312S13FEx-solns

# Similarly b 0 and c 0 e if the matrix a is both

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Similarly b = 0 and c = 0. e) If the matrix A is both invertible and diagonalizable, then A 1 is diagonalizable. Solution: TRUE. If A diagonalizable, then A = SDS 1 , where D is a diagonal matrix. Now take the inverse of both sides. A–4. Consider the matrix A = bracketleftbigg 2 1 2 2 bracketrightbigg . If vectorx R 2 is a unit vector, what is the largest that bardbl Avectorx bardbl could possibly be? Solution: The key fact is that if M is a self-adjoint n × n with eigenvalues λ 1 λ 2 ≤ · · · ≤ λ n (necessarily real) then for any vectorx we have λ 1 bardbl vectorx bardbl 2 ≤ ( vectorx, Mvectorx ) ≤ λ n bardbl vectorx bardbl 2 . Since bardbl Avectorx bardbl 2 = ( Avectorx, Avectorx ) = ( vectorx, A Avectorx ) , the matrix M := A A is self adjoint and positive semi-definite, the above expression is largest if vectorx is an eigenvector of A A corresponding to its largest eigenvalue, λ max . Thus bardbl Avectorx bardbl 2 λ max bardbl vectorx bardbl 2 . Because the problem specifies that vectorx is a unit vector, then bardbl Avectorx bardbl 2 λ max . We now compute λ max . The computation is routine. Note that we use the matrix M = A A , not A . A A = parenleftbigg 2 2 2 2 parenrightbigg parenleftbigg 2 2 2 2 parenrightbigg = parenleftbigg 8 2 2 5 parenrightbigg . Then the characteristic polynomial is det( A A λI ) = λ 2 13 λ + 36 = ( λ 9)( λ 4) . Consequently λ max = 9 so bardbl Avectorx bardbl ≤ 3. 2

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A–5. Let A be an m × n matrix, and suppose vectorv and vectorw are orthogonal eigenvectors of A T A . Show that Avectorv and Avectorw are orthogonal. Solution: Say A T Avectorw = λvectorw . Then ( Avectorv, Avectorw ) = ( vectorv, A T Avectorw ) = ( vectorv, λvectorw ) = λ ( vectorv, vectorw ) . But vectorv and vectorw are given to be orthogonal so the result follows. Note that in this we never needed to use that vectorv is also an eigenvector of A T A Part B Six questions, 15 points each (so 90 points total). B–1. Find an orthogonal matrix R that diagonalizes A := 1 1 0 1 1 0 0 0 2 Solution: The characteristic polynomial is det( A λI ) = (2 λ ) det parenleftbigg 1 λ 1 1 1 λ parenrightbigg = (2 λ )[(1 λ ) 2 1] = λ ( λ 2) 2 . Thus the eigenvalues are λ 1 = 0, λ 2 = λ 3 = 2 For λ 1 we find vectorv 1 = 1 1 0 For λ 2 = λ 3 = 2 we want the kernel of the matrix A 2 I = 1 1 0 = 1 1 0 0 0 0 . It is easy to see that this kernel consists of all vectors of the form a a c . One simple orthogonal basis is vectorv 2 = 1 1 0 and vectorv 3 = 0 0 1 . The orthogonal matrix R that diagonalizes A has orthonormal eigenvectors as its columns. The orthogonal eigenvectors vectorv 1 , vectorv 2 , and vectorv 3 do the job simply by making them into unit vectors. Thus R = 1 / 2 1 / 2 0 1 / 2 1 / 2 0 0 0 1 B–2. Let V be the vector space spanned by the two functions e x and e x , considered only on the interval [ 1 , 1]. Give V the L 2 inner product: ( f, g ) = integraldisplay 1 1 f ( x ) g ( x ) dx.
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