hw6-solutions

in matlaboctave we get the eigenvalues are b using

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Unformatted text preview: ing and a. Using the same approach as previous part we get is basis for [ eigenspace corresponding to eigenvalue b. [ . is basis for eigenspace corresponding to eigenvalue [ [ . . Problem 3. (a) Another solution: ( () (b) () () ( )→ () ( ( ( ) ) Problem 4. (a) Since A is diagonalizable then: ) () s.t. () is diagonal and ) ( )→ since the inverse of a diagonal matrix is also diagonal then is also diagonal (it’s easy to prove that since exists then inverse of also exists). (b) Since A is diagonalizable then: ( ( )→ ) s.t. ( ) is diagonal and ( ) . Problem 5....
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