Question

parts a, b, c and d

(a) T(31,72) = (2x - 20, -2.61 +5r) on V =R? (b) T(31,12,13) = (-11 + 12,562, 46, -212 +503) on V =R3. (c) T(21, 12) = (2x1 + ix2, 21 +222) on V = C. d] on V = M2x2(R) (with the Frobenius Inner Product). с (d) T ] 3. For each linear operator T in Exercise 1 find an ON basis 3 for V consisting of eigenvectors of T (if possible).

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Answer 1

By defo there exists a basis of eigenvectors of T off I s (1) T1,0) = (2,-5) = 24-2ez, T(ez)= T(0,1) = (-2,5) = -24 + so [] = 1.2 diagonalisable. 5 e -2 cs L-2 X-2 2 2 x-5 2 2 26 [0] 2 -4 ( 2) - are Characteristic polynomeal of [17: = (x-2)(2-5)-4 x² - 7x+6. Roots of Roots of this polynomial nomial 2= 7+ √49-24 715 Ź 1,6 régenvectors: ?,=1, 2,16 = -x+2y = 0 > x=24 eigenvector is 9,76; (1 31637 [:) » Ax+y=0 y = -23 so eigenverfor is (2) 0,= ( 2 ) = (2) othogonal 11 0,11 = 5 = || , || 43 A Us = 1 / ( ) is an ON basis for R² Ü) TIC,)= TU..0,0)=(-1,0,4); TE) (1, 5,-2) Tle) = (0,0,5) [1] = Again have to calculate envalues eigenvectors d check if T is diagonalisa lisable. Voing online software SO 2 VE -1 I so o 05 ec 4 -2 5 eigenvalues

) (d)t ( [i :)) - [ ] T () = [o il - en Tle) = ei , TCG)= po 1- 1 so [T] - Characteristic O polymerocal 0 L 0 1 O 0 - 1 0 Ο Υ -1 7 x(ce(2²) *1 11-21-x) -1 (1)) of 17 stk = 24 2x²+1 = (x2) 2-2x²+1 (e²_1² 1=1,-l n=1 Fingerwolves 9 N2=-) eigenvettors 9 o clearly these four vectors weators are are orthogonal = et lg = lztly, Ig=l, tez Oy = - et la Their norma & Tz (Frobenious norm). ONB is 1/2 o , /2,ol to set So o / J2 - Ya O 6762 0752 1