Question #168139

Show that T:R^3 →R^2: T(x,y,z)= (2x+y-z,x+z) is a linear transformation. Verify that T satisfies the Rank-nullity theorem


1
Expert's answer
2021-03-03T04:21:48-0500

Let us show that T:R3R2,  T(x,y,z)=(2x+yz,x+z),T: \mathbb R^3 \to \mathbb R^2,\ \ T(x,y,z) = (2x+y-z,x+z), is a linear transformation.

Let a,bR, (x1,y1,z1),(x2,y2,z2)R3.a,b\in \mathbb R,\ (x_1,y_1,z_1),(x_2,y_2,z_2)\in\mathbb R^3. 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+by2az1bz2,ax1+bx2+az1+bz2)=(a(2x1+y1z1)+b(2x2+y2z2),a(x1+z1)+b(x2+z2))=(a(2x1+y1z1),a(x1+z1))+(b(2x2+y2z2),b(x2+z2))=a(2x1+y1z1,x1+z1)+b(2x2+y2z2,x2+z2)=aT(x1,y1,z1)+bT(x2,y2,z2)T[a(x_1,y_1,z_1)+b(x_2,y_2,z_2)]=T[(ax_1+bx_2, ay_1+by_2, az_1+bz_2)]= (2(ax_1+bx_2)+ay_1+by_2-(az_1+bz_2), ax_1+bx_2+az_1+bz_2)= (2ax_1+2bx_2+ay_1+by_2-az_1-bz_2, ax_1+bx_2+az_1+bz_2)= (a(2x_1+y_1-z_1)+b(2x_2+y_2-z_2), a(x_1+z_1)+b(x_2+z_2))= (a(2x_1+y_1-z_1), a(x_1+z_1))+(b(2x_2+y_2-z_2), b(x_2+z_2))= a(2x_1+y_1-z_1, x_1+z_1)+b(2x_2+y_2-z_2, x_2+z_2)= aT(x_1,y_1,z_1)+bT(x_2,y_2,z_2)

Therefore, TT 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+yz,x+z)=(0,0)}={(x,y,z)R3  2x+yz=0,x+z=0}={(x,y,z)R3  y=z2x,x=z}={(x,y,z)R3  y=3z,x=z}={(z,3z,z)  zR}={z(1,3,1)  zR}\ker(T)=\{(x,y,z)\in\mathbb R^3\ |\ T(x,y,z)=(0,0)\}= \{(x,y,z)\in\mathbb R^3\ |\ (2x+y-z,x+z)=(0,0)\}= \{(x,y,z)\in\mathbb R^3\ |\ 2x+y-z=0, x+z=0\}= \{(x,y,z)\in\mathbb R^3\ |\ y=z-2x, x=-z\}= \{(x,y,z)\in\mathbb R^3\ |\ y=3z, x=-z\}= \{(-z,3z,z)\ |\ z\in\mathbb R\}=\{z(-1,3,1)\ |\ z\in\mathbb R\}


It follows that dim(kerT)=1.\dim(\ker T)=1.


Let us find Im(T).Im(T). Let (a,b)R2(a,b)\in\mathbb R^2 be arbitrary. Then T[(b,a2b,0)]=(2b+a2b+0,b+0)=(a,b)T[(b, a-2b, 0)]=(2b+a-2b+0, b+0)=(a,b), and therefore, TT is surjective. It followws that dim(ImT)=dim(R2)=2\dim(Im T)=\dim(\mathbb R^2)=2


Consequently, TT satisfies the Rank-nullity theorem:


dim(kerT)+dim(ImT)=1+2=3=dim(R3).\dim(\ker T)+\dim(Im T)=1+2=3=\dim(\mathbb R^3).



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