Algebra 2 - Quiz 2 sol PDF

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Quiz 2 sol...

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University of Dundee

Mathematics Division

MA22001 Mathematics 2B Algebra Quiz 2 Name: 1. Classify whether the following statement as True or False. If false, justify your answer. Let V = {[x, y, 0]T |x, y ∈ R} and W = {[w, 0, z]T |w, z ∈ R}. Then R3 = V ⊕ W . [3 Marks] Solution: False. [1] V ∩ W 6= {0} since V ∩ W = {[α, 0, 0]T |α ∈ R}. Thus R3 = V + W , but not V ⊕ W . [2]

2. Use the Gram-Schmidt process to obtain an orthogonal basis {v1 , v2 , v3 } for the subspace of R3 spanned by the linearly independent vectors [10 Marks]       2 1 1 u1 =  1  , u3 =  1  . u2 =  0  , −1 1 1 Solution: Let v1 = u1 = [1, 1, 1]T . [2] Next we choose v2 such that span(u1 , u2 ) = span(v1 , v2 ), v2 ⊥ v1 and so the vector v2 is v2



 hu2 , v1 i = u2 − v1 hv1 , v1 i       1 1 2   =  0 − 1 3 1 1   1 1 −2  = 3 1



 1 ′ We replace v2 by v2 =  −2  to simplify our calculation. [3] 1 Next we choose v3 such that

′

span(u1 , u2 , u3 ) = span(v1 , v2 , v3 ) = span(v1 , v2 , v3 ), v3 ⊥ v1 , and v3 ⊥ v2 and so the vector v3 is    hu3 , v2 i hu3 , v1 i v1 − v2 v3 = u3 − hv1 , v1 i hv2 , v2 i         1   2 1 2   −1  =  1 − 1 − −2  3 6 −1 1 1     1 9 3 1 0 =  0  = 2 6 −1 −9   1 ′ ′ ′ We rescale v3 and use v 3 =  0 . [5] Thus, {v1 , v2 , v3 } is an orthogonal basis. −1 

2

3. Let

 1 0 2 A =  3 −1 3  . 2 0 1 

Compute the following: (a) The eigenvalues of A.

[4 Marks]

(b) A basis for eigenspace of a repeated eigenvalue of A.

[6 Marks]

(c) The algebraic and geometric multiplicity of the repeated eigenvalue.

[2 Marks]

Solution: (a) The characteristic equation is 0 = |A − λI|   1−λ 0 2   3 −1 − λ 3 =   2 0 1−λ   1−λ 2 = (−1 − λ)  2 1−λ = (−1 − λ){(1 − λ)2 − 4}

         

= (−1 − λ)((1 − λ) + 2)((1 − λ) − 2) = (−1 − λ)2 (3 − λ) [3] Hence, the eigenvalues are λ1 = λ2 = −1 (repeated) and λ3 = 3 [1]. (b) For λ1 = λ2 = −1, the associated eigenvectors are the solution of the linear system     0 x1    (A − λ1 I) x2 = 0  0 x3 which give the augmented matrix       1 0 1 0 1 0 1 0 2 0 2 0 [A + I|0] =  3 0 3 0  −→  3 0 3 0  −→  0 0 0 0  0 0 0 0 2 0 2 0 2 0 2 0   x1 from which it follows that an eigenvector x =  x2  in an eigenspace E−1 satisfies x3 x1 + x3 = 0 [3]. Therefore, both x2 and x3 are free. Setting x2 = s and x3 = t, we have            −1 0  −1 0  −t   E−1 =  s  = s  0  + t  1  = span   0  ,  1      1 0 1 0 t     0   −1    Since these two vectors are linearly independent, E−1 has a basis 0 , 1  .   0 1 [3] (c) The algebraic multiplicity of λ1 = λ2 = −1 is 2 also 2 [1]. 3

[1] and geometric multiplicity is

4...