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In Exercises 19 and $20,$ find $(a)$ the largest eigenvalue and $(b)$ the eigenvalue closest to zero. In each case, set $\mathbf{x}_{0}=(1,0,0,0)$ and carry out approximations until the approximating sequence seems accurate to four decimal places. Include the approximate eigenvector.$$A=\left[\begin{array}{cccc}{10} & {7} & {8} & {7} \\ {7} & {5} & {6} & {5} \\ {8} & {6} & {10} & {9} \\ {7} & {5} & {9} & {10}\end{array}\right]$$

an estimate for the eigenvalue to five decimal places is .01015 . The actual eigenvalue is .01015005 to eight decimal places. An estimate for the corresponding eigenvector is $\left[\begin{array}{c}{1} \\ {-.251135} \\ {.148953}\end{array}\right]$

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 8

Iterative Estimates for Eigenvalues

Vectors

Campbell University

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

Lectures

02:56

In mathematics, a vector (…

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In Exercises 19 and $20,$ …

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In Exercises $1-4,$ the ma…

02:20

Let $A$ be as in Exercise …

06:02

Diagonalize the matrices i…

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00:59

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Okay, So in this question, we want to basically approximate that again vectors. And I'm also Ivan values questions era and life spied in about. So in this example right here, I've only given the I've been very close to zero closest to zero. If you want the Ivan value life value, you can just simply said athletic 20. So just change this line of code right here, and you will gets the lives talking about, which should be 19.18 So what I've done here is basically I've cut it up the invest power method to solve this to to find the ivy value. So in this case, just to go through it in case you don't understand what I did or how it works. So we start off without initial X and we we set an Alfa and perform this step. So first of that we do is we have to solve a minus Alfa identity. Why is equal to X? So we want to sell for Why so moving a minus? I after invest, I got to decide you have Why Easy to a minus off a after I Inglis X. So what? The sepia is simply this stuff. You're eight minus alpha times. Why? Taken in vessel exit? Inglis and we have multiplied by X. Then we want to find about the entreating. Why? Which absolute value. The absolute value is as light as possible. So then we compute new alcohol, plus one of them you'll and then we compute x, which is why divided five new. So that's what this step, that's what this whole thing does here. So the printed out results is I'm here as a concerned then the wagon value close to zero is 0.122

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