1. Consider transposing a 2 × 2 matrix (we\'ve mentioned that transposing is lin
ID: 3117033 • Letter: 1
Question
1. Consider transposing a 2 × 2 matrix (we've mentioned that transposing is linear : L : M2 2 M2 2 given by.(:/)_:/T (a) Find the eigenvalues of L. State the multiplicity of each eigenvalue. Hint: Using the standard basis B of M2,2 for both the domain and codomain, first rewrite the linear transformation in the form [L(F)B = AF B. The matrix A should be 4 × 4, From here, find the eigenvalues of A. (b) For each eigenvalue you obtained in (a), find a basis for the corresponding eigenspace. State the dimension of each eigenspace Hint: First find the eigenvectors of A, which will be column vectors in R4 since these eigenvectors are in the form B. Rewrite in the form t to return to 2 x 2 matrices. For each eigenvalue, the basis of the eigenspace must be a set of 2 × 2 matrices.Explanation / Answer
1.Let {E11,E12,E21,E22} denoter the standard basis of M2,2 where Eij has 1 as the ijth entry, the other entries being 0. Then L(E11) = (E11)T = E11, L(E12) = (E12)T = E21, L(E21) = (E21)T = E12 and L(E22) = (E22)T = E22 so that the standard matrix of L is A =
1
0
0
0
0
0
1
0
0
1
0
0
0
0
0
1
It may be observed that the entries in the columns of A are the coefficients of E11,E12,E21,E22 in L(E11), L(E12), L(E21) and L(E22). Now, the eigenvalues of L are same as the eigenvalues of A. The characteristic equation of A is det(A-I4)= 0 or, 4-23+2-1 = 0 or,(-1)3(+1)= 0. Thus the eigenvalues of L are 1 (of algebraic multiplicity 3)and -1 (of algebraic multiplicity 1).
(b).The eigenvectors of A associated with the eigenvalue 1 are the solutions to the equation (A-I4)X = 0. Further, A-I4 =
0
0
0
0
0
-1
1
0
0
1
-1
0
0
0
0
0
Thus, if X = (x,y,z,w)T, then this equation is equivalent to y-z = 0 or, y =z. Then X = (x,z,z,w)T= x(1,0,0,0)T +z(0,1,1,0)T +w(0,0,0,1)T. Thus, the eigenbasis of A associated with the eigenvalue 1 is {(1,0,0,0)T , (0,1,1,0)T, (0,0,0,1)T}. Similarly, the eigenbasis of A associated with the eigenvalue -1 is {(0,-1,1,0)T}. Then the eigenbasis of L associated with the eigenvalue 1 is {E11,P,E22} where P =
0
1
1
0
and the eigenbasis of A associated with the eigenvalue -1 is {Q}, where Q =
0
-1
1
0
1
0
0
0
0
0
1
0
0
1
0
0
0
0
0
1
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