Answer to Question #226354 in Linear Algebra for Nikhil

Question

Given any n∈N, is possible to define a linear transformation whose kernel has dimension n.
True or false with full explanation

Answer 1

Consider the map $T:\R^n\to\R^n, T(x_1,x_2,...,x_n)=(0,0,...,0)$ for any $(x_1,x_2,...,x_n)\in\R^n.$ Then

$T(a(x_1,x_2,...,x_n)+b(y_1,y_2,...,y_n))=0=\\=a\cdot 0+b\cdot 0=a\cdot T(x_1,x_2,...,x_n)+b\cdot T(y_1,y_2,...,y_n),$ and hence $T$ is a linear transformation. Its kernel is

$ker (T)=\{(x_1,x_2,..,x_n)\in\R^n:T(x_1,x_2,..,x_n)=(0,0,...,0)\}=R^n.$

Since $dim(\R^n)=n,$ we conclude for any $n∈\N$, is possible to define a linear transformation whose kernel has dimension $n$ .

Answer: true