quiz3_sol

quiz3_sol - ) = 2. (d) (5 points) Find an orthonormal basis...

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Math 33A, Section 3, Fall 2009 Quiz #3 – Solutions A = ± - 1 2 0 0 1 0 2 - 1 (a) (5 points) Find image( A ) and write it as a linear span of as few vectors as possible, i.e. without any redundant vectors. Solution. image( A ) = span { ± - 1 1 , ± 2 0 , ± 0 2 , ± 0 - 1 } = span { ± - 1 1 , ± 2 0 } (b) (5 points) Find kernel( A ) and write it as a linear span of as few vectors as possible. Solution. We have to solve the linear system: ± - 1 2 0 0 | 0 1 0 2 - 1 | 0 . Using Gauss-Jordan elimination we get the following RREF: ± 1 0 2 - 1 | 0 0 1 1 - 1 2 | 0 . By taking x 4 = t , x 3 = s , x 2 = - s + 1 2 t , x 1 = - 2 s + t , we obtain: kernel( A ) = span { - 2 - 1 1 0 , 1 1 2 0 1 } (c) (5 points) Find rank( A ) and nullity( A ). Solution. From parts (a) and (b) we have: rank( A ) = 2, nullity( A
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Unformatted text preview: ) = 2. (d) (5 points) Find an orthonormal basis for kernel( A ). Solution. We apply the Gram-Schmidt process to vectors: ~v 1 = 1 1 / 2 1 , ~v 2 = -2-1 1 1 ~u 1 = ~v 1 k ~v 1 k = 2 / 3 1 / 3 2 / 3 ~v || 2 = ( ~v 2 ~u 1 ) ~u 1 = -10 / 9-5 / 9-10 / 9 ~v 2 = ~v 2-~v || 2 = -8 / 9-4 / 9 1 10 / 9 ~u 2 = ~v 2 k ~v 2 k = 1 261 -8-4 9 10 2...
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quiz3_sol - ) = 2. (d) (5 points) Find an orthonormal basis...

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