Question

4. (S.20). Find eigen values and eigen spaces of the matrix A = [ 2 3 1 0 -1 2 0 0 3 ].

          4. (S.20). Find eigen values and eigen spaces of the matrix

A = [
2 3 1
0 -1 2
0 0 3
].
        
4. (S.20). Find eigen values and eigen spaces of the matrix

A = [
2 3 1
0 -1 2
0 0 3
].

Added by Seda K.

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Elementary Differential Equations and Boundary Value Problem
Elementary Differential Equations and Boundary Value Problem
William E. Boyce, Richard C. DiPrima 7th Edition
Chapter 7
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Transcript

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00:01 We are going to find the eigenvalues of the corresponding eigenspaces of the matrix a equal to 3 1 0 negative 1 2 and 0 0 3 so the first thing to note here is that this matrix is upper triangular and when a matrix is upper triangular also when it is lower triangular or diagonal we know the eigenvalues of that matrix are the elements on its main diagonal so this implies that the eigenvalues of a are the entries on the main diagonal of a so the eigenvalues of this matrix a are 2 negative 1 3 or taking them in an ascending order lambda 1 equal negative 1 lambda 2 equal 2 and 3 equal 3 which are the elements on the main diagonal of this 3 by 3 matrix that's the same thing for lower triangular matrix or a diagonal matrix good so for each of those eigenvalues we are going to find the eigenspaces so we have already here the eigenvalues now we're going to talk about eigenspaces and of course this means working with each eigenvalue separately so for lambda 1 equal negative 1 first so we got to solve the equation ax equal lambda 1 x for x different from 0 that's just the definition of eigenvalue that is we must find a non -zero vector x such that this non -zero vector satisfies the equation ax plus ax equal lambda 1 x and then this is matrix a is 2 3 1 0 negative 1 2 and 0 0 3 times so vector x is unknown let's say x1 x2 x3 are its coordinates equal negative 1 x that is negative x or in other words negative x1 negative x2 negative x3 and so we write this system of equations here we get 2x1 plus 3x2 plus x3 equal negative x1 then negative x2 plus 2x3 equal negative x2 and the last equation 3x3 equal negative x3 and so we simplify these equations first one for x simplify for x1 the second one for x2 and the third for x3 so we get 3x1 plus 3x2 plus x3 equal 0 and then 2x3 equal 0 because negative x1 cancel out both sides of the equation and 3x3 plus x3 is 4x3 equal 0 now from the last two equations 2x3 equal 0 and 4x3 equal 0 we get x3 equal 0 so the third component of any eigenvector associated to eigenvalue lambda 1 equal negative 1 gotta be 0 let's see what happened with the other two components so now from the first equation 3x1 plus 3x2 plus x3 equal 0 and knowing now that x3 is 0 we get 3x1 plus 3x2 equal 0 which means x1 plus x3 equal 0 and that means that x1 is negative x2 okay that's another equation so the third component gotta be 0 and the first and second components gotta be one the opposite of the other in sign so let's say that the general form or let's say already that the eigenspace for lambda 1 equal negative 1 is okay let me arrange this bit for lambda 1 better we know already this is negative 1 here so this eigenspace is a set of vectors of the form we put x2 any value the second component let's call it alpha then x1 gotta be negative x2 that is negative alpha and the third one gotta be 0 and that for alpha any real number different from 0 because if we put alpha equals 0 we get 0 vector which is not an eigenvector so alpha gotta be different from 0 and this is the form of any eigenvector associated to eigenvalue lambda 1 equal negative 1 or in other words that's the description of the eigenspace for lambda 1 good so do the same for i think i didn't use any index to talk about lambda 1 for example here let's see no index only the phrase so for lambda 2 equal 2 so again the equation is ax equal lambda 2x for x different from 0 and that leads to the linear system i can write the linear system directly because we know matrix system here we only change the right hand side which will be 2 times x or 2x1 2x2 2x3 so i'm going to write directly the linear system coming from that that is 2x1 plus 3x2 plus x3 equal 2x1 then negative x2 plus 2x3 equal 2x2 and last equation is 3x3 equal 2x3 so from the last equation let's say from 3x3 equal 2x3 by subtracting both sides 2x3 we get x3 equals 0 directly so again the third component of any eigenvector associated to…
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