Given that T:R3→R3 is linear operator given by
T(x1,x2,x3)=(x1,x3,−2x2−x3) Consider the standard basis of R3 which are e1,e2,e3 , thus
T(e1)=a1e1+a2e2+a3e3=(a1,a2,a3)=(1,0,0)T(e2)=b1e1+b2e2+b3e3=(b1,b2,b3)=(0,0,−2)T(e3)=c1e1+c2e2+c3e3=(c1,c2,c3)=(0,1,−1) Thus, matrix of T with respect to standard basis is
M=M(T)vv=⎣⎡1000010−2−1⎦⎤ Where v=(e1,e2,e3) .
Now, polynomial of T is given f(x)=−x3+2 , thus operator
f(T)=−T3+3I⟺−M3+2I3×3
Now,
M3=⎝⎛10002−1023⎠⎞⟹−M3=⎝⎛−1000−210−2−3⎠⎞⟹−M3+2I3×3=⎝⎛1000010−2−1⎠⎞ Thus,
f(T)=⎝⎛1000010−2−1⎠⎞
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