Let us show that T:R3→R2, T(x,y,z)=(2x+y−z,x+z), is a linear transformation.
Let a,b∈R, (x1,y1,z1),(x2,y2,z2)∈R3. Then T[a(x1,y1,z1)+b(x2,y2,z2)]=T[(ax1+bx2,ay1+by2,az1+bz2)]=(2(ax1+bx2)+ay1+by2−(az1+bz2),ax1+bx2+az1+bz2)=(2ax1+2bx2+ay1+by2−az1−bz2,ax1+bx2+az1+bz2)=(a(2x1+y1−z1)+b(2x2+y2−z2),a(x1+z1)+b(x2+z2))=(a(2x1+y1−z1),a(x1+z1))+(b(2x2+y2−z2),b(x2+z2))=a(2x1+y1−z1,x1+z1)+b(2x2+y2−z2,x2+z2)=aT(x1,y1,z1)+bT(x2,y2,z2)
Therefore, T is a linear transformation.
Let us find its kernel:
ker(T)={(x,y,z)∈R3 ∣ T(x,y,z)=(0,0)}={(x,y,z)∈R3 ∣ (2x+y−z,x+z)=(0,0)}={(x,y,z)∈R3 ∣ 2x+y−z=0,x+z=0}={(x,y,z)∈R3 ∣ y=z−2x,x=−z}={(x,y,z)∈R3 ∣ y=3z,x=−z}={(−z,3z,z) ∣ z∈R}={z(−1,3,1) ∣ z∈R}
It follows that dim(kerT)=1.
Let us find Im(T). Let (a,b)∈R2 be arbitrary. Then T[(b,a−2b,0)]=(2b+a−2b+0,b+0)=(a,b), and therefore, T is surjective. It followws that dim(ImT)=dim(R2)=2
Consequently, T satisfies the Rank-nullity theorem:
dim(kerT)+dim(ImT)=1+2=3=dim(R3).
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