Answer to Question #206117 in Linear Algebra for Bafana

Question

Prove that there does not exist a linear map T : R5->R5 such that

range T = null T.

Answer 1

Let us prove that there does not exist a linear map $T: \R^5\to\R^5$ such that $range\ T = null\ T$ using the method by contradiction. Suppose that $T: \R^5\to\R^5$ is such a linear map that $range\ T = null\ T$. Then $\dim(range\ T) = \dim(null\ T).$ According to Rank–nullity theorem, $\dim(range\ T) +\dim(null\ T)=\dim \R^5.$ It follows that $2\dim(range\ T) =5,$ that is impossible because the left side of the last equality is an even number, but on the right side there is an odd number. This contradiction proves the statement.