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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.
Another solution: (
() () ( )→ () (
( ( ) ) 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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- Fall '13