Answer to Question #168139 in Linear Algebra for Nikhil

Question

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

Answer 1

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

Let $a,b\in \mathbb R,\ (x_1,y_1,z_1),(x_2,y_2,z_2)\in\mathbb R^3.$ Then $T[a(x_1,y_1,z_1)+b(x_2,y_2,z_2)]=T[(ax_1+bx_2, ay_1+by_2, az_1+bz_2)]= (2(ax_1+bx_2)+ay_1+by_2-(az_1+bz_2), ax_1+bx_2+az_1+bz_2)= (2ax_1+2bx_2+ay_1+by_2-az_1-bz_2, ax_1+bx_2+az_1+bz_2)= (a(2x_1+y_1-z_1)+b(2x_2+y_2-z_2), a(x_1+z_1)+b(x_2+z_2))= (a(2x_1+y_1-z_1), a(x_1+z_1))+(b(2x_2+y_2-z_2), b(x_2+z_2))= a(2x_1+y_1-z_1, x_1+z_1)+b(2x_2+y_2-z_2, x_2+z_2)= aT(x_1,y_1,z_1)+bT(x_2,y_2,z_2)$

Therefore, $T$ is a linear transformation.

Let us find its kernel:

$\ker(T)=\{(x,y,z)\in\mathbb R^3\ |\ T(x,y,z)=(0,0)\}= \{(x,y,z)\in\mathbb R^3\ |\ (2x+y-z,x+z)=(0,0)\}= \{(x,y,z)\in\mathbb R^3\ |\ 2x+y-z=0, x+z=0\}= \{(x,y,z)\in\mathbb R^3\ |\ y=z-2x, x=-z\}= \{(x,y,z)\in\mathbb R^3\ |\ y=3z, x=-z\}= \{(-z,3z,z)\ |\ z\in\mathbb R\}=\{z(-1,3,1)\ |\ z\in\mathbb R\}$

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

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

Consequently, $T$ satisfies the Rank-nullity theorem:

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