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Without calculation, find one eigenvalue and two linearly independent eigenvectors of $A=\left[\begin{array}{ccc}{5} & {5} & {5} \\ {5} & {5} & {5} \\ {5} & {5} & {5}\end{array}\right] .$ Justify your answer.

The vectors $(2,2,-4),(2,-2,0)$ are Eigenvectors corresponds to the Eigenvalue 0

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 1

Eigenvectors and Eigenvalues

Vectors

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Okay, so I'll try to keep this video pretty short. So without any calculation, you want to find one Aiken value. And two, I invent this off this matrix here. So three by three matrix. Who's entries are all five. So Okay, the best way to look at it is some The columns. You noticed that the some off the columns are all constant, namely 15 sir. Lambda equals 15. Must be, and I can value. There's a little property off island values that you could use if you find that if if the some off the columns ends up, being constant than that will be an item of value. Now to get the argon vectors, What we can do is there just okay, so it's pretty obvious that you can You can see that 111 is a night, in fact, because if you add the roads so I'm gonna do it a different color If you add the rose up, which really corresponds to multiplication off this matrix by 111 you get 15 15 50 which is really just 15 times 111 So these two things here are the bargain bit value. I convict, period. Okay, Now, the other types of I convicted you get are the ones that sort of map. So we can think you think of this as a lenient map. Then you can think off. What is the colonel off this liniment? In other words, what defectors say X Wasit make this zero. So that gives you two other I can. Vectors. Namely one monos 10 and 10 Modest. 1001 modest one. However you want to write it. Well, we've got more than we bargained for anyway. But these should all be I inventors. I've got at least two Aiken. Victor's without really doing much calculation. A lot of it is by inspection as well as just observing certain properties of the Matrix, So I'm gonna face there.

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