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Diagonalize the matrices in Exercises $7-20,$ if possible. The eigenvalues for Exercises $11-16$ are as follows: $(11) \lambda=1,2,3$ (12) $\lambda=2,8 ;(13) \lambda=5,1 ;(14) \lambda=5,4 ;(15) \lambda=3,1 ;(16)$ $\lambda=2,1 .$ For Exercise $18,$ one eigenvalue is $\lambda=5$ and one eigenvector is $(-2,1,2) .$$\left[\begin{array}{ll}{3} & {-1} \\ {1} & {5}\end{array}\right]$

diagonalizable

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 3

Diagonalization

Vectors

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So we have a matrix A that we want to Dad, analyze. Let's get started. So how first step to diagonals matrix is to find the egg in Victor's at the age and values we started again by finding again values? Well, Ford it determinant of a minus them. The times identity equals zero. So we solve this equation for Lambda. We have three minus then, uh, minus 115 minus than the court zero. That gives us a nice in your non linear equation of the form three minus than the times five minus lambda, minus one times mine. This one equals zero. We want to solve this equation. We have 15 minus three Ramdas minus five. Lamb does purse rammed the square. Thus one call zero. We rewrite this equation by simplifying everything's we get them to square minus eight. Them does 1st 16 he court zero Oh, we get than the minus four square. Its course is you. So you if you can see it directly, good. Otherwise you can use the formula to solve. Ah, second degree polynomial. The important thing is that you figure out that render one is the same as them. The to which is four. So we have one alien values with a multiplicity, too. And now we're gonna find So Step two, we're gonna find the Eggen vectors of them. Duh equals for so how to find? Okay, Hagen Victors, we solve this system so a minus four times and then to d. C for being Oregon value, times X equals zero. So zero is a vector for zero and X is it, uh, what we want to solve for? It's a vector. So we have this linear system, these linear equation, and we'll have something that looks like this. So minus one minus one 11 kinds x one x two equal 00 So that's the system for this problem. What does it mean to solve that systems? It means that minus x one minus x two equals zero and x one plus X two equals zero. And that means that x one equal minus X two. Because those to a question are the same, you can simply multiply. Either of those two is equation by minus one and you'll get the other one. So that means that X, which looks probably theirs so x two times not equal X two times minus 11 is our Eggen vector. So any multiple of minus 11 is our again Victor, So minus want one? Is is the Onley Eggen sector? Uh huh. Hey, so that means we can not agonize A So we can't agonize a dia gone No lights, Okay, Because a means a needs and linearly independent Angan vectors to be diagonal Izabal dia Gone lies and it obviously doesn't have And linearly independent again vectors. It only has one. And in our case, fan equals two because a is a two by two square matrix. So this is an example of a two by two matrix that we can not bag on arise because it has Onley one Eggen vector So we can maybe have it little conclusion. Therefore, a isn't died otherwise and there you have it. So we found we find we found are a invent values by looking at the determinant of a minus lambda times identity and refined that we found that there's only one pagan value for a Eggen vector for a Therefore it's not final Diagonal Izabal

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