i. The operator T(x,y) can be written in the form: (−yx)=(01−10)(xy). (01−10) is the matrix of the operator in the standard ordered basis. The matrix can be easily found by considering equality: (−yx)=(a11a21a12a22)(xy). From the latter we get: −y=a11x+a12y and x=a21x+a22y. We immediately receive: a11=0,a12=−1 and a21=1,a22=0.
ii. In order to find the matrix in the ordered basis B={α1,α2}, we consider the vector (xy) in the basis B. We receive: (xy)=(12)t1+(1−1)t2. We receive: x=t1+t2, y=2t1−t2. Thus, t1=3x+y, t2=32x−y. In the same way we consider the vector (−yx)=(12)s1+(1−1)s2. We receive: s1=3−y+x, s2=3−2y−x. The aim is to find the matrix satisfying: (3−y+x3−2y−x)=(b11b21b12b22)(3x+y32x−y). We receive: 3−y+x=b113x+y+b1232x−y, 3−2y−x=b213x+y+b2232x−y. We set y=−x in the first equation and receive: b12=32. Setting y=2x yields: b11=−31. We set y=−x in the second equation and get: b22=31. Setting y=2x provides: b21=−35. Thus, matrix T in the ordered bases B has the form: (−31−353231).
iii. Matrix T−cI has the form: T−cI=(−c1−1−c). det(T−cI)=c2+1>0. Thus, the matrix is invertible for all real c. It means, that the operator is invertible.
Answer: i. (01−10); ii. (−31−353231); iii. det(T−cI)=c2+1>0. Therefore, the operator is invertible.
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