hw5-solutions

Each point in the plane rotates into that is each

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Unformatted text preview: ane rotates into . That is, each point is an eigenvector of corresponding to eigenvalue (you will get the point if you don’t consider this situation) Problem 4. (a) Setting up | (b) Setting up | (c) Setting up | | | | gives us: gives us: gives us: . . . Problem 5. (a) Using Matlab/Octave: eig(A) gives us: -3, 3, 3, 3 Manually: | | | | | ( ) ( ) | Simplifying this formula gives us: ( ) ( ) . (b) For : [ [ [ [ ] No we can use Matlab/Octave to solve this system using rref: ([ ) [ Which means: [ So [ ] [ ] [ ] is the set of base vectors for the eigenspace corresponding to For : [ [ [ [ [ No we can use Matlab/Octave to solve this system using rref: ([ ) [ . Which means: { [ So [ ] is the only base vector for the eigenspace corresponding to . (c) Algebraic multiplicity of 3 is three and algebraic multiplicity of -3 is one. Geometric multiplicity of 3 is three (since the dimension of corresponding eigenspace is three) and geometric multiplicity of eigenvalue -3 is one....
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