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Class 22 : Answer 1 (D). Clearly since 3 is the associated eigenvalue for both x and y , Ax=3x and Ay=3y then A(cx+dy)=Acx+Ady=c(Ax)+c(Ay)=c(3x)+d(3x)=3(cx+dy) Thus the answer to 22.1 is D. Class 22 : Answer 2 (D).
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Class 22: Answer 1 (D) Clearly since 3 is the associated eigenvalue for both x and y, Ax=3x and Ay=3y then A(cx+dy)=Acx+Ady=c(Ax)+c(Ay)=c(3x)+d(3x)=3(cx+dy) Thus the answer to 22.1 is D.
Class 22: Answer 2 (D) If w is an eigenvector of A, how does the vector Aw compare geometrically to the vector w? By definition (the eigenequation) Aw=lw means that the vector Aw is actually a constant multiple of w so that Aw is in the same direction but if (w is not 1) it will have a different length Thus the answer to 22.2 is D.
Class 22: Answer 3 (B) By definition one finds eigenvectors by solving the homogeneous equation (A-lambdaI)x=0 which is equal to finding the nullspace of A-lambdaI Thus the answer to 22.3 is B.
Class 22: Answer 4 (E) Which of the following statements is correct? • The set of eigenvectors of a matrix A forms the eigenspace of A. • The set of eigenvectors of a matrix A spans the eigenspace of A. • Since any multiple of an eigenvector is also an eigenvector, the eigenspace always has infinite dimension. • More than one of the above statements are correct. • None of the above statements are correct. There is no such thing as “an eigenspace of A” but there IS an eigenspace of A corresponding to a particular eigenvalue, usually written as Elambda Eigenspaces can never be “infinite dimensional.” Thus all the statements are false. Thus the answer to 22.4 is E.