling 00. So 12 IA -4 B equals 0 and 36 A+ 12 IB equals 0. Once again, I just always try to divide out to get it more simplistic so I can see three IA minus b = 0. Remember I ^2 is -1, so an answer to this one would be negative I3. So three I * -I is really going to be -3 I squared, which turns into +3. Positive 3 -, 3 is 0. When Lambda equal 12 I, we'd get -12 I -436 negative 12 I times AB equaling 00 so -12 AI -4 B equals zero 36 A -12 IB equals 0. Dividing through I can get three IA plus b = 0. So a is I when B is 3 and the second equations are really just multiples of the 1st. Thank you and have a wonderful day.