Exercises. Linear Algebra for Differential Equations. Exercise 1. Find the eigenvalues and eigenvectors of the following matrix:

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1 Exercises Exercise Find the eigenvalues and eigenvectors of the following matrix: λ det (λ )(λ 5) + 3 λ 2 6λ + 8 (λ 2)(λ 4) 3 λ So the eigenvalues are λ 2 and λ 2 4 Now we find the eigenspaces (that is, the eigenvectors) corresponding to each eigenvalue First, for λ we find (non zero) solutions to: 3 x 3 The solutions are 3x or 3x That is, vectors of the form: t 3t 3 So we may takev 3) Next, for λ 2 we find (non zero) solutions to: x 3 3 The solutions are x or x That is, vectors of the form: t So we may take v 2 ) 22 TAMU Math Learning Center of 9

2 Exercise 2 Find the eigenvalues and eigenvectors of the following matrix: λ det (λ + 2) 2 2 λ So the eigenvalues are λ 3 and λ 2 Now we find the eigenspaces (that is, the eigenvectors) corresponding to each eigenvalue First, for λ we find (non zero) solutions to: x The solutions are x + or x That is, vectors of the form: t t So we may takev ) Next, for λ 2 we find (non zero) solutions to: x The solutions are x or x That is, vectors of the form: t So we may take v 2 ) 22 TAMU Math Learning Center 2 of 9

3 Exercise 3 Find the eigenvalues and eigenvectors of the following matrix: We find the zeros of the characteristic polynomial: 3 λ 2 det (λ + )(λ 3) + 8 λ 2 2λ + 5 (λ ) λ So the eigenvalues are λ + 2i and λ 2 2i To find the eigenvectors, we solve: 3 ( + 2i) 2 x 2 2i 2 4 ( + 2i) 4 2 2i That is: ( i)x or ( i)x so the solutions are vectors of the form: t t t( i) i So then a basis for the eigenspace associated to + 2i is { } So we say v i i Since the matrix has real entries, we know that ( the ) eigenvector associated to λ 2 is the complex conjugate of the eigenvector associated to λ So we take v 2 + i 22 TAMU Math Learning Center 3 of 9

4 Exercise 4 Find the eigenvalues and eigenvectors of the following matrix: det 2 λ 5 (λ 2)(λ + 2) + 5 λ λ So the eigenvalues are λ i and λ 2 i To find v we solve: 2 i 5 x 2 i That is (2 i)x 5x ) 2 If we multiply both sides by 2 + i we get 5x 5(2 + i) or x (2 + i) So then we 2 + i can take v as the eigenvector associated with λ i 2 + i Similar to the previous problem, v 2 22 TAMU Math Learning Center 4 of 9

5 Exercise 5 Find the eigenvalues and eigenvectors of the following matrix: ( ) 4 4 λ det λ(λ 4) + 4 λ 2 4λ + 4 (λ 2) λ So λ 2 is the only eigenvalue and has algebraic multiplicity 2 Now we find the eigenvectors by solving: 2 x 4 2 That is 2x + or 2x So then v is the only linearly independent eigenvector associated to 2) λ 2 Next we find a generalized eigenvector by solving: 2 x or 2x + or + 2 So any vector of the form: t v 2 + t is a generalized eigenvector associated to λ 2 We ll take the one with t so that w ( ) 22 TAMU Math Learning Center 5 of 9

6 Exercise 6 Find the eigenvalues and eigenvectors of the following matrix: First, we find the eigenvalues: det λ 3 2 λ 3 2 λ 3 ( λ) det ( λ)(λ + 2)(λ ) λ λ So the eigenvalues are λ 2 and λ 2,3 That is, is an eigenvalue with algebraic multiplicity 2 First, we find the eigenvectors associated to 2 by row reducing: So x and x 3 are basic variables and is free This says x, x 3 and is free So any non zero vector of the form: t is an eigenvector So we take v Next we find the eigenvectors associated to by row reducing: This is already in echelon form It says x is the basic variable and and x 3 are free So x + x 3 and s and x 3 t So any non zero vector of the form: x s + t s s + t x 3 t So then we can take v 2 and v 3 Note: these vectors are slightly different than the ones found in the video; but both answers are correct Indeed, v 2 + v 3 which is the one found in the video 22 TAMU Math Learning Center 6 of 9

7 Exercise 7 Find the eigenvalues and eigenvectors of the following matrix: det 3 λ λ 6 3 λ λ (3 λ) det 6 det (3 λ)(λ + 3) 2 3 λ 3 λ So the eigenvalues are λ,2 3 and λ 3 3 That is, 3 is an eigenvalue with algebraic multiplicity 2 We first find the eigenvectors associated to the eigenvalue 3: Then s and x 3 t are free This says x x 3 t and s So any non zero vector of the form: x t s s t x 3 t So we take: v, v 2 Now we find the eigenvector associated to 3: So x and x 3 are basic variables and t is free This matrix says that x t and x 3 so any vector non zero vector of the form: x t t t So we take: v 3 22 TAMU Math Learning Center 7 of 9

8 Exercise 8 Find the eigenvalues and eigenvectors of the following matrix: λ det 3 λ λ λ λ λ det + det + det 3 5 λ 5 5 λ 5 3 λ(λ(λ 5) + 6) + 3(λ 5) λ λ(λ 2 5λ + 6) ( 2λ + 4) λ(λ 3)(λ 2) + 2(λ 2) (λ 2)(λ 2 3λ + 2) (λ 2)(λ )(λ 2) So λ and λ 2,3 2 We find the eigenvectors associated to first: /2 / So, x and are basic and x 3 t is free Then any non zero vector of the form: x t/2 /2 t/2 t /2 t So we can take: x 3 v 2 Now for the eigenvalue 2: 2 /2 /2 /2 /2 /2 / /2 / /2 /2 So x 3 t is free So x and x 3 t So any non zero vector of the form: x t t That is, we can take v 2 There is only one linearly independent eigenvector associated to 2 So now we find an eigenvector by solving (M 2I)w v 2 We do this by row reducing: 2 /2 / /2 /2 /2 / /2 /2 22 TAMU Math Learning Center 8 of 9

9 So x and are basic and x 3 are free So x and 2 + x 3 and x 3 t So any non zero vector of the form: x 2 + t 2 + t x 3 t So we can take the generalized eigenvector to be: w 2 22 TAMU Math Learning Center 9 of 9

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